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

Using matrix method, solve the following system of equations:

$x - 2 y + 3 z = 6$

$x + 4 y + z = 12$

$x - 3 y + 4 z = 1$ .

Accepted Answer

Given system of equations,

$x - 2 y + 3 z = 6$

$x + 4 y + z = 12$

$x - 3 y + 4 z = 1$ .

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

$[\begin{matrix} 1 & - 2 & 3 \\ 1 & 4 & 1 \\ 1 & - 3 & 2 \end{matrix}] [\begin{matrix} x \\ y \\ z \end{matrix}] = [\begin{matrix} 6 \\ 12 \\ 1 \end{matrix}] A X = B$

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

where

$A^{- 1} = \frac{Adj A}{|A|}$

$|A| = 1 (8 + 3) + 2 (2 - 1) + 3 (- 3 - 4)$

$= 11 + 2 - 21$

$= - 8 \neq 0$

Since $A$ is a non-singular matrix, therefore the system of equations has a unique solution.

Co-factors of elements of coefficient matrix $A$ are as follows:

$C_{11} = {(- 1)}^{1 + 1} M_{11} = (1) \times |\begin{matrix} 4 & 1 \\ - 3 & 2 \end{matrix}| = 8 + 3 = 11$

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

$C_{13} = {(- 1)}^{1 + 3} M_{13} = (1) \times |\begin{matrix} 1 & 4 \\ 1 & - 3 \end{matrix}| = - 3 - 4 = - 7$

$C_{21} = {(- 1)}^{2 + 1} M_{21} = (- 1) \times |\begin{matrix} - 2 & 3 \\ - 3 & 2 \end{matrix}| = (- 1) (- 4 + 9) = - 5$

$C_{22} = {(- 1)}^{2 + 2} M_{22} = (1) \times |\begin{matrix} 1 & 3 \\ 1 & 2 \end{matrix}| = 2 - 3 = - 1$

$C_{23} = {(- 1)}^{2 + 3} M_{23} = (- 1) \times |\begin{matrix} 1 & - 2 \\ 1 & - 3 \end{matrix}| = - (- 3 + 2) = 1$

$C_{31} = {(- 1)}^{3 + 1} M_{31} = (1) \times |\begin{matrix} - 2 & 3 \\ 4 & 1 \end{matrix}| = - 2 - 12 = - 14$

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

$C_{33} = {(- 1)}^{3 + 3} M_{33} = (1) \times |\begin{matrix} 1 & - 2 \\ 1 & 4 \end{matrix}| = 4 + 2 = 6$

Therefore, the Co-factor matrix $C$ of the matrix $A$ is

$C = [\begin{matrix} 11 & - 1 & - 7 \\ - 5 & - 1 & 1 \\ - 14 & 2 & 6 \end{matrix}]$

$Adj A = {[\begin{matrix} 11 & - 1 & - 7 \\ - 5 & - 1 & 1 \\ - 14 & 2 & 6 \end{matrix}]}^{'} = [\begin{matrix} 11 & - 5 & - 14 \\ - 1 & - 1 & 2 \\ - 7 & 1 & 6 \end{matrix}]$

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

$A^{- 1} = \frac{1}{- 8} [\begin{matrix} 11 & - 5 & - 14 \\ - 1 & - 1 & 2 \\ - 7 & 1 & 6 \end{matrix}]$

Put in equation $(1)$ , we get

$X = A^{- 1} B$

$[\begin{matrix} x \\ y \\ z \end{matrix}] = \frac{1}{- 8} [\begin{matrix} 11 & - 5 & - 14 \\ - 1 & - 1 & 2 \\ - 7 & 1 & 6 \end{matrix}] [\begin{matrix} 6 \\ 12 \\ 1 \end{matrix}]$

$[\begin{matrix} x \\ y \\ z \end{matrix}] = - \frac{1}{8} [\begin{matrix} 66 - 60 - 14 \\ - 6 - 12 + 2 \\ - 42 + 12 + 6 \end{matrix}]$

$[\begin{matrix} x \\ y \\ z \end{matrix}] = - \frac{1}{8} [\begin{matrix} - 8 \\ - 16 \\ - 24 \end{matrix}]$

$[\begin{matrix} x \\ y \\ z \end{matrix}] = [\begin{matrix} 1 \\ 2 \\ 3 \end{matrix}]$

Using the property of equality of matrices, we get

$x = 1, y = 2, z = 3 .$

Using matrix method, solve the following system of equations:

$x - 2 y + 3 z = 6$

$x + 4 y + z = 12$

$x - 3 y + 4 z = 1$ .

Important Questions on Determinants

Learn and Prepare for any exam you want !

Using matrix method, solve the following system of equations: x-2y+3z=6 x+4y+z=12 x-3y+4z=1.

Important Questions on Determinants

Learn and Prepare for any exam you want !

Using matrix method, solve the following system of equations:

$x - 2 y + 3 z = 6$

$x + 4 y + z = 12$

$x - 3 y + 4 z = 1$ .