Name: Properties of Determinants (Part 3)
Uploaded: 2019-07-19
Duration: 3 min 53 s
Description: Suppose we multiply a row of a determinant with a particular value, then what would be the change in the value of the determinant? Watch this video to learn...

Question

Solve the system of linear equations by matrix method

$x - y + z = 4$

$2 x + y - 3 z = 0$

$x + y + z = 2$

Accepted Answer

Given

$x - y + z = 4$

$2 x + y - 3 z = 0$

$x + y + z = 2$

The matrix form of the system of linear equations is as follows

$[\begin{matrix} 1 & - 1 & 1 \\ 2 & 1 & - 3 \\ 1 & 1 & 1 \end{matrix}] [\begin{matrix} x \\ y \\ z \end{matrix}] = [\begin{matrix} 4 \\ 0 \\ 2 \end{matrix}] A X = B$

$\Rightarrow X = A^{- 1} B$

where, $A - 1 = \frac{Adj A}{|\begin{matrix} A \end{matrix}|}$

$|\begin{matrix} A \end{matrix}| = 1 |\begin{matrix} 1 & - 3 \\ 1 & 1 \end{matrix}| - (- 1) |\begin{matrix} 2 & - 3 \\ 1 & 1 \end{matrix}| + |\begin{matrix} 2 & 1 \\ 1 & 1 \end{matrix}| = 1 [1 \times 1 - (- 3) \times 1] + 1 [2 \times 1 - 1 \times (- 3)] + 1 (2 \times 1 - 1 \times 1) = 1 (1 + 3) + 1 (2 + 3) + 1 (2 - 1) = 4 + 5 + 1 = 10$

Since it is a nonsingular matrix, therefore, it has a unique solution.

Minors of the elements of $A$ are as follows

$M_{11} = |\begin{matrix} 1 & - 3 \\ 1 & 1 \end{matrix}| = 1 \times 1 - (- 3) \times 1 = 1 + 3 = 4$

$M_{12} = |\begin{matrix} 2 & - 3 \\ 1 & 1 \end{matrix}| = 2 \times 1 - (- 3) \times 1 = 2 + 3 = 5$

$M_{13} = |\begin{matrix} 2 & 1 \\ 1 & 1 \end{matrix}| = 2 \times 1 - 1 \times 1 = 2 - 1 = 1$

$M_{21} = |\begin{matrix} - 1 & 1 \\ 1 & 1 \end{matrix}| = (- 1) \times 1 - 1 \times 1 = - 1 - 1 = - 2$

$M_{22} = |\begin{matrix} 1 & 1 \\ 1 & 1 \end{matrix}| = 1 \times 1 - 1 \times 1 = 1 - 1 = 0$

$M_{23} = |\begin{matrix} 1 & - 1 \\ 1 & 1 \end{matrix}| = 1 \times 1 - 1 \times (- 1) = 1 + 1 = 2$

$M_{31} = |\begin{matrix} - 1 & 1 \\ 1 & - 3 \end{matrix}| = (- 1) \times (- 3) - 1 \times 1 = 3 - 1 = 2$

$M_{32} = |\begin{matrix} 1 & 1 \\ 2 & - 3 \end{matrix}| = 1 \times (- 3) - 2 \times 1 = - 3 - 2 = - 5$

$M_{33} = |\begin{matrix} 1 & - 1 \\ 2 & 1 \end{matrix}| = 1 \times 1 - (- 1) \times 2 = 1 + 2 = 3$

Co-factors of the elements of $A$ are as follows

$C_{11} = (- 1)^{1 + 1} M_{11} = 1 \times 4 = 4$

$C_{12} = (- 1)^{1 + 2} M_{12} = - 1 \times 5 = - 5$

$C_{13} = (- 1)^{1 + 3} M_{13} = 1 \times 1 = 1$

$C_{21} = (- 1)^{2 + 1} M_{21} = - 1 \times (- 2) = 2$

$C_{22} = (- 1)^{2 + 2} M_{22} = 1 \times 0 = 0$

$C_{23} = (- 1)^{2 + 3} M_{23} = (- 1) \times 2 = - 2$

$C_{31} = (- 1)^{3 + 1} M_{31} = 1 \times 2 = 2$

$C_{32} = (- 1)^{3 + 2} M_{32} = (- 1) \times (- 5) = 5$

$C_{33} = (- 1)^{3 + 3} M_{33} = 1 \times 3 = 3$

The Co-factor $C$ matrix of $A$ is

$C = [\begin{matrix} 4 & - 5 & 1 \\ 2 & 0 & - 2 \\ 2 & 5 & 3 \end{matrix}]$

$Adj A = C^{'} = [\begin{matrix} 4 & 2 & 2 \\ - 5 & 0 & 5 \\ 1 & - 2 & 3 \end{matrix}]$

$A^{- 1} = \frac{Adj A}{|\begin{matrix} A \end{matrix}|} = \frac{1}{10} [\begin{matrix} 4 & 2 & 2 \\ - 5 & 0 & 5 \\ 1 & - 2 & 3 \end{matrix}]$

Put in equation (i)

$X = A^{- 1} B$

$[\begin{matrix} x \\ y \\ z \end{matrix}] = \frac{1}{10} [\begin{matrix} 4 & 2 & 2 \\ - 5 & 0 & 5 \\ 1 & - 2 & 3 \end{matrix}] [\begin{matrix} 4 \\ 0 \\ 2 \end{matrix}] = \frac{1}{10} [\begin{matrix} 4 \times 4 + 2 \times 0 + 2 \times 2 \\ (- 5) \times 4 + 0 \times 0 + \times 5 \times 2 \\ 1 \times 4 + - 2 \times 0 + 3 \times 2 \end{matrix}] = \frac{1}{10} [\begin{matrix} 20 \\ - 10 \\ 10 \end{matrix}]$

$[\begin{matrix} x \\ y \\ z \end{matrix}] = [\begin{matrix} 2 \\ - 1 \\ 1 \end{matrix}]$

Using the property of equality of matrices, we get

$x = 2, y = - 1, z = 1$

Answer.

Solve the system of linear equations by matrix method

$x - y + z = 4$

$2 x + y - 3 z = 0$

$x + y + z = 2$

Important Questions on Determinants

Learn and Prepare for any exam you want !

Solve the system of linear equations by matrix method x-y+z=4 2x+y-3z=0 x+y+z=2

Important Questions on Determinants

Learn and Prepare for any exam you want !

Solve the system of linear equations by matrix method

$x - y + z = 4$

$2 x + y - 3 z = 0$

$x + y + z = 2$