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CBSE Class 12 Maths Revision Notes Chapter 3 Matrices

Matrices are ordered rectangular arrays of numbers or functions arranged in rows and columns. For CBSE Class 12 Maths, this chapter covers matrix types, operations, transpose, symmetric and skew-symmetric matrices, and invertible matrices.

Matrices help organise numerical data in a compact form. They are used in solving equations, representing transformations, handling data tables and studying different mathematical systems. In Class 12 Maths, this chapter builds the foundation for matrix algebra.

Use these CBSE Class 12 Maths Revision Notes Chapter 3 for the 2026–27 academic year to revise definitions, types, formulas, operations and properties. These Class 12 Mathematics Chapter 3 notes follow the NCERT flow and help students revise Matrices in a clear, exam-ready format.

Key Takeaways

Matrix: A matrix is an ordered rectangular array of numbers or functions.
Order: A matrix with m rows and n columns has order m × n.
Addition rule: Two matrices can be added only when they have the same order.
Multiplication rule: AB is defined only when columns of A equal rows of B.

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Access Class 12 Maths Chapter 3 Matrices Notes in 30 Minutes

Matrices become easier when you revise the chapter in order. Start with matrix meaning and order, then move to types, operations, transpose and inverse concepts.

Revision Area	What to Revise
Matrix Meaning	Rectangular arrangement of numbers or functions
Rows and Columns	Horizontal and vertical arrangements
Order of Matrix	m × n form
Types of Matrices	Row, column, square, diagonal, scalar, identity, zero
Equality	Same order and equal corresponding elements
Operations	Addition, subtraction, scalar multiplication
Matrix Multiplication	Row-by-column multiplication
Transpose	Rows changed into columns
Special Matrices	Symmetric and skew-symmetric
Invertible Matrices	Matrix whose inverse exists

Matrices Class 12 Notes: Meaning, Rows, Columns and Order

Matrices Class 12 Notes begin with the definition of a matrix. A matrix arranges numbers or functions in rows and columns.

What Is a Matrix?

A matrix is an ordered rectangular array of numbers or functions.

The numbers or functions inside a matrix are called elements or entries.

Matrices are usually denoted by capital letters such as A, B, C or P.

Example:

A = [1 2; 3 4]

This matrix has two rows and two columns.

Rows and Columns of a Matrix

Rows are horizontal arrangements of elements. Columns are vertical arrangements of elements.

For example:

A = [1 2; 0 1]

Part	Elements
Row 1	1, 2
Row 2	0, 1
Column 1	1, 0
Column 2	2, 1

Order of a Matrix

If a matrix has m rows and n columns, its order is m × n.

A matrix of order m × n has mn elements.

Example:

A = [1 2 3; 4 5 6]

This matrix has 2 rows and 3 columns.

So, its order is 2 × 3.

Number of elements = 2 × 3 = 6.

General Form of a Matrix

The general form of a matrix A of order m × n is:

A = [aij]m×n

Here:

Symbol	Meaning
i	Row number
j	Column number
aij	Element in ith row and jth column
m × n	Order of the matrix

The element a23 means the element in the second row and third column.

Types of Matrices in Class 12 Mathematics Chapter 3 Notes

Different matrices are named according to their rows, columns and element positions.

Row Matrix

A matrix is a row matrix if it has only one row.

Its order is 1 × n.

Example:

A = [2 5 7]

This is a row matrix of order 1 × 3.

Column Matrix

A matrix is a column matrix if it has only one column.

Its order is m × 1.

Example:

A = [2; 5; 7]

This is a column matrix of order 3 × 1.

Square Matrix

A matrix is a square matrix if the number of rows equals the number of columns.

Its order is n × n.

Example:

A = [1 2; 3 4]

This is a square matrix of order 2.

Diagonal Matrix

A square matrix is a diagonal matrix if all non-diagonal elements are zero.

Example:

A = [2 0 0; 0 5 0; 0 0 7]

Here, only diagonal elements may be non-zero.

Scalar Matrix

A scalar matrix is a diagonal matrix in which all diagonal elements are equal.

Example:

A = [4 0 0; 0 4 0; 0 0 4]

Here, all diagonal elements are 4.

Identity Matrix

An identity matrix is a square matrix in which all diagonal elements are 1 and all non-diagonal elements are 0.

It is denoted by I.

Example:

I = [1 0 0; 0 1 0; 0 0 1]

An identity matrix works like 1 in matrix multiplication.

Zero Matrix

A zero matrix has all elements equal to zero.

It is also called a null matrix.

Example:

O = [0 0; 0 0]

A zero matrix can have any order.

Types of Matrices Quick Table

Type of Matrix	Condition	Example Order
Row Matrix	One row only	1 × n
Column Matrix	One column only	m × 1
Square Matrix	Rows = columns	n × n
Diagonal Matrix	Non-diagonal elements are zero	n × n
Scalar Matrix	Equal diagonal elements	n × n
Identity Matrix	Diagonal elements are 1	n × n
Zero Matrix	All elements are zero	Any order

Equality of Matrices

Two matrices are equal only when they have the same order and their corresponding elements are equal.

Conditions for Equality

Let A = [aij] and B = [bij].

A = B if:

A and B have the same order.
aij = bij for every corresponding element.

Example:

If:

A = [2 3; 4 5]
B = [2 3; 4 5]

Then A = B.

But if:

A = [2 3; 4 5]
B = [3 2; 4 5]

Then A ≠ B because the corresponding elements are not equal.

How to Solve Unknowns Using Equal Matrices

When two matrices are equal, compare corresponding entries.

Example:

If:

[x 4; 5 y] = [2 4; 5 7]

Then:

x = 2
y = 7

This method is used in questions where unknowns are placed inside matrices.

Operations on Matrices

Operations on matrices include addition, subtraction, scalar multiplication and multiplication of matrices. Each operation has a condition.

Addition of Matrices

Two matrices can be added only when they have the same order.

If A = [aij] and B = [bij], then:

A + B = [aij + bij]

Example:

A = [1 2; 3 4]
B = [5 6; 7 8]

A + B = [6 8; 10 12]

Properties of Matrix Addition

Property	Rule
Commutative Law	A + B = B + A
Associative Law	(A + B) + C = A + (B + C)
Additive Identity	A + O = A
Additive Inverse	A + (-A) = O

These properties hold when all matrices have the same order.

Subtraction of Matrices

Two matrices can be subtracted only when they have the same order.

If A = [aij] and B = [bij], then:

A - B = [aij - bij]

Example:

A = [6 8; 10 12]
B = [1 2; 3 4]

A - B = [5 6; 7 8]

Negative of a Matrix

The negative of a matrix A is written as -A.

It is obtained by multiplying every element of A by -1.

If:

A = [1 -2; 3 4]

Then:

-A = [-1 2; -3 -4]

Multiplication of a Matrix by a Scalar

Scalar multiplication means multiplying every element of a matrix by a constant.

If k is a scalar and A = [aij], then:

kA = [kaij]

Example:

If A = [1 2; 3 4], then:

2A = [2 4; 6 8]

Properties of Scalar Multiplication

Property	Rule
Distributive over matrix addition	k(A + B) = kA + kB
Distributive over scalar addition	(k + l)A = kA + lA
Associative with scalars	k(lA) = (kl)A
Multiplicative identity	1A = A

Matrix Multiplication in Class 12 Maths Revision Notes

Multiplication of matrices is one of the most important parts of Maths Notes for Chapter 3 Matrices Class 12. It is different from ordinary multiplication because the order matters.

Condition for Matrix Multiplication

Let A be a matrix of order m × n and B be a matrix of order p × q.

The product AB is defined only when:

n = p

This means:

Number of columns in A = Number of rows in B

The order of AB will be:

m × q

Matrix A	Matrix B	AB Defined?	Order of AB
2 × 3	3 × 4	Yes	2 × 4
3 × 2	2 × 1	Yes	3 × 1
2 × 3	2 × 2	No	Not defined
1 × 3	3 × 1	Yes	1 × 1

How to Find AB

To multiply matrices, multiply each row of A with each column of B.

If:

A = [a b]
B = [c; d]

Then:

AB = [ac + bd]

For larger matrices, each entry of AB is found by row-column multiplication.

Properties of Matrix Multiplication

Property	Rule
Associative Law	A(BC) = (AB)C
Left Distributive Law	A(B + C) = AB + AC
Right Distributive Law	(A + B)C = AC + BC
Multiplicative Identity	AI = IA = A
Non-Commutative in general	AB ≠ BA

Matrix multiplication is generally not commutative. This means AB and BA may not be equal.

Sometimes one product may be defined while the other may not be defined.

Important Notes on Matrix Multiplication

AB may be defined even when BA is not defined.
AB and BA can have different orders.
Even when both are defined, AB may not equal BA.
If AB = BA, then A and B are called commuting matrices.
The zero product property does not always work like numbers. AB = O does not always mean A = O or B = O.

Transpose of a Matrix

The transpose of a matrix is obtained by changing rows into columns and columns into rows.

It is denoted by A′ or AT.

Meaning of Transpose

If A is of order m × n, then AT is of order n × m.

Example:

A = [1 2 3; 4 5 6]

Then:

AT = [1 4; 2 5; 3 6]

The first row of A becomes the first column of AT.

Properties of Transpose

Property	Rule
Transpose of transpose	(AT)T = A
Transpose of scalar multiple	(kA)T = kAT
Transpose of sum	(A + B)T = AT + BT
Transpose of product	(AB)T = BTAT

The order reverses in the transpose of a product.

So, (AB)T = BTAT, not ATBT.

Symmetric and Skew-Symmetric Matrices

Symmetric and skew-symmetric matrices are special square matrices linked with transpose.

Symmetric Matrix

A square matrix A is symmetric if:

AT = A

This means:

aij = aji for all i and j.

Example:

A = [1 2; 2 3]

Here, AT = A. So, A is symmetric.

Skew-Symmetric Matrix

A square matrix A is skew-symmetric if:

AT = -A

This means:

aij = -aji for all i and j.

In a skew-symmetric matrix, all diagonal elements are zero.

Example:

A = [0 2; -2 0]

Here, AT = -A. So, A is skew-symmetric.

Symmetric and Skew-Symmetric Matrix Difference Table

Point	Symmetric Matrix	Skew-Symmetric Matrix
Main condition	AT = A	AT = -A
Element relation	aij = aji	aij = -aji
Diagonal elements	Can be any value	Always zero
Matrix type	Square matrix	Square matrix

Expressing a Matrix as Sum of Symmetric and Skew-Symmetric Matrices

Every square matrix A can be expressed as:

A = 1/2(A + AT) + 1/2(A - AT)

Here:

Part	Matrix Type
1/2(A + AT)	Symmetric matrix
1/2(A - AT)	Skew-symmetric matrix

This result is important for proof-based questions.

Invertible Matrices and Elementary Operations

The later part of Class 12 Mathematics Chapter 3 notes introduces elementary operations and inverse of a matrix.

Elementary Row Operations

Elementary row operations are transformations done on rows of a matrix.

The three row operations are:

Operation	Meaning
Ri ↔ Rj	Interchange two rows
Ri → kRi	Multiply a row by a non-zero scalar
Ri → Ri + kRj	Add a multiple of one row to another row

Elementary Column Operations

Elementary column operations are transformations done on columns of a matrix.

The three column operations are:

Operation	Meaning
Ci ↔ Cj	Interchange two columns
Ci → kCi	Multiply a column by a non-zero scalar
Ci → Ci + kCj	Add a multiple of one column to another column

Invertible Matrix

A square matrix A is invertible if there exists another matrix B such that:

AB = BA = I

Here, B is called the inverse of A and is written as A⁻¹.

So:

AA⁻¹ = A⁻¹A = I

Key Inverse Properties

Property	Rule
Inverse of inverse	(A⁻¹)⁻¹ = A
Inverse of product	(AB)⁻¹ = B⁻¹A⁻¹
Inverse of transpose	(AT)⁻¹ = (A⁻¹)T
Identity inverse	I⁻¹ = I

The inverse exists only for a square matrix that satisfies the required condition.

Quick Revision Tables for Matrices

These tables help revise Matrices Class 12 Notes before solving NCERT exercises.

Matrix Operations Table

Operation	Condition	Result
A + B	Same order	Add corresponding elements
A - B	Same order	Subtract corresponding elements
kA	Any matrix A	Multiply each element by k
AB	Columns of A = rows of B	Row-column product
AT	Any matrix A	Rows become columns
A⁻¹	A is invertible square matrix	AA⁻¹ = A⁻¹A = I

Matrix Types Table

Matrix Type	Quick Check
Row Matrix	One row
Column Matrix	One column
Square Matrix	Rows = columns
Diagonal Matrix	Non-diagonal entries are zero
Scalar Matrix	Equal diagonal entries
Identity Matrix	Diagonal entries are 1
Zero Matrix	All entries are zero
Symmetric Matrix	AT = A
Skew-Symmetric Matrix	AT = -A

Formula and Property Table

Concept	Formula or Rule
Order of Matrix	m × n
Number of Elements	mn
Matrix Equality	Same order and corresponding elements equal
Matrix Addition	[aij] + [bij] = [aij + bij]
Scalar Multiplication	k[aij] = [kaij]
Matrix Multiplication	AB defined if columns of A = rows of B
Transpose of Product	(AB)T = BTAT
Symmetric Part of A	1/2(A + AT)
Skew-Symmetric Part of A	1/2(A - AT)

Important Terms in Matrices

Term	Meaning
Matrix	Ordered rectangular array of numbers or functions
Element	Entry inside a matrix
Row	Horizontal arrangement of elements
Column	Vertical arrangement of elements
Order	Number of rows × number of columns
Square Matrix	Matrix with equal rows and columns
Diagonal Elements	Elements a11, a22, a33 and so on
Identity Matrix	Square matrix with diagonal 1 and other entries 0
Transpose	Matrix obtained by interchanging rows and columns
Invertible Matrix	Matrix whose inverse exists

Common Mistakes in Class 12 Mathematics Chapter 3 Notes

Mistake	Correct Point
Adding matrices of different orders	Addition needs same order
Multiplying matrices without checking order	Columns of first matrix must equal rows of second
Assuming AB = BA	Matrix multiplication is generally not commutative
Writing (AB)T = ATBT	Correct formula is (AB)T = BTAT
Calling every diagonal matrix an identity matrix	Identity matrix has all diagonal entries equal to 1
Forgetting zero diagonal in skew-symmetric matrix	Diagonal elements must be zero
Confusing row matrix and column matrix	Row matrix has one row, column matrix has one column

Useful Links for Class 12 Maths

Section	Useful Links
Syllabus	CBSE Class 12 Maths Syllabus
Revision Notes	CBSE Class 12 Maths Revision Notes
Maths Notes	CBSE Class 12 Maths Revision Notes Chapter 1
NCERT Solutions	NCERT Solutions Class 12 Maths
Sample Papers	CBSE Sample Papers for Class 12 Maths
Important Questions	Important Questions Class 12 Maths
NCERT Books	NCERT Books for Class 12 Maths
Previous Year Papers	CBSE Maths Question Paper Class 12

Q.1

$\begin{array}{l} Obtain the inverse of the matrix A = [\begin{array}{l} 1 0 4 \\ 1 1 6 \\ - 3 0 - 10 \end{array}] \\ using row operatins . \end{array}$

Ans

$\begin{array}{l} Write A = IA, ie ., \\ [\begin{array}{l} 1 0 4 \\ 1 1 6 \\ - 3 0 - 10 \end{array}] = [\begin{array}{l} 1 0 0 \\ 0 1 0 \\ 0 0 1 \end{array}] A \\ One applying R_{2} \to R_{2} - R_{1} and R_{3} \to R_{3} - R_{1}, we get \\ [\begin{array}{l} 1 0 4 \\ 0 1 2 \\ 0 0 2 \end{array}] = [\begin{array}{l} 1 0 0 \\ - 1 1 0 \\ 3 0 1 \end{array}] A \\ On applying R_{3} \to \frac{1}{2} R_{3}, \\ [\begin{array}{l} 1 0 4 \\ 0 1 2 \\ 0 0 1 \end{array}] = [\begin{array}{l} 1 0 0 \\ - 1 1 0 \\ \frac{3}{2} 0 \frac{1}{2} \end{array}] A \\ On applying R_{1} \to R_{1} - 4 R_{3} and R_{2} \to R_{2} - 2 R_{3}, we get \\ [\begin{array}{l} 1 0 0 \\ 0 1 0 \\ 0 0 1 \end{array}] = [\begin{array}{l} - 5 0 - 2 \\ - 4 1 - 1 \\ \frac{3}{2} 0 \frac{1}{2} \end{array}] A \\ Hence, A^{- 1} = [\begin{array}{l} - 5 0 - 2 \\ - 4 1 - 1 \\ \frac{3}{2} 0 \frac{1}{2} \end{array}] \end{array}$

Q.2 Show that for any square matrix A, A + A’ is a symmetric matrix and A – A’ is a skew symmetric matrix.

Ans

Let B = A + A’
B’ = (A + A’)’
= A’ + (A’)’ [as (A + B)’ = A’ + B’]
= A’ + A [ (A’)’ = A]
= B
∴ B = A + A’ is a symmetric matrix.

Now let C = A – A’
C’ = (A – A’)’
= A’ – (A’)’
= A’ – A
= -(A – A’)
∴ C = A – A’ is a skew symmetric matrix.

Q.3

$Find the matrix x so that x [\begin{array}{l} 1 2 3 \\ 4 5 6 \end{array}] = [\begin{array}{l} - 7 - 8 - 9 \\ 2 4 6 \end{array}]$

Ans

$\begin{array}{l} Let x = [\begin{array}{l} a b \\ c d \end{array}] \\ ∴ [\begin{array}{l} a b \\ c d \end{array}] [\begin{array}{l} 1 2 3 \\ 4 5 6 \end{array}] = [\begin{array}{l} - 7 - 8 - 9 \\ 2 4 6 \end{array}] \\ \Rightarrow [\begin{array}{l} a \times 1 + b \times 4 a \times 2 + b \times 5 a \times 3 + b \times 6 \\ c \times 1 + d \times 4 c \times 2 + d \times 5 c \times 3 + d \times 6 \end{array}] = [\begin{array}{l} - 7 - 8 - 9 \\ 2 4 6 \end{array}] \\ \Rightarrow [\begin{array}{l} a + 4 b 2 a + 5 b 3 a + 6 b \\ c + 4 b 2 c + 5 b 3 c + 6 b \end{array}] = [\begin{array}{l} - 7 - 8 - 9 \\ 2 4 6 \end{array}] \\ By using equality and solving \\ \begin{array}{l} a + 4 b = - 7 \\ 2 a + 5 b = -8 \end{array}\} . \dots\dots (1) \begin{array}{l} c + 4 d = 2 \\ 2 c + 5 d = 4 \end{array}\} . \dots\dots (2) \\ Now, by (1) a = 1 b = - 2 \\ by (2) c = 2 d = 0 \\ Hence x = [\begin{array}{l} 1 - 2 \\ 2 0 \end{array}] \end{array}$

Q.4

$If A = [\begin{array}{l} 3 - 4 \\ 1 - 1 \end{array}], then prove that A^{n} = [\begin{array}{l} 1 + 2 n - 4 n \\ n 1 - 2 n \end{array}] .$

Ans

$\begin{array}{l} Let P (n) be the statement . \\ A^{n} = [\begin{array}{l} 1 + 2 n - 4 n \\ n 1 - 2 n \end{array}] \\ For n = 1, A^{1} = A = [\begin{array}{l} 3 - 4 \\ 1 - 1 \end{array}] = [\begin{array}{l} 1 + 2 .1 - 4 .1 \\ 1 1 - 2 .1 \end{array}] \\ \Rightarrow P (1) is true . \\ Let P (m) be true, i . e ., \\ A^{m} = [\begin{array}{l} 1 + 2 m - 4 m \\ m 1 - 2 m \end{array}] \\ Now, A^{m + 1} = {AA}^{m} = [\begin{array}{l} 3 - 4 \\ 1 - 1 \end{array}] [\begin{array}{l} 1 + 2 m - 4 m \\ m 1 - 2 m \end{array}] \\ = [\begin{array}{l} 3 + 2 m - 4 m - 4 \\ 1 + m - 2 m - 1 \end{array}] \\ = [\begin{array}{l} 1 + 2 (m + 1) - 4 (m + 1) \\ m + 1 1 - 2 m - 1 \end{array}] \\ \Rightarrow P (m + 1) is ture . \\ Hence by mathematical induction, P (n) is true for all \end{array}$

Q.5 If A, B are symmetric matrices of same order, then show that AB – BA is a skew- symmetric matrix.

Ans

Here, (AB – BA)’ = (AB)’ – (BA)’
= B’A’ – A’B’
= –(A’B’ – B’A’)
= –(AB – BA) [ A’ = A, B’ = B]
⇒ (AB – BA)’ = –(AB – BA)
Hence, AB – BA is a skew-symmetric matrix.

Q.6

$\begin{array}{l} Express A = [\begin{array}{l} 3 3 - 1 \\ - 2 - 2 1 \\ - 4 - 5 2 \end{array}] as the sum of a symmetric \\ and a skew - symmetric matrix . \end{array}$

Ans

$\begin{array}{l} Here A ‘ = [\begin{array}{l} 3 - 2 - 4 \\ 3 - 2 - 5 \\ - 1 1 2 \end{array}] \\ Let P = \frac{1}{2} (A + A ‘) \\ ∴ \frac{1}{2} (A + A ‘) = \frac{1}{2} [\begin{array}{l} 3 3 - 1 \\ - 2 - 2 1 \\ - 4 - 5 2 \end{array}] + [\begin{array}{l} 3 - 2 - 4 \\ 3 - 2 - 5 \\ - 1 1 2 \end{array}] \\ = \frac{1}{2} [\begin{array}{l} 6 1 - 5 \\ 1 - 4 - 4 \\ - 5 - 4 4 \end{array}] = [\begin{array}{l} 3 \frac{1}{2} \frac{- 5}{2} \\ \frac{1}{2} - 2 - 2 \\ \frac{- 5}{2} - 2 2 \end{array}] \\ Now P ‘ = [\begin{array}{l} 3 \frac{1}{2} \frac{- 5}{2} \\ \frac{1}{2} - 2 - 2 \\ \frac{- 5}{2} - 2 2 \end{array}] = P \\ Thus P = \frac{1}{2} [A - A ‘] \\ \frac{1}{2} [A - A ‘] = \frac{1}{2} [[\begin{array}{l} 3 3 - 1 \\ - 2 - 2 1 \\ - 4 - 5 2 \end{array}] - [\begin{array}{l} 3 - 2 - 4 \\ 3 - 2 - 5 \\ - 1 1 2 \end{array}]] \\ = \frac{1}{2} [[\begin{array}{l} 0 5 3 \\ - 2 0 6 \\ - 3 - 6 0 \end{array}] = [\begin{array}{l} 0 \frac{5}{2} \frac{3}{2} \\ - \frac{5}{2} - 3 0 \\ - \frac{3}{2} - 3 0 \end{array}]] \\ Then Q ‘ = [\begin{array}{l} 0 - \frac{5}{2} - \frac{3}{2} \\ \frac{5}{2} 0 - 3 \\ \frac{3}{2} 3 0 \end{array}] = - Q \\ Thus Q = \frac{1}{2} (A - A ‘) is a skew symmetric matrix . \\ Now, P + Q = [\begin{array}{l} 3 \frac{1}{2} \frac{- 5}{2} \\ \frac{1}{2} - 2 - 2 \\ \frac{- 5}{2} - 2 2 \end{array}] + [\begin{array}{l} 0 \frac{5}{2} \frac{3}{2} \\ - \frac{5}{2} 0 3 \\ - \frac{3}{2} - 3 0 \end{array}] = [\begin{array}{l} 3 3 - 1 \\ - 2 - 2 1 \\ - 4 - 5 2 \end{array}] = A \end{array}$

Q.7

$If A = [\begin{array}{l} - 2 \\ 4 \\ 5 \end{array}], B = [1 3 - 6], verify that (AB) ‘ = B ‘ A ‘ .$

Ans

$\begin{array}{l} Here A = [\begin{array}{l} - 2 \\ 4 \\ 5 \end{array}], B = [1 3 - 6] \\ AB = [\begin{array}{l} - 2 \\ 4 \\ 5 \end{array}] [1 3 - 6] = [\begin{array}{l} - 2 - 6 12 \\ 4 12 - 24 \\ 5 15 - 30 \end{array}] \\ (AB ‘) = [\begin{array}{l} - 2 4 5 \\ - 6 12 15 \\ 12 - 24 - 30 \end{array}] \\ Now, A ‘ = [- 2 4 5], B ‘ = [\begin{array}{l} 1 \\ 3 \\ - 6 \end{array}] \\ B ‘ A ‘ = [\begin{array}{l} 1 \\ 3 \\ - 6 \end{array}] [- 2 4 5] = [\begin{array}{l} - 2 4 5 \\ - 6 12 15 \\ 12 - 24 - 30 \end{array}] = (AB) ‘ \\ Hence (AB) ‘ = B ‘ A ‘ \end{array}$

Q.8

$If A = [\begin{array}{l} 1 0 \\ - 1 7 \end{array}] and I = [\begin{array}{l} 1 0 \\ 0 1 \end{array}], then find k so that A^{2} = 8 A + kI .$

Ans

$\begin{array}{l} A^{2} = AA = [\begin{array}{l} 1 0 \\ - 1 7 \end{array}] [\begin{array}{l} 1 0 \\ - 1 7 \end{array}] \\ = [\begin{array}{l} 1 + 0 0 + 0 \\ - 1 - 7 0 + 49 \end{array}] = [\begin{array}{l} 1 0 \\ - 8 49 \end{array}] \\ 8 A + KI = 8 [\begin{array}{l} 1 0 \\ - 1 7 \end{array}] + K [\begin{array}{l} 1 0 \\ 0 1 \end{array}] = [\begin{array}{l} 8 + k 0 \\ - 8 56 + k \end{array}] \\ Given A^{2} = 8 A + kI \\ \Rightarrow [\begin{array}{l} 1 0 \\ - 8 49 \end{array}] = [\begin{array}{l} 8 + k 0 \\ - 8 56 + k \end{array}] \\ \Rightarrow 8 + k = 1 and 56+ k =49 \\ \Rightarrow k = - 7 \end{array}$

Q.9

$Find x and y if x + y = [\begin{array}{l} 5 2 \\ 0 9 \end{array}] and x - y = [\begin{array}{l} 3 6 \\ 0 - 1 \end{array}]$

Ans

$\begin{array}{l} We have \\ (x + y) + (x - y) = [\begin{array}{l} 5 2 \\ 0 9 \end{array}] + [\begin{array}{l} 3 6 \\ 0 - 1 \end{array}] \\ 2 x = [\begin{array}{l} 5 + 3 6 + 2 \\ 0 + 0 9 - 1 \end{array}] = [\begin{array}{l} 8 8 \\ 0 8 \end{array}] \\ \Rightarrow x = \frac{1}{2} [\begin{array}{l} 8 8 \\ 0 8 \end{array}] \\ \Rightarrow x = [\begin{array}{l} 4 4 \\ 0 4 \end{array}] \\ Now (x + y) (x - y) = [\begin{array}{l} 5 2 \\ 0 9 \end{array}] - [\begin{array}{l} 3 6 \\ 0 1 \end{array}] \\ 2 y = [\begin{array}{l} 5 - 3 2 - 6 \\ 0 - 0 9 - 1 \end{array}] \\ \Rightarrow y = \frac{1}{2} [\begin{array}{l} 2 - 4 \\ 0 8 \end{array}] \\ \Rightarrow y = [\begin{array}{l} 1 - 2 \\ 0 4 \end{array}] \\ Hence x = [\begin{array}{l} 4 4 \\ 0 4 \end{array}] y = \frac{1}{2} [\begin{array}{l} 1 - 2 \\ 0 4 \end{array}] \end{array}$

Q.10

$\begin{array}{l} If A = [\begin{matrix} \begin{matrix} 8 \\ 4 \\ 3 \end{matrix} & \begin{matrix} 0 \\ - 2 \\ 6 \end{matrix} \end{matrix}] and B = [\begin{matrix} \begin{matrix} 2 \\ 4 \\ - 5 \end{matrix} & \begin{matrix} - 2 \\ 2 \\ 1 \end{matrix} \end{matrix}], X \\ such that A + X = B . \end{array}$

Ans

$\begin{array}{l} We have A + X = B \\ \Rightarrow [\begin{matrix} \begin{matrix} 8 \\ 4 \\ 3 \end{matrix} & \begin{matrix} 0 \\ - 2 \\ 6 \end{matrix} \end{matrix}] + X = [\begin{matrix} \begin{matrix} 2 \\ 4 \\ - 5 \end{matrix} & \begin{matrix} - 2 \\ 2 \\ 1 \end{matrix} \end{matrix}] \\ \Rightarrow X = [\begin{matrix} \begin{matrix} 2 \\ 4 \\ - 5 \end{matrix} & \begin{matrix} - 2 \\ 2 \\ 1 \end{matrix} \end{matrix}] - [\begin{matrix} \begin{matrix} 8 \\ 4 \\ 3 \end{matrix} & \begin{matrix} 0 \\ - 2 \\ 6 \end{matrix} \end{matrix}] \\ = [\begin{matrix} \begin{matrix} 2 - 8 \\ 4 - 4 \\ - 5 - 3 \end{matrix} & \begin{matrix} - 2 - 0 \\ 2 + 2 \\ 1 - 6 \end{matrix} \end{matrix}] \\ = [\begin{matrix} \begin{matrix} - 6 \\ 0 \\ - 8 \end{matrix} & \begin{matrix} - 2 \\ 4 \\ - 5 \end{matrix} \end{matrix}] \end{array}$

Q.11

$Given A = [\begin{array}{l} 5 - 1 0 \\ 7 3 4 \end{array}] and B = [\begin{array}{l} - 3 2 0 \\ - 5 6 2 \end{array}] . Find A + B, A - B .$

Ans

$\begin{array}{l} Here, A + B = [\begin{array}{l} 5 - 1 0 \\ 7 3 4 \end{array}] + [\begin{array}{l} - 3 2 0 \\ - 5 6 2 \end{array}] \\ = [\begin{array}{l} 2 1 0 \\ 2 9 6 \end{array}] \\ A - B = [\begin{array}{l} 5 - 1 0 \\ 7 3 4 \end{array}] - [\begin{array}{l} - 3 2 0 \\ - 5 6 2 \end{array}] \\ = [\begin{array}{l} 5 + 3 - 1 - 2 0 - 0 \\ 7 + 3 3 - 6 4 - 2 \end{array}] \\ = [\begin{array}{l} 8 - 3 0 \\ 12 - 3 2 \end{array}] \end{array}$

Q.12

$If [\begin{array}{l} x + y 4 \\ 7 xy \end{array}] = [\begin{array}{l} 6 4 \\ 7 8 \end{array}], find the values of x and y .$

Ans

$\begin{array}{l} x + y = 6 \\ xy = 8 \Rightarrow y = \frac{8}{x} \\ so, x + \frac{8}{x} = 6 \\ \Rightarrow x^{2} - 6 x + 6 = 0 \\ \Rightarrow (x - 2) (x - 4) = 4 \\ when x = 2, y = 4 \\ or x = 4, y = 2 \end{array}$

Q.13

$Construct a 22 matrix A = [a_{ij}] whose elements are given by a_{ij} = \frac{{(i + j)}^{2}}{2}$

Ans

$\begin{array}{l} Here A = [\begin{array}{l} a_{11} a_{12} \\ a_{21} a_{22} \end{array}] \\ a_{11} = \frac{{(1 + 1)}^{2}}{2} = \frac{2^{2}}{2} = \frac{4}{2} = 2 a_{12} = \frac{{(1 + 2)}^{2}}{2} = \frac{3^{2}}{2} = \frac{9}{2} \\ a_{21} = \frac{{(2 + 1)}^{2}}{2} = \frac{3^{2}}{2} = \frac{9}{2} a_{22} = \frac{{(2 + 2)}^{2}}{2} = \frac{16}{2} = 8 \\ Hence A = [\begin{array}{l} 2 \frac{9}{2} \\ \frac{9}{2} 8 \end{array}] \end{array}$

Q.14

$\begin{array}{l} Consider the matrix \\ A = [\begin{array}{l} 2 5 \sqrt{7} 3 \\ 35 6 \frac{5}{3} - 1 \\ \sqrt{3} 1 6 0 \end{array}] \end{array}$

Ans

(i) Order of the matrix = 3

(ii) a₁₃ = 7, a₂₁ = 35

a₂₃ = 5/3, a₃₄ = 0

Q.15

$\begin{array}{l} If A = [\begin{array}{l} 31 \\ 7 5 \end{array}], find x and y such that A^{2} + XI = YA . \\ Hence find A^{- 1} . \end{array}$

Ans

$\begin{array}{l} Given A = [\begin{array}{l} 31 \\ 7 5 \end{array}] \\ ∴ A^{2} = A . A = + [\begin{array}{l} 31 \\ 7 5 \end{array}] [\begin{array}{l} 31 \\ 7 5 \end{array}] \\ \Rightarrow A^{2} = [\begin{array}{l} 9 +7 3+5 \\ 21 + 35 7 + 25 \end{array}] \\ \Rightarrow A^{2} = [\begin{array}{l} 168 \\ 56 32 \end{array}] . \dots.. (1) \\ and xI = x [\begin{array}{l} 10 \\ 0 1 \end{array}] = [\begin{array}{l} x 0 \\ 0 x \end{array}] \\ RHS = yA = y [\begin{array}{l} 31 \\ 7 5 \end{array}] = [\begin{array}{l} 3 y y \\ 7 y 5 y \end{array}] . \dots.. (2) \\ LHS = A^{2} + xI = [\begin{array}{l} 168 \\ 56 32 \end{array}] + [\begin{array}{l} x 0 \\ 0 x \end{array}] \\ By using equality of matrices \\ 16+ x = 3 y \\ \Rightarrow 56 = 7 y \\ \Rightarrow y = 8 \\ And, x = 8 \\ Substituting the values of x and y, we get \\ A^{2} + 8 I = 8 A \\ \Rightarrow 8 I = 8 A - A^{2} \\ \Rightarrow I = \frac{1}{8} [8 A - A^{2}] \\ \Rightarrow {IA}^{- 1} = \frac{1}{8} [8 {AA}^{- 1} - A ({AA}^{- 1})] \\ \Rightarrow {IA}^{- 1} = \frac{1}{8} [8 I - AI] \\ \Rightarrow {IA}^{- 1} = \frac{1}{8} [8 {AA}^{- 1} - A ({AA}^{- 1})] \\ \Rightarrow {IA}^{- 1} = \frac{1}{8} [8 I - AI] \\ \Rightarrow = \frac{1}{8} [8 I - A] \\ \Rightarrow = \frac{1}{8} [[\begin{array}{l} 8 0 \\ 08 \end{array}] - [\begin{array}{l} 31 \\ 75 \end{array}]] \\ \Rightarrow = \frac{1}{8} [\begin{array}{l} 5 -1 \\ - 73 \end{array}] \\ \Rightarrow x = 8, y = 8 and A^{- 1} = \frac{1}{8} [\begin{array}{l} 5 -1 \\ - 73 \end{array}] \end{array}$

Q.16 Find the order of matrix A = [a ub>ij]_{m x n}.

AnsOrder of the matrix is m x n.

Q.17 A matrix has 24 elements. What re the possible orders it can have?

AnsThe possible orders of the matrix are:

1 24, 2 12, 3 8, 4 6, 6 4, 8 3, 12 2 and 24 1.

Q.18

$\begin{array}{l} If A and B are two equal square matrices such that \\ A = [\begin{array}{l} x y \\ 23 \end{array}] and B = [\begin{array}{l} 14 \\ 2 3 \end{array}], then find the values of x and y . \end{array}$

Ans

x = 1 and y = 4 (Equality matrices property)

Q.19 If A and B are two square atrices and K is a scalar quantity then K(A+B) = ______.

Ans K(A+B) = KA+KB

Q.20

$Write the general formula for elements of matrix A= [\begin{array}{l} \frac{1}{2} \frac{1}{2} \\ \frac{2}{1} \frac{2}{2} \end{array}] = [\begin{array}{l} 1 \frac{1}{2} \\ 2 1 \end{array}] .$

Ans

$A = [aij] is a 2 \times 2 matrix whose elements are given by = aij = \frac{i}{j}$

Q.21 A square matrix in which elements in the diagonal are all 1 and rest are all zero is called an ________.

Ans

Identity Matrix

Q.22 Define identity matrix.

Ans

A square matrix in which elements in the diagonal are all 1 and rest all are zero is called an identity matrix.

Q.23

$If A = [\begin{array}{l} 10 \\ 24 \end{array}] and B = [\begin{array}{l} 2 -1 \\ 37 \end{array}], find A + B .$

Ans

$\begin{array}{l} A + B = [\begin{array}{l} 10 \\ 24 \end{array}] + [\begin{array}{l} 2 -1 \\ 37 \end{array}] \\ = [\begin{array}{l} 1 +20 - 7 \\ 2 +34 +7 \end{array}] \\ = [\begin{array}{l} 3 - 1 \\ 5 11 \end{array}] \end{array}$

Q.24

$If A = [\begin{array}{l} 1 3 5 \\ 2 4 6 \\ 3 5 7 \end{array}], find the diagonal elements of A .$

Ans

Diagonal elements of A are (1, 4, 7).

Q.25

$If A = [\begin{array}{l} a 0 \\ 0 a \end{array}] and B = [\begin{array}{l} 0 a \\ a 0 \end{array}], find the value of AB .$

Ans

$\begin{array}{l} AB = [\begin{array}{l} a 0 \\ 0 - a \end{array}] + [\begin{array}{l} a 0 \\ 0 - a \end{array}] = [\begin{array}{l} a \times 0 + 0 \times a a \times a + 0 \times 0 \\ 0 \times 0 + (- a) \times a 0 \times a + (- a) \times 0 \end{array}] \\ = [\begin{array}{l} 0 a^{2} \\ - a^{2} 0 \end{array}] \end{array}$

Q.26

$\begin{array}{l} If A = [\begin{array}{l} 10 \\ 0 -1 \end{array}] and B = [\begin{array}{l} a 0 \\ 0 - a \end{array}] satisfying 2 A +2 B = 0, \\ then find the value of a . \end{array}$

Ans

$\begin{array}{l} 2 A + B =2 [\begin{array}{l} 10 \\ 0 -1 \end{array}] + [\begin{array}{l} a 0 \\ 0 - a \end{array}] = [\begin{array}{l} 2 + a 0 \\ 0 -2 - a \end{array}] \\ Given, 2 A + B = 0 \\ = [\begin{array}{l} 2 + a 0 \\ 0 - 2 a \end{array}] \\ \Rightarrow a = - 2 \end{array}$

Q.27

$If A = [\begin{array}{l} a - a \\ b - b \end{array}] and B = [\begin{array}{l} b a \\ - b - a \end{array}], find the value of 2 A -2 B .$

Ans

$\begin{array}{l} 2 A - 2 B = [\begin{array}{l} 2 a -2 a \\ 2 b -2 b \end{array}] - [\begin{array}{l} 2 b 2 a \\ - 2 b -2 a \end{array}] \\ = [\begin{array}{l} 2 a -2 b -2 a -2 a \\ 2 b - (- 2 b) -2 b - (- 2 a) \end{array}] \\ = [\begin{array}{l} 2 a -2 b -4 a \\ 4 b 2 a -2 b \end{array}] \end{array}$

Q.28

$\begin{array}{l} Let A = [\begin{array}{l} 0 - \tan \frac{α}{2} \\ \tan \frac{α}{2} 0 \end{array}] and I be the identity matrix of order 2. \\ show that : I + A (I - A) [\begin{array}{l} cosα - sinα \\ sinα cosα \end{array}] \end{array}$

Ans

$\begin{array}{l} I + A = [\begin{array}{l} 1 0 \\ 0 1 \end{array}] + [\begin{array}{l} 0 - \tan \frac{α}{2} \\ \tan \frac{α}{2} 0 \end{array}] = [\begin{array}{l} 1 - \tan \frac{α}{2} \\ \tan \frac{α}{2} 1 \end{array}] \\ I - A = [\begin{array}{l} 1 0 \\ 0 1 \end{array}] - [\begin{array}{l} 0 - \tan \frac{α}{2} \\ \tan \frac{α}{2} 0 \end{array}] = [\begin{array}{l} 1 \tan \frac{α}{2} \\ - \tan \frac{α}{2} 1 \end{array}] \\ ∴ (I - A) [\begin{array}{l} cosα - sinα \\ sinα cosα \end{array}] = [\begin{array}{l} 1 \tan \frac{α}{2} \\ - \tan \frac{α}{2} 1 \end{array}] [\begin{array}{l} cosα - sinα \\ sinα cosα \end{array}] \\ \Rightarrow (I - A) [\begin{array}{l} cosα - sinα \\ sinα cosα \end{array}] = [\begin{array}{l} 1 \tan \frac{α}{2} \\ - \tan \frac{α}{2} 1 \end{array}] [\frac{\frac{1 - \tan^{2} \frac{α}{2}}{1 + \tan^{2} \frac{α}{2}} - \frac{2 \tan \frac{α}{2}}{1 + \tan^{2} \frac{α}{2}}}{\frac{2 \tan \frac{α}{2}}{1 + \tan^{2} \frac{α}{2}} \frac{1 - \tan^{2} \frac{α}{2}}{1 + \tan^{2} \frac{α}{2}}}] \\ \Rightarrow (I - A) [\begin{array}{l} cosα - sinα \\ sinα cosα \end{array}] = [\begin{array}{l} 1 t \\ - t 1 \end{array}] [\frac{\frac{1 - t^{2}}{1 + t^{2}} - \frac{2 t}{1 + t^{2}}}{\frac{2 t}{1 + t^{2}} \frac{1 - t^{2}}{1 + t^{2}}}], where t = \frac{tanα}{2} \\ \Rightarrow (I - A) [\begin{array}{l} cosα - sinα \\ sinα cosα \end{array}] = [\frac{\frac{1 - t^{2} + 2 t^{2}}{1 + t^{2}} \frac{- 2 t + t - t^{3}}{1 + t^{2}}}{\frac{- t + t^{3} + 2 t}{1 + t^{2}} \frac{2 t^{2} + 1 - t^{2}}{1 + t^{2}}}] \\ \Rightarrow (I - A) [\begin{array}{l} cosα - sinα \\ sinα cosα \end{array}] = [\frac{\frac{1 - t^{2}}{1 + t^{2}} \frac{- t + (1 - t^{2})}{1 + t^{2}}}{\frac{t (1 + t^{2})}{1 + t^{2}} \frac{1 + t^{2}}{1 + t^{2}}}] = [\begin{array}{l} 1 - t \\ t 1 \end{array}] \\ \Rightarrow (I - A) [\begin{array}{l} cosα - sinα \\ sinα cosα \end{array}] = [\begin{array}{l} 1 - \tan \frac{α}{2} \\ \tan \frac{α}{2} 1 \end{array}] = I + A \end{array}$

Q.29

$Find the matrix x so that x [\begin{array}{l} 1 2 3 \\ 4 5 6 \end{array}] = [\begin{array}{l} - 7 -8 -9 \\ 4 5 6 \end{array}] .$

Ans

$\begin{array}{l} Let x = [\begin{array}{l} a b \\ c d \end{array}] \\ ∴ [\begin{array}{l} a b \\ c d \end{array}] [\begin{array}{l} 1 2 3 \\ 4 5 6 \end{array}] = [\begin{array}{l} - 7 -8 -9 \\ 4 5 6 \end{array}] \\ \Rightarrow [\begin{array}{l} a \times 1 + b \times 4 a \times 2 + b \times 5 a \times 3 + b \times 6 \\ c \times 1 + d \times 4 c \times 2 + d \times 5 c \times 3 + d \times 6 \end{array}] = [\begin{array}{l} - 7 -8 -9 \\ 4 5 6 \end{array}] \\ \Rightarrow [\begin{array}{l} a + 4 b 2 a + 5 b 3 a + 6 b \\ c + 4 d 2 c + 5 d 3 c + 6 d \end{array}] = [\begin{array}{l} - 7 -8 -9 \\ 2 4 6 \end{array}] \\ By using equality and solving \\ \begin{array}{l} a +4 b = -7 \\ 2 a +5 b = - 8 \end{array}\} . \dots\dots.. (1) \begin{array}{l} c +4 d = 2 \\ 2 c +5 d = 4 \end{array}\} . \dots\dots.. (2) \\ Now, by (1) a = 1 b = - 2 \\ by (2) c = 2 d = 0 \\ Hence x = [\begin{array}{l} 1 - 2 \\ 2 0 \end{array}] \end{array}$

Q.30 Show that for any square matrix A, A + A’ is a symmetric matrix and A – A’ is a skew symmetric matrix.

Ans

Let B = A + A’
B’ = (A + A’)’
= A’ + (A’)’ [as (A + B)’ = A’ + B’]
= A’ + A [ (A’)’ = A]
= B
∴ B = A + A’ is a symmetric matrix.

Now let C = A – A’
C’ = (A – A’)’
= A’ – (A’)’
= A’ – A
= -(A – A’)
∴ C = A – A’ is a skew symmetric matrix.

Q.31 Find the order of matrix A = [_ij]_{m x n}.

AnsOrder of the matrix is m x n.

Q.32

$\begin{array}{l} If A and B are two equal square matrices such that \\ A = [\begin{array}{l} x y \\ 2 3 \end{array}] and B = [\begin{array}{l} 1 4 \\ 2 3 \end{array}], then find the values of x and y . \end{array}$

Ans

$x = 1 and y = 4 (Equality of matrices property) .$

Q.33

$Construct a 2 ×2 matrix A = [aij] whose elements are given by aij \frac{i}{j} .$

Ans

$A = [\begin{array}{l} \frac{1}{1} \frac{1}{2} \\ \frac{2}{1} \frac{2}{2} \end{array}] = [\begin{array}{l} 1 \frac{1}{2} \\ 2 1 \end{array}]$

Q.34

$\begin{array}{l} If A = [\begin{array}{l} 1 2 3 \\ 3 -2 1 \\ 4 2 1 \end{array}], that show that A^{3} - 23 A - 40 I = 0 . \end{array}$

Ans

$\begin{array}{l} We have A^{2} = A . A = [\begin{array}{l} 1 2 3 \\ 3 -2 1 \\ 4 2 1 \end{array}] [\begin{array}{l} 1 2 3 \\ 3 -2 1 \\ 4 2 1 \end{array}] \\ = [\begin{array}{l} 19 4 8 \\ 1 12 8 \\ 14 6 15 \end{array}] \\ So, A^{3} = A^{2} . A = [\begin{array}{l} 19 4 8 \\ 1 12 8 \\ 14 6 15 \end{array}] [\begin{array}{l} 1 2 3 \\ 3 -2 1 \\ 4 2 1 \end{array}] \\ = [\begin{array}{l} 63 46 69 \\ 69 -6 23 \\ 92 46 63 \end{array}] \\ Now, \\ A^{3} -23 A -40 I = [\begin{array}{l} 63 46 69 \\ 69 -6 23 \\ 92 46 63 \end{array}] - 23 [\begin{array}{l} 1 2 3 \\ 3 -2 1 \\ 4 2 1 \end{array}] - 40 [\begin{array}{l} 1 0 0 \\ 0 1 0 \\ 0 0 1 \end{array}] \\ = [\begin{array}{l} 63 46 69 \\ 69 -6 23 \\ 92 46 63 \end{array}] + [\begin{array}{l} - 23 -46 -69 \\ - 69 46 -23 \\ - 92 -46 -23 \end{array}] + [\begin{array}{l} - 40 0 0 \\ 0 -40 0 \\ 0 0 -40 \end{array}] \\ = [\begin{array}{l} 63 -23 -40 46-46+0 69-69+0 \\ 69 -69 +0 -6+46-40 23-23+0 \\ 92 -92 +0 46-46+0 63-23-40 \end{array}] \\ = [\begin{array}{l} 0 0 0 \\ 0 0 0 \\ 0 0 0 \end{array}] = 0 = RHS \end{array}$

Q.35

$If A = [\begin{array}{l} 1 0 2 \\ 0 2 1 \\ 2 0 3 \end{array}], prove that A^{3} - 6 A^{2} + 7 A + 2 I = 0 .$

Ans

$\begin{array}{l} We have, A = [\begin{array}{l} 1 0 2 \\ 0 2 1 \\ 2 0 3 \end{array}] \\ A^{2} = A . A = [\begin{array}{l} 1 0 2 \\ 0 2 1 \\ 2 0 3 \end{array}] . [\begin{array}{l} 1 0 2 \\ 0 2 1 \\ 2 0 3 \end{array}] = [\begin{array}{l} 5 0 8 \\ 2 4 5 \\ 8 0 13 \end{array}] \\ A^{3} = A^{2} . A = [\begin{array}{l} 5 0 8 \\ 2 4 5 \\ 8 0 13 \end{array}] . [\begin{array}{l} 1 0 2 \\ 0 2 1 \\ 2 0 3 \end{array}] = [\begin{array}{l} 21 0 34 \\ 12 8 23 \\ 34 0 55 \end{array}] \\ LHS : \\ A^{3} - 6 A^{2} + 7 A + 2 I \\ = [\begin{array}{l} 21 0 34 \\ 12 8 23 \\ 34 0 55 \end{array}] - 6 [\begin{array}{l} 5 0 8 \\ 2 4 5 \\ 8 0 13 \end{array}] + 7 [\begin{array}{l} 1 0 2 \\ 0 2 1 \\ 2 0 3 \end{array}] + 2 [\begin{array}{l} 1 0 0 \\ 0 1 0 \\ 0 0 1 \end{array}] \\ = [\begin{array}{l} 21 0 34 \\ 12 8 23 \\ 34 0 55 \end{array}] + [\begin{array}{l} 30 0 -48 \\ - 12 -24 -30 \\ - 48 0 -78 \end{array}] + [\begin{array}{l} 7 0 14 \\ 0 14 7 \\ 14 0 21 \end{array}] + [\begin{array}{l} 2 0 0 \\ 0 2 0 \\ 0 0 2 \end{array}] \\ = [\begin{array}{l} 21 -30+7+2 0+0+0+0 34-48+14+0 \\ 12 -12+0+0 8-24+14+2 23-30+7+0 \\ 34 -48+14+0 0+0+0+0 55-78+21+2 \end{array}] \\ = [\begin{array}{l} 0 0 0 \\ 0 0 0 \\ 0 0 0 \end{array}] \\ = RHS \end{array}$

Q.36

$\begin{array}{l} Express the following matrix as the sum of a symmetric and a skew symmetric matrix : \\ [\begin{array}{l} 6 -2 2 \\ - 2 3 -1 \\ 2 -1 3 \end{array}] \end{array}$

Ans

$\begin{array}{l} P = \frac{1}{2} (A + A ‘) \\ = \frac{1}{2} \{[\begin{array}{l} 6 -2 2 \\ - 2 3 -1 \\ 2 -1 3 \end{array}] + [\begin{array}{l} 6 -2 2 \\ - 2 3 -1 \\ 2 -1 3 \end{array}]\} \\ = \frac{1}{2} [\begin{array}{l} 12 -4 4 \\ - 4 6 -2 \\ 4 -2 6 \end{array}] = [\begin{array}{l} 6 -2 2 \\ - 2 3 -1 \\ 2 -1 3 \end{array}] \\ Thus, P = \frac{1}{2} (A - A ‘) is symmetric matrix . \\ Q = \frac{1}{2} (A - A ‘) \\ = \frac{1}{2} \{[\begin{array}{l} 6 -2 2 \\ - 2 3 -1 \\ 2 -1 3 \end{array}] + [\begin{array}{l} 6 -2 2 \\ - 2 3 -1 \\ 2 -1 3 \end{array}]\} = [\begin{array}{l} 0 0 0 \\ 0 0 0 \\ 0 0 0 \end{array}] \\ Q ‘ = [\begin{array}{l} 0 0 0 \\ 0 0 0 \\ 0 0 0 \end{array}] \\ Thus, Q = \frac{1}{2} (A - A ‘) is skew symmetric matrix . \\ P + Q = \frac{A + A ‘}{2} + \frac{A + A ‘}{2} = \frac{2 A}{2} = A \\ Thus, A is expressed as a sum of a symmetric and a skew \\ symmetric matrix . \end{array}$

Q.37

$\begin{array}{l} Let A = [\begin{array}{l} 0 1 \\ 0 0 \end{array}], show that {(aI + bA)}^{n} = a^{n} I + {na}^{n - 1} bA, where I is the identity \\ matrix of order 2 and n \in N . \end{array}$

Ans

$\begin{array}{l} We shall prove the result by using the principle of mathematical induction . \\ For n = 1, we have \\ P (1) : (al + bA) = al + {ba}^{0} A \\ = al + bA \\ Therefore, the result is true for n = 1. \\ Let the result be true for n = k \\ so, \\ P {(al + bA)}^{k} = a^{k} l + {ka}^{k - 1} bA Now, we will prove that it is true for n = k +1 \\ {(al + bA)}^{k + 1} = {(al + bA)}^{k} (al + bA) \\ = (a^{k} + {ka}^{k - 1} bA) (al + bA) \\ = a^{k + 1} l + {ka}^{k} bA + a^{k} bAl + {ka}^{k - 1} b^{2} A^{2} \\ {(al + bA)}^{k + 1} = a^{k + 1} + (k + 1) a^{k} bA + {ka}^{k - 1} b^{2} A^{2} . \dots.. (1) \\ Now, A^{2} = [\begin{array}{l} 0 1 \\ 0 0 \end{array}] [\begin{array}{l} 0 1 \\ 0 0 \end{array}] \\ = [\begin{array}{l} 0 1 \\ 0 0 \end{array}] = 0 \\ From equation (1), we have \\ {(al + bA)}^{k + 1} = a^{k + 1} + (k + 1) a^{k} bA + {ka}^{k - 1} b^{2} (0) \\ = a^{k + 1} + (k + 1) a^{k} bA \\ Therefore, the result is true for n = k +1. \\ Thus, by the Principle of Mathematical Induction, we have \\ {(aI + bA)}^{n} = a^{n} I + {na}^{n - 1} bA, for all n \in N . \end{array}$

Q.38

$\begin{array}{l} Find the inverse of the following matrix, if it exists . \\ [\begin{matrix} \begin{array}{l} 2 - 3 3 \\ 2 2 3 \end{array} \\ 3 - 2 2 \end{matrix}] \end{array}$

Ans

$\begin{array}{l} Let A = [\begin{matrix} \begin{array}{l} 2 - 3 3 \\ 2 2 3 \end{array} \\ 3 - 2 2 \end{matrix}] \\ We know that A = IA \\ ∴ [\begin{matrix} \begin{array}{l} 2 - 3 3 \\ 2 2 3 \end{array} \\ 3 - 2 2 \end{matrix}] = [\begin{matrix} \begin{array}{l} 1 0 0 \\ 0 1 0 \end{array} \\ 0 0 1 \end{matrix}] A \\ \Rightarrow [\begin{matrix} \begin{array}{l} 2 - 3 3 \\ 0 5 0 \end{array} \\ 3 - 2 2 \end{matrix}] = [\begin{matrix} \begin{array}{l} 1 0 0 \\ - 1 1 0 \end{array} \\ 0 0 1 \end{matrix}] A (R_{2} \to R_{2} - R_{1}) \\ \Rightarrow [\begin{matrix} \begin{array}{l} 2 - 3 3 \\ 0 1 0 \end{array} \\ 3 - 2 2 \end{matrix}] = [\begin{matrix} \begin{array}{l} 1 0 0 \\ - \frac{1}{5} \frac{1}{5} 0 \end{array} \\ 0 0 1 \end{matrix}] A (R_{2} \to \frac{1}{5} R_{2}) \\ \Rightarrow [\begin{matrix} \begin{array}{l} - 1 - 1 1 \\ 0 1 0 \end{array} \\ 3 - 2 2 \end{matrix}] = [\begin{matrix} \begin{array}{l} 1 0 - 1 \\ - \frac{1}{5} \frac{1}{5} 0 \end{array} \\ 0 0 1 \end{matrix}] A (R_{1} \to R_{1} - R_{3}) \\ \Rightarrow [\begin{matrix} \begin{array}{l} - 1 0 1 \\ 0 1 0 \end{array} \\ 3 0 2 \end{matrix}] = [\begin{matrix} \begin{array}{l} \frac{4}{5} \frac{1}{5} - 1 \\ - \frac{1}{5} \frac{1}{5} 0 \end{array} \\ - \frac{2}{5} \frac{2}{5} 1 \end{matrix}] A (\begin{array}{l} R_{1} \to R_{1} + R_{2} and \\ R_{3} \to R_{3} + 2 R_{2} \end{array}) \\ \Rightarrow [\begin{matrix} \begin{array}{l} - 1 0 1 \\ 0 1 0 \end{array} \\ 0 0 5 \end{matrix}] = [\begin{matrix} \begin{array}{l} \frac{4}{5} \frac{1}{5} - 1 \\ - \frac{1}{5} \frac{1}{5} 0 \end{array} \\ 2 1 - 2 \end{matrix}] A (R_{3} \to R_{3} + 3 R_{1}) \\ \Rightarrow [\begin{matrix} \begin{array}{l} - 1 0 1 \\ 0 1 0 \end{array} \\ 0 0 1 \end{matrix}] = [\begin{matrix} \begin{array}{l} \frac{4}{5} \frac{1}{5} - 1 \\ - \frac{1}{5} \frac{1}{5} 0 \end{array} \\ \frac{2}{5} \frac{1}{5} \frac{- 2}{5} \end{matrix}] A (R_{3} \to \frac{1}{5} R_{3}) \\ \Rightarrow [\begin{matrix} \begin{array}{l} - 1 0 1 \\ 0 1 0 \end{array} \\ 0 0 1 \end{matrix}] = [\begin{matrix} \begin{array}{l} \frac{2}{5} 0 \frac{- 3}{5} \\ - \frac{1}{5} \frac{1}{5} 0 \end{array} \\ \frac{2}{5} \frac{1}{5} \frac{- 2}{5} \end{matrix}] A (R_{3} \to R_{1} - R_{3}) \\ \Rightarrow [\begin{matrix} \begin{array}{l} 1 0 0 \\ 0 1 0 \end{array} \\ 0 0 1 \end{matrix}] = [\begin{matrix} \begin{array}{l} - \frac{2}{5} 0 \frac{3}{5} \\ - \frac{1}{5} \frac{1}{5} 0 \end{array} \\ \frac{2}{5} \frac{1}{5} \frac{- 2}{5} \end{matrix}] A (R_{1} \to (- 1) R_{1}) \\ A^{- 1} = [\begin{matrix} \begin{array}{l} - \frac{2}{5} 0 \frac{3}{5} \\ - \frac{1}{5} \frac{1}{5} 0 \end{array} \\ \frac{2}{5} \frac{1}{5} \frac{- 2}{5} \end{matrix}] \end{array}$

Q.39

$\begin{array}{l} Find the values of x, y, z if the matrix \\ A = [\begin{matrix} 0 2 y z \\ x y - z \\ x - y z \end{matrix}] satisfy the equation A ‘ A = I_{3} . \end{array}$

Ans

$\begin{array}{l} A = [\begin{matrix} 0 2 y z \\ x y - z \\ x - y z \end{matrix}] and A ‘ = [\begin{matrix} 0 x x \\ 2 y y - y \\ z - z z \end{matrix}] \\ Now, \\ A ‘ A = I_{3} \Rightarrow [\begin{matrix} 0 x x \\ 2 y y - y \\ z - z z \end{matrix}] . [\begin{matrix} 0 2 y z \\ x y - z \\ x - y z \end{matrix}] = [\begin{matrix} 1 0 0 \\ 0 1 0 \\ 0 0 1 \end{matrix}] \\ [\begin{matrix} 0 + x^{2} + x^{2} 0 + xy - xy 0 - xz + xz \\ 0 + yx - yx 4 y^{2} + y^{2} + y^{2} 2 yz - yz - yz \\ 0 - zx + zx 2 yz - yz - yz z^{2} + z^{2} + z^{2} \end{matrix}] = [\begin{matrix} 1 0 0 \\ 0 1 0 \\ 0 0 1 \end{matrix}] \\ [\begin{matrix} 2 x^{2} 0 0 \\ 0 6 y^{2} 0 \\ 0 0 3 z^{2} \end{matrix}] = [\begin{matrix} 1 0 0 \\ 0 1 0 \\ 0 0 1 \end{matrix}] \\ \Rightarrow 2 x^{2} = 1, 6 y^{2} = 1 and 3 z^{2} = 1 \\ \Rightarrow x = \pm \frac{1}{\sqrt{6}}, y = \pm \frac{1}{\sqrt{6}} and z = \pm \frac{1}{\sqrt{3}} \end{array}$

Q.40

$If A = [\begin{matrix} \begin{array}{l} 1 2 2 \\ 2 1 2 \end{array} \\ 2 2 1 \end{matrix}], prove that A^{2} - 4 A - 5 I = 0. Hence, find A^{- 1} .$

Ans

$\begin{array}{l} Given A = If A = [\begin{matrix} 1 2 2 \\ 2 1 2 \\ 2 2 1 \end{matrix}] \\ \Rightarrow A^{2} = [\begin{matrix} \begin{array}{l} 1 2 2 \\ 2 1 2 \end{array} \\ 2 2 1 \end{matrix}] . [\begin{matrix} \begin{array}{l} 1 2 2 \\ 2 1 2 \end{array} \\ 2 2 1 \end{matrix}] \\ A^{2} = [\begin{array}{l} 1 + 4 + 4 2 + 2 + 4 2 + 4 + 2 \\ 2 + 2 + 4 4 + 1 + 4 4 + 2 + 2 \\ 2 + 4 + 2 4 + 2 + 2 4 + 4 + 1 \end{array}] \\ = [\begin{matrix} \begin{array}{l} 9 8 8 \\ 8 9 8 \end{array} \\ 8 8 9 \end{matrix}] \\ Hence, A^{2} - 4 A - 5 I = [\begin{matrix} \begin{array}{l} 9 8 8 \\ 8 9 8 \end{array} \\ 8 8 9 \end{matrix}] - 4 [\begin{matrix} \begin{array}{l} 1 2 2 \\ 2 1 2 \end{array} \\ 2 2 1 \end{matrix}] - 5 [\begin{matrix} \begin{array}{l} 1 0 0 \\ 0 1 0 \end{array} \\ 0 0 1 \end{matrix}] \\ = [\begin{matrix} \begin{array}{l} 9 8 8 \\ 8 9 8 \end{array} \\ 8 8 9 \end{matrix}] + [\begin{matrix} - 4 & - 8 & - 8 \\ - 8 & - 4 & - 8 \\ - 8 & - 8 & - 4 \end{matrix}] + [\begin{matrix} \begin{array}{l} - 5 0 0 \\ 0 - 5 0 \end{array} \\ 0 0 5 \end{matrix}] \\ \Rightarrow A^{2} - 4 A -5 I = 0 \\ A^{2} - 4 A =5 I \\ A (A - 4 I) = 5 I \\ A^{- 1} = \frac{1}{5} \{[\begin{array}{l} 1 - 4 2 - 02 - 0 \\ 2 - 01 - 42 - 0 \\ 2 - 02 - 01 - 4 \end{array}]\} \\ A^{- 1} = \frac{1}{5} [\begin{matrix} \begin{array}{l} - 3 2 2 \\ 2 - 3 2 \end{array} \\ 2 2 - 3 \end{matrix}] \end{array}$

Q.41

$A = [\begin{matrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{matrix}] and f (x) = x^{3} - 6 x^{2} + 7 x + 2, verify that f (A) =0.$

Ans

$\begin{array}{l} We have A = [\begin{matrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{matrix}] \\ So, A^{2} = A, A [\begin{matrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{matrix}] . [\begin{matrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{matrix}] \\ = [\begin{array}{l} 1 + 0 + 4 0 + 0 + 0 2 + 0 + 6 \\ 0 + 0 + 2 0 + 4 + 0 0 + 2 + 3 \\ 2 + 0 + 6 0 + 0 + 0 4 + 0 + 9 \end{array}] \\ A^{2} = [\begin{matrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{matrix}] \\ A^{3} = A^{2}, A [\begin{matrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{matrix}] . [\begin{matrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{matrix}] = [\begin{matrix} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{matrix}] \\ Since, f (x) = x^{3} - 6 x^{2} + 7 x + 2 \\ Putting x = A, we get \\ f (A) = A^{3} - 6 A^{2} + 7 A +2 I \\ = [\begin{matrix} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{matrix}] - 6 [\begin{matrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{matrix}] + 7 [\begin{matrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{matrix}] + 2 [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] \\ = [\begin{matrix} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{matrix}] + [\begin{matrix} 30 & 0 & - 48 \\ 12 & - 24 & - 30 \\ - 48 & 0 & - 78 \end{matrix}] + [\begin{matrix} 7 & 0 & 14 \\ 0 & 14 & 7 \\ 14 & 0 & 21 \end{matrix}] + [\begin{matrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{matrix}] \\ = [\begin{matrix} 21 - 30 + 7 + 2 0 + 0 + 0 + 0 34 - 48 +14+0 \\ 12 - 12 + 0 + 0 8 - 24 + 14 + 2 23 - 30 + 7 + 0 \\ 34 - 48 + 14 + 0 0 + 0 + 0 + 0 55 - 78 + 21 + 2 \end{matrix}] \\ = [\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}] \\ ∴ f (A) = 0 \end{array}$

Q.42

$\begin{array}{l} Find A^{- 1}, when A = [\begin{matrix} 1 & 2 & - 3 \\ 2 & 3 & 2 \\ 3 & - 3 & - 4 \end{matrix}] Hence, solve the following system of equations : \\ x +2 y - 3 z =-4 \\ 2 x +3 y +2 z = 2 \\ 3 x -3 y -4 z =11 \end{array}$

Ans

$\begin{array}{l} Since A = IA, So \\ [\begin{matrix} 1 & 2 & - 3 \\ 2 & 3 & 2 \\ 3 & - 3 & - 4 \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] A \\ [\begin{matrix} 1 & 2 & - 3 \\ 0 & - 1 & 8 \\ 0 & - 9 & 5 \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 0 & 1 \end{matrix}] A [\begin{array}{l} R_{2} \to R_{2} + (- 2) R_{1} \\ R_{3} \to R_{3} + (- 3) R_{1} \end{array}] \\ [\begin{matrix} 1 & 2 & - 3 \\ 0 & - 1 & 8 \\ 0 & - 9 & - 5 \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ 2 & - 1 & 0 \\ 3 & 0 & - 1 \end{matrix}] A [\begin{array}{l} R_{2} \to (- 2) R_{2} \\ R_{3} \to (- 3) R_{3} \end{array}] \\ [\begin{matrix} 1 & 0 & 13 \\ 0 & 1 & - 8 \\ 0 & 0 & 67 \end{matrix}] = [\begin{matrix} - 3 & 2 & 0 \\ 2 & - 1 & 0 \\ - 15 & 9 & - 1 \end{matrix}] A [\begin{array}{l} R_{1} \to R_{1} + (- 2) R_{2} \\ R_{3} \to R_{3} + (- 9) R_{2} \end{array}] \\ [\begin{matrix} 1 & 0 & 13 \\ 0 & 1 & - 8 \\ 0 & 0 & 1 \end{matrix}] = [\begin{matrix} - 3 & 2 & 0 \\ 2 & - 1 & 0 \\ \frac{- 15}{67} & \frac{9}{67} & \frac{- 1}{67} \end{matrix}] A [R_{3} \to \frac{1}{67} R_{3}] \\ [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] = [\begin{matrix} \frac{- 3}{67} & \frac{17}{67} & \frac{13}{67} \\ \frac{14}{67} & \frac{5}{67} & \frac{- 8}{67} \\ \frac{- 15}{67} & \frac{9}{67} & \frac{- 1}{67} \end{matrix}] A [\begin{array}{l} R_{1} \to R_{1} + (- 13) R_{3} \\ R_{2} \to R_{2} + 8 R_{3} \end{array}] \\ ∴ A^{- 1} = [\begin{matrix} \frac{- 6}{67} & \frac{17}{67} & \frac{13}{67} \\ \frac{14}{67} & \frac{5}{67} & \frac{- 8}{67} \\ \frac{- 15}{67} & \frac{9}{67} & \frac{- 1}{67} \end{matrix}] \end{array}$ $\begin{array}{l} x +2 y - 3 z =-4 \\ 2 x +3 y +2 z = 2 \\ 3 x -3 y -4 z =11 \\ A = [\begin{matrix} 1 & 2 & - 3 \\ 2 & 3 & 2 \\ 3 & - 3 & - 4 \end{matrix}], x = [\begin{matrix} x \\ y \\ z \end{matrix}] and B [\begin{matrix} - 4 \\ 2 \\ 11 \end{matrix}] \\ So, \\ x = A^{- 1} B = [\begin{matrix} \frac{- 6}{67} & \frac{17}{67} & \frac{13}{67} \\ \frac{14}{67} & \frac{5}{67} & \frac{- 8}{67} \\ \frac{- 15}{67} & \frac{9}{67} & \frac{- 1}{67} \end{matrix}] [\begin{matrix} - 4 \\ 2 \\ 11 \end{matrix}] \\ = \frac{1}{67} [\begin{matrix} - 6 & 17 & 13 \\ 14 & 5 & - 8 \\ - 15 & 9 & - 1 \end{matrix}] [\begin{matrix} - 4 \\ 2 \\ 11 \end{matrix}] \\ = \frac{1}{67} [\begin{array}{l} 201 \\ - 134 \\ 67 \end{array}] [\begin{matrix} \frac{201}{67} \\ \frac{- 134}{67} \\ \frac{67}{67} \end{matrix}] \\ [\begin{matrix} x \\ y \\ z \end{matrix}] = [\begin{matrix} 3 \\ - 2 \\ 1 \end{matrix}] \Rightarrow x = 3, y = - 2 and z = 1 \end{array}$

Q.43

$\begin{array}{l} Find A^{- 1} by using elementary transformations, where \\ A = [\begin{matrix} 2 & - 1 & 4 \\ 4 & 0 & 2 \\ 3 & - 2 & 7 \end{matrix}] \end{array}$

Ans

$\begin{array}{l} Since A = IA, So \\ [\begin{matrix} 2 & - 1 & 4 \\ 4 & 0 & 2 \\ 3 & - 2 & 7 \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] A \\ [\begin{matrix} 3 & - 2 & 7 \\ 4 & 0 & 2 \\ 2 & - 1 & 4 \end{matrix}] = [\begin{matrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{matrix}] A [R_{1} \leftrightarrow R_{3}] \\ [\begin{matrix} 1 & - 1 & 3 \\ 4 & 0 & 2 \\ 2 & - 1 & 4 \end{matrix}] = [\begin{matrix} - 1 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{matrix}] A [R_{1} \to R_{3} - R_{3}] \\ [\begin{matrix} 1 & - 1 & 3 \\ 0 & 4 & - 10 \\ 0 & 1 & - 2 \end{matrix}] = [\begin{matrix} - 1 & 0 & 1 \\ 4 & 1 & - 4 \\ 3 & 0 & - 2 \end{matrix}] A [\begin{array}{l} R_{2} \to R_{2} + (- 4) R_{1} \\ R_{3} \to R_{3} + (- 2) R_{1} \end{array}] \\ [\begin{matrix} 1 & - 1 & 3 \\ 0 & 1 & - 2 \\ 0 & 4 & - 10 \end{matrix}] = [\begin{matrix} - 1 & 0 & 1 \\ 3 & 0 & - 2 \\ 4 & 1 & - 4 \end{matrix}] A [R_{2} \leftrightarrow R_{3}] \\ [\begin{matrix} 1 & 0 & 1 \\ 0 & 1 & - 2 \\ 0 & 0 & - 2 \end{matrix}] = [\begin{matrix} 2 & 0 & - 1 \\ 3 & 0 & - 2 \\ - 8 & 1 & 4 \end{matrix}] A [\begin{array}{l} R_{1} \to R_{1} + R_{2} \\ R_{3} \to R_{3} + (- 4) R_{2} \end{array}] \\ [\begin{matrix} 1 & 0 & 1 \\ 0 & 1 & - 2 \\ 0 & 0 & 1 \end{matrix}] = [\begin{matrix} 2 & 0 & - 1 \\ 3 & 0 & - 2 \\ 4 & \frac{- 1}{2} & - 2 \end{matrix}] A [R_{3} \to (- \frac{1}{2}) R_{3}] \\ [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] = [\begin{matrix} - 2 & \frac{1}{2} & - 1 \\ 11 & - 1 & - 6 \\ 4 & \frac{1}{2} & - 2 \end{matrix}] A [\begin{array}{l} R_{1} \to R_{1} - R_{3} \\ R_{2} \to R_{2} + 2 R_{3} \end{array}] \\ Hence, A^{- 1} = [\begin{matrix} - 2 & \frac{1}{2} & - 1 \\ 11 & - 1 & - 6 \\ 4 & \frac{- 1}{2} & - 2 \end{matrix}] . \end{array}$

Q.44

$If F (x) = [\begin{matrix} cosx & - sinx & 0 \\ sinx & cosx & 0 \\ 0 & 0 & 1 \end{matrix}], show that F (x) F (y) = F (x + y) .$

Ans

$\begin{array}{l} Given : F (x) = F (x) = [\begin{matrix} cosx & - sinx & 0 \\ sinx & cosx & 0 \\ 0 & 0 & 1 \end{matrix}] \\ F (y) = [\begin{matrix} cosy & - siny & 0 \\ siny & cosy & 0 \\ 0 & 0 & 1 \end{matrix}] \\ L . H . S . : \\ F (x), F (y) = [\begin{matrix} cosx & - sinx & 0 \\ sinx & cosx & 0 \\ 0 & 0 & 1 \end{matrix}] [\begin{matrix} cosy & - siny & 0 \\ siny & cosy & 0 \\ 0 & 0 & 1 \end{matrix}] \\ = [\begin{matrix} cosxcosy - sinx csiny - sinycosx - sinxcosy 0 \\ sinxcosy + cosx siny - sinxsiny + cosxcosy 0 \\ 0 0 1 \end{matrix}] \\ = [\begin{matrix} \cos (x + y) - \sin (x + y) 0 \\ \sin (x + y) - \cos (x + y) 0 \\ 0 0 1 \end{matrix}] \\ = F (x + y) \\ = R . H . S . \end{array}$

Q.45

$If x [\begin{matrix} 5 \\ 6 \end{matrix}] + y [\begin{matrix} - 1 \\ 1 \end{matrix}] = [\begin{matrix} 15 \\ 7 \end{matrix}], find the value of x and y .$

Ans

$\begin{array}{l} Given : \\ x [\begin{matrix} 5 \\ 6 \end{matrix}] + y [\begin{matrix} - 1 \\ 1 \end{matrix}] = [\begin{matrix} 15 \\ 7 \end{matrix}] \\ \Rightarrow [\begin{matrix} 5 x \\ 6 x \end{matrix}] + y [\begin{matrix} - y \\ y \end{matrix}] = [\begin{matrix} 15 \\ 7 \end{matrix}] \\ \Rightarrow [\begin{matrix} 5 x - y \\ 6 x + y \end{matrix}] = [\begin{matrix} 15 \\ 7 \end{matrix}] \\ \Rightarrow 5 x - y = 15, 6 x + y = 7 \\ \Rightarrow x = 2 and y = - 5 \end{array}$

Q.46

$Find X, if Y = [\begin{matrix} \begin{array}{l} 5 \\ 1 \end{array} & \begin{array}{l} 7 \\ 3 \end{array} \end{matrix}] and 2 X + Y = [\begin{matrix} \begin{array}{l} 1 \\ - 3 \end{array} & \begin{array}{l} 0 \\ 2 \end{array} \end{matrix}] .$

Ans

$\begin{array}{l} Given, Y = [\begin{matrix} \begin{array}{l} 5 \\ 1 \end{array} & \begin{array}{l} 7 \\ 4 \end{array} \end{matrix}] \\ and 2 X + Y = [\begin{matrix} \begin{array}{l} 1 \\ - 3 \end{array} & \begin{array}{l} 0 \\ 2 \end{array} \end{matrix}] . \dots\dots (i) \\ Putting value of Y in equation (i), we get \\ 2 X + [\begin{matrix} \begin{array}{l} 5 \\ 1 \end{array} & \begin{array}{l} 7 \\ 4 \end{array} \end{matrix}] = [\begin{matrix} \begin{array}{l} 1 \\ - 3 \end{array} & \begin{array}{l} 0 \\ 2 \end{array} \end{matrix}] \\ 2 X = [\begin{matrix} \begin{array}{l} 1 \\ - 3 \end{array} & \begin{array}{l} 0 \\ 2 \end{array} \end{matrix}] - [\begin{matrix} \begin{array}{l} 5 \\ 1 \end{array} & \begin{array}{l} 7 \\ 3 \end{array} \end{matrix}] \\ = [\begin{matrix} 1 & - 5 \\ - 3 & - 1 \end{matrix} \begin{matrix} 0 & - 7 \\ 2 & - 4 \end{matrix}] \\ X = \frac{1}{2} [\begin{matrix} \begin{array}{l} - 4 \\ - 4 \end{array} & \begin{array}{l} - 7 \\ - 2 \end{array} \end{matrix}] \\ = [\begin{matrix} - 2 & - \frac{7}{2} \end{matrix}] \end{array}$

Q.47

$\begin{array}{l} Using elementary transformations, find the inverse \\ of the matrix [\begin{matrix} \begin{array}{l} 1 \\ 2 \end{array} & \begin{array}{l} - 2 \\ 3 \end{array} \end{matrix}] if it exists . \end{array}$

Ans

$\begin{array}{l} Let A = [\begin{array}{l} \begin{matrix} 1 \end{matrix} - 5 \\ 3 3 \end{array}] \\ Since, A = IA \\ [\begin{matrix} \begin{array}{l} 1 \\ 3 \end{array} & \begin{array}{l} - 5 \\ 3 \end{array} \end{matrix}] = [\begin{matrix} \begin{array}{l} 1 \\ 0 \end{array} & \begin{array}{l} 0 \\ 1 \end{array} \end{matrix}] A \\ Apply R_{2} \to R_{2} - 3 R_{1} \\ [\begin{matrix} \begin{array}{l} 1 \\ 3 \end{array} & \begin{array}{l} - 5 \\ 18 \end{array} \end{matrix}] = [\begin{matrix} \begin{array}{l} 1 \\ - 3 \end{array} & \begin{array}{l} 0 \\ 1 \end{array} \end{matrix}] A \\ R_{2} \to \frac{1}{18} R_{2} \\ ∴ [\begin{matrix} \begin{array}{l} 1 \\ 0 \end{array} & \begin{array}{l} - 5 \\ 1 \end{array} \end{matrix}] = [\begin{matrix} \begin{array}{l} 1 \\ - \frac{1}{6} \end{array} & \begin{array}{l} 0 \\ \frac{1}{18} \end{array} \end{matrix}] A \\ Apply R_{1} \to R_{1} + 5 R_{2} \\ [\begin{matrix} \begin{array}{l} 1 \\ 0 \end{array} & \begin{array}{l} 0 \\ 1 \end{array} \end{matrix}] = [\begin{matrix} \begin{array}{l} \frac{1}{6} \\ - \frac{1}{6} \end{array} & \begin{array}{l} \frac{5}{18} \\ \frac{1}{18} \end{array} \end{matrix}] A \\ A^{- 1} = [\begin{matrix} \begin{array}{l} \frac{1}{6} \\ - \frac{1}{6} \end{array} & \begin{array}{l} \frac{5}{18} \\ \frac{1}{18} \end{array} \end{matrix}] \end{array}$

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FAQs (Frequently Asked Questions)

The order of a matrix decides which operations are possible. Addition and subtraction need the same order. Matrix multiplication needs columns of the first matrix to match rows of the second matrix.

Two matrices are equal when they have the same order and all corresponding elements are equal. If even one corresponding element is different, the matrices are not equal.

Matrix multiplication depends on row-column order. AB and BA may have different orders or different values. Sometimes AB may be defined while BA may not be defined.

Check whether AT = -A. Also check the diagonal entries. In every skew-symmetric matrix, all diagonal elements must be zero.

Remember that transpose changes rows into columns. For sums, transpose distributes normally: (A + B)T = AT + BT. For products, the order reverses: (AB)T = BTAT.

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