# Solving Systems of Linear Equations

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• Q1

Write the given linear system in augmented matrix.

Marks:3

The augmented matrix of the given linear system is:

• Q2

What are the steps taken in Gauss Elimination Method to solve linear system?

Marks:3

The steps taken in Gauss Elimination Method are:
1. Construct augmented matrix for the system.

2. Use elementary row operations to transform augmented matrix into a triangular one.
3. Write down the new linear system for which triangular matrix is the associated augmented matrix.
4. Solve the new system. You may need to assign parametric values to some unknowns and then apply the method of back substitution to solve the new system.

• Q3

What will be the first step to reduce the given equation in lower triangular form to eliminate the first column?

Marks:3

The first step in Gaussian elimination is to use the top row to eliminate the first column from each subsequent row. For each, we have to carry out 1 division to calculate the multiple of row 1 to be subtracted, followed by n multiplication-subtractions. This makes n + 1 operations for each of the n 1 rows below the top one, leading to a total of (n 1)(n + 1) = n2 1 operations needed to eliminate the first column.

• Q4

Use back substitution to find the solution of the problem given below.

Marks:5

The last row in the augmented matrix represents the equation –

Therefore,

The second row of the augmented matrix represents the equation 2y + z = 4.

Substituting the value of z into this equation and solving for y gives us:

Finally, the first row of the augmented matrix represents the equation x + y + 2z = 6.
Substituting the values of y and z into the first equation yields:

• Q5

Solve the linear system given below.
x+yz=1 (1)
8x+3y−6z=1 (2)
−4xy+3z=1 (3)

Marks:5

The linear system is:

x+yz=1 (1)

8x+3y−6z=1 (2)

−4xy+3z=1 (3)

By Elimination Method: This method tries to eliminate variables until there's only 1 variable left.

Let us find 2 equations having same coefficient (plus or minus) for same variables. For example, see equations (1) and (3). The coefficient for y is 1 and -1 respectively. We will add the two equations to eliminate y and to get equation (4).

 x + y − z = 1 (1) −4x − y + 3z = 1 (3) + ------------------------ −3x + 2z = 2 (4)

Note that equation (4) consists of variables x and z. Now we need another equation having same variables as equation (4). To get this we'll eliminate y from equation (1) and (2). In equation (1) and (2), the coefficients of y are 1 and 3 respectively. In order to eliminate y, we multiply equation (1) by 3 and then subtract equation (2) from equation (1).

 x + y − z = 1 (1) × 3 3x + 3y − 3z = 3 (1) −8x + 3y − 6z = 1 (2) −8x + 3y − 6z = 1 (2) ------------------------- − −5x + 3z = 2 (5)

With equations (4) and (5), let's try to eliminate z.

 −3x + 2z = 2 (4) × 3 −9x + 6z = 6 (4) −5x + 3z = 2 (5) × 2 −10x + 6z = 4 (5) ------------------------- − x = 2 (6)

From equation (6), we get x = 2. Now, we can substitute this value of x to equation (4) to get the value for z.

 −3(2) + 2z = 2 (4) −6 + 2z = 2 2z = 8 z = 8 ÷ 2 z = 4

Finally, we can substitute the value of z to equation (1) to get y.

 2 + y − 4 = 1 (1) y = 1 − 2 + 4 y = 3

Hence, the solution to the system of linear equations is

x = 2, y = 3, z = 4.