What uses do determinants have in real life? Explain its concept.

Chapter 4 Determinants of Class 12 Mathematics explains the fundamental concepts and applications of determinants. Learning Determinants is important because they have numerous applications. A Determinant is a function that can be used to connect any square matrix to a real or complex integer. This principle is used to determine whether an equation system has a unique solution. In addition, Determinants have numerous applications in engineering, Science, Economics and Social Science because it is useful for understanding 'cross- product' and to find a solution of a system of equations.

What are the properties of Determinants?

Reflection, all zero, proportionality, switching, factor, scaler, multiple, sum, triangle, cofactor matrix and invariance are the properties of a Determinant. There are various applications of Determinants in all of these properties, such as reflection, which depicts that it does not change with changes in rows and columns, and zero, which depicts that if rows and columns are similar to others, then the determinant will also be zero.

What are the uses of determinants?

Determinants are very important in algebra. To solve linear equations, you must first understand determinants. This is a significant concept that is also useful in calculating area, volume and other integrals. Determinants can also be used to calculate the values in square matrices.

Home > CBSE > CBSE Important Questions Class 12 Maths Chapter 4 – Determinants

CBSE Important Questions Class 12 Maths Chapter 4 – Determinants

Class 12 Maths Chapter 4 Determinants is part of the CBSE Class 12 Maths syllabus. Every square matrix has a number associated with it called its determinant. Students learn how to evaluate determinants of order 2 and 3, the properties that make evaluation easier, minors and cofactors, the adjoint and inverse of a matrix, and how to solve a system of linear equations using matrices.

These CBSE important questions for Class 12 Maths Chapter 4 cover finding the area of a triangle using determinants, the determinant of an inverse matrix, properties of determinants, minors and cofactors, singular and non-singular matrices, finding the inverse using the adjoint, and solving a system of equations by the matrix method. Every question has a clear step-by-step solution in simple English, so students can practise on their own, parents can check the working, and teachers can use them for class tests.

Q1. Find the area of the triangle whose vertices are (1, 0), (6, 0) and (4, 3).

Answer: The area of a triangle with vertices (x₁, y₁), (x₂, y₂) and (x₃, y₃) is
Δ = ½ |x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)|
= ½ |1(0 − 3) + 6(3 − 0) + 4(0 − 0)|
= ½ |−3 + 18 + 0|
= ½ × 15
= 15/2 = 7.5 square units

Q2. If A is an invertible matrix of order 2, then det(A⁻¹) is equal to
(a) det(A)
(b) 1/det(A)
(c) 1
(d) 0

Answer: (b) 1/det(A)
We know AA⁻¹ = I.
Taking determinants of both sides and using det(AB) = det(A) × det(B):
det(A) × det(A⁻¹) = det(I) = 1
So det(A⁻¹) = 1/det(A).
(This also shows why a matrix must be non-singular, that is det(A) ≠ 0, to have an inverse.)

Q3. Write the main properties of determinants.

Answer: • The value of a determinant does not change if rows and columns are interchanged, that is |A′| = |A|.
• If any two rows (or columns) are interchanged, the sign of the determinant changes.
• If any two rows (or columns) are identical, the determinant is zero.
• If every element of a row (or column) is multiplied by k, the determinant gets multiplied by k.
• If every element of a row (or column) is zero, the determinant is zero.
• The determinant is unchanged if a multiple of one row (or column) is added to another.

Q4. Define minor and cofactor. Find the cofactor of the element a₁₂ in the determinant |[1, 2, 3], [4, 5, 6], [7, 8, 9]|.

Answer: The minor Mᵢⱼ of an element aᵢⱼ is the determinant obtained by deleting the ith row and the jth column.
The cofactor is Aᵢⱼ = (−1)^i+j Mᵢⱼ.
For a₁₂ = 2: Delete row 1 and column 2 to get the minor
M₁₂ = |[4, 6], [7, 9]| = (4)(9) − (6)(7) = 36 − 42 = −6
Cofactor A₁₂ = (−1)¹⁺² × (−6) = (−1)(−6) = 6

Q5. What is a singular matrix? Find the value of x for which the matrix [[x, 2], [8, x]] is singular.

Answer: A square matrix is singular if its determinant is zero. A singular matrix has no inverse.
For the given matrix:
det = (x)(x) − (2)(8) = x² − 16
Setting the determinant to zero:
x² − 16 = 0
x² = 16
x = 4 or x = −4

Q6. Find the adjoint of the matrix A = [[2, 3], [1, 4]] and hence find A⁻¹.

Answer: For a 2 × 2 matrix [[a, b], [c, d]], the adjoint is [[d, −b], [−c, a]].
So adj A = [[4, −3], [−1, 2]]
Determinant: det A = (2)(4) − (3)(1) = 8 − 3 = 5
Since det A ≠ 0, the inverse exists.
A⁻¹ = (1/det A) × adj A
= (1/5) × [[4, −3], [−1, 2]]
= [[4/5, −3/5], [−1/5, 2/5]]

Q7. Evaluate the determinant |[1, 2, 3], [2, 3, 1], [3, 1, 2]|.

Answer: Expand along the first row:
= 1 × |[3, 1], [1, 2]| − 2 × |[2, 1], [3, 2]| + 3 × |[2, 3], [3, 1]|
= 1(6 − 1) − 2(4 − 3) + 3(2 − 9)
= 1(5) − 2(1) + 3(−7)
= 5 − 2 − 21
= −18

Q8. Show that the points A(a, b + c), B(b, c + a) and C(c, a + b) are collinear.

Answer: Three points are collinear if the area of the triangle they form is zero, that is the determinant
|[a, b + c, 1], [b, c + a, 1], [c, a + b, 1]| = 0
Apply the column operation C₂ → C₂ + C₁:
The second column becomes (a + b + c), (a + b + c), (a + b + c).
Now take the common factor (a + b + c) out of column 2:
= (a + b + c) × |[a, 1, 1], [b, 1, 1], [c, 1, 1]|
Columns 2 and 3 are now identical, so the determinant is zero.
Hence the area is zero and the three points are collinear.

Q9. Solve the system of equations using the matrix method: 2x + 3y = 5 and 3x − 2y = 1.

Answer: Write the system as AX = B, where
A = [[2, 3], [3, −2]], X = [x, y]′ and B = [5, 1]′
det A = (2)(−2) − (3)(3) = −4 − 9 = −13
Since det A ≠ 0, a unique solution exists.
adj A = [[−2, −3], [−3, 2]]
A⁻¹ = (1/−13) × [[−2, −3], [−3, 2]]
X = A⁻¹B:
x = (1/−13)[(−2)(5) + (−3)(1)] = (1/−13)(−13) = 1
y = (1/−13)[(−3)(5) + (2)(1)] = (1/−13)(−13) = 1
So x = 1 and y = 1.

Q10. Write the condition for a system of linear equations AX = B to have (i) a unique solution, (ii) no solution and (iii) infinitely many solutions.

Answer: (i) Unique solution (consistent): det A ≠ 0, that is A is non-singular. Then X = A⁻¹B.
(ii) and (iii): If det A = 0, compute (adj A)B.
• If (adj A)B ≠ O, the system has no solution and is inconsistent.
• If (adj A)B = O, the system has either infinitely many solutions or no solution, and must be examined further.

Q11. If A is a square matrix of order 3 and |A| = 5, find (i) |2A| and (ii) |adj A|.

Answer: (i) |kA| = kⁿ|A|, where n is the order of the matrix.
Here k = 2 and n = 3:
|2A| = 2³ × 5 = 8 × 5 = 40
(ii) |adj A| = |A|ⁿ⁻¹
= 5³⁻¹ = 5² = 25

Q12. If A = [[3, 1], [−1, 2]], show that A² − 5A + 7I = O.

Answer: First compute A²:
Row 1: (3×3 + 1×−1) = 8 and (3×1 + 1×2) = 5
Row 2: (−1×3 + 2×−1) = −5 and (−1×1 + 2×2) = 3
A² = [[8, 5], [−5, 3]]
Now 5A = [[15, 5], [−5, 10]]
And 7I = [[7, 0], [0, 7]]
Putting it together:
A² − 5A + 7I
= [[8 − 15 + 7, 5 − 5 + 0], [−5 + 5 + 0, 3 − 10 + 7]]
= [[0, 0], [0, 0]]
= O. Hence proved.

Related Study Material on Extramarks

Q.1-

$If value of determinant |\begin{matrix} \sin 80 ° & \cos 80 ° \\ \sin 10 ° & \cos 10 ° \end{matrix}| is$

Opt-

0.
$\frac{1}{\sqrt{2}}$
1/2.
1.

Ans –

$\begin{array}{l} We have, \\ = |\begin{matrix} \sin 80 ° & \cos 80 ° \\ \sin 10 ° & \cos 10 ° \end{matrix}| \\ = \sin 80 ° \cos 10 ° + \cos 80 ° \sin 10 ° \\ = \sin (80 ° + 10 °) = \sin 90 ° = 1 . \end{array}$

Q.2-

$\begin{array}{l} Using properties of determinants, show that \\ |\begin{array}{l} 1 + a 1 1 \\ 1 1 + b 1 \\ 1 1 1 + c \end{array}| = abc (1 + \frac{1}{a} + \frac{1}{b} + \frac{1}{c}) = abc + bc + ca + ab . \end{array}$

Opt-

$\begin{array}{l} Let = |\begin{array}{l} 1 + a 1 1 \\ 1 1 + b 1 \\ 1 1 1 + c \end{array}| \\ By applying c_{1} c_{2} and c \end{array}$

Ans-

$\begin{array}{l} Let = |\begin{array}{l} 1 + a 1 1 \\ 1 1 + b 1 \\ 1 1 1 + c \end{array}| \\ By applying c_{1} c_{2} and c_{3} c_{3} c_{2}, we get \\ |\begin{array}{l} a 1 0 \\ b 1 + b b \\ 0 1 c \end{array}| \\ Expanding along R_{1}, we obtain \\ = a |\begin{array}{l} 1 + b b \\ 1 c \end{array}| 1 |\begin{array}{l} b b \\ 0 c \end{array}| + 0 \\ = a (c + bc + b) (bc) \\ = ac + abc + ab + bc \\ = abc + bc + ca + ab \\ = abc [a + \frac{1}{a} + \frac{1}{b} + \frac{1}{c}] \end{array}$

Q.3

$\begin{array}{l} If A = |\begin{array}{l} 1 -1 2 \\ 0 2 -3 \\ 3 -2 4 \end{array}| and B |\begin{array}{l} - 2 0 1 \\ 9 2 -3 \\ 6 1 -2 \end{array}| then find AB . \\ Using AB, solve the folowing system of equations : \\ x y +2 z =1 \\ 2 y 3 z =1 \\ 3 x 2 y +4 z = 2 \end{array}$

Ans-

$\begin{array}{l} Here, AB = |\begin{array}{l} 1 -1 2 \\ 0 2 -3 \\ 3 -2 4 \end{array}| |\begin{array}{l} - 2 0 1 \\ 9 2 -3 \\ 6 1 -2 \end{array}| \\ = |\begin{array}{l} 2 9 +1 2 0 2 + 2 1+3 4 \\ 0 +18 18 0+4 3 0 6 + 6 \\ 6 18 + 24 0 4 +43 +6 8 \end{array}| \\ = |\begin{array}{l} 1 0 0 \\ 0 1 0 \\ 0 0 1 \end{array}| \\ Then AB = I \\ Thus A^{1} = B \\ {|\begin{array}{l} 1 -1 2 \\ 0 2 -3 \\ 3 -2 4 \end{array}|}^{1} = |\begin{array}{l} - 2 0 1 \\ 9 2 -3 \\ 6 1 -2 \end{array}| \\ Now given equations can be written as A_{1} X = B_{1} \\ where, A_{1} = |\begin{array}{l} 1 -1 2 \\ 0 2 -3 \\ 3 -2 4 \end{array}| X = [\begin{array}{l} x \\ y \\ z \end{array}] and B_{1} €‹ = [\begin{array}{l} 1 \\ 1 \\ 2 \end{array}] \\ Now A_{1} = A \\ ˆ´ {A_{1}}^{1} = A^{1} \\ ‡’ {A_{1}}^{1} = |\begin{array}{l} - 2 0 1 \\ 9 2 -3 \\ 6 1 -2 \end{array}| \\ as X = {A_{1}}^{1} B_{1} \\ ˆ´ X = |\begin{array}{l} - 2 0 1 \\ 9 2 -3 \\ 6 1 -2 \end{array}| [\begin{array}{l} 1 \\ 1 \\ 2 \end{array}] \\ X = |\begin{array}{l} - 2 + 0 + 1 \\ 9 +2 -6 \\ 6 +1 -4 \end{array}| [\begin{array}{l} 0 \\ 5 \\ 3 \end{array}] \\ Hence, x = 0, y = 5 and z = 3 \end{array}$

Q.4-

$\begin{array}{l} Show that |\begin{array}{l} a b c \\ a + 2 x b +2 y c +2 z \\ x y z \end{array}| = 0 by \\ using properties of determinants . \end{array}$

Ans-

$\begin{array}{l} |\begin{array}{l} a b c \\ a + 2 x b +2 y c +2 z \\ x y z \end{array}| = |\begin{array}{l} a b c \\ a b c \\ x y z \end{array}| + |\begin{array}{l} a b c \\ 2 x 2 y 2 z \\ x y z \end{array}| \\ (\begin{array}{l} All elements of R_{2} are expressed as sum of \\ two terms . Therefore, the determinant can \\ be expressed as sum of two determinant . \end{array}) \\ = 0 + 2 |\begin{array}{l} a b c \\ a b c \\ x y z \end{array}| (2 is taken out common from R_{2} .) \\ (\begin{array}{l} If any two rows or columns of a determinant are \\ identical then the value of determinant is zero . \end{array}) \\ = 0 + 0 = 0 \end{array}$

CBSE Important Questions Class 12 Maths Chapter 4 – Determinants

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