Question

If $A = [\begin{matrix} 2 & - 3 & 5 \\ 3 & 2 & - 4 \\ 1 & 1 & - 2 \end{matrix}]$ , find $A^{- 1}$ and hence solve the system of equations

$2 x - 3 y + 5 z = 11, 3 x + 2 y - 4 z = - 5, x + y - 2 z = - 3$ .

Accepted Answer

Given matrix

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

We have

$|A| = 2 (- 4 + 4) + 3 (- 6 + 4) + 5 (3 - 2)$

$= 2 (0) + 3 (- 2) + 5 (1)$

$= 0 - 6 + 5$

$|A| = - 1 \neq 0$

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:

$C_{11} = {(- 1)}^{1 + 1} M_{11} = (1) \times |\begin{matrix} 2 & - 4 \\ 1 & - 2 \end{matrix}| = - 4 + 4 = 0$

$C_{12} = {(- 1)}^{1 + 2} M_{12} = (- 1) \times |\begin{matrix} 3 & - 4 \\ 1 & - 2 \end{matrix}| = - (- 6 + 4) = 2$

$C_{13} = {(- 1)}^{1 + 3} M_{13} = (1) \times |\begin{matrix} 3 & 2 \\ 1 & 1 \end{matrix}| = 3 - 2 = 1$

$C_{21} = {(- 1)}^{2 + 1} M_{21} = (- 1) \times |\begin{matrix} - 3 & 5 \\ 1 & - 2 \end{matrix}| = (- 1) (6 - 5) = - 1$

$C_{22} = {(- 1)}^{2 + 2} M_{22} = (1) \times |\begin{matrix} 2 & 5 \\ 1 & - 2 \end{matrix}| = - 4 - 5 = - 9$

$C_{23} = {(- 1)}^{2 + 3} M_{23} = (- 1) \times |\begin{matrix} 2 & - 3 \\ 1 & 1 \end{matrix}| = - (2 + 3) = - 5$

$C_{31} = {(- 1)}^{3 + 1} M_{31} = (1) \times |\begin{matrix} - 3 & 5 \\ 2 & - 4 \end{matrix}| = 12 - 10 = 2$

$C_{32} = {(- 1)}^{3 + 2} M_{32} = (- 1) \times |\begin{matrix} 2 & 5 \\ 3 & - 4 \end{matrix}| = (- 1) (- 8 - 15) = 23$

$C_{33} = {(- 1)}^{3 + 3} M_{33} = (1) \times |\begin{matrix} 2 & - 3 \\ 3 & 2 \end{matrix}| = 4 + 9 = 13$

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

$C = [\begin{matrix} 0 & 2 & 1 \\ - 1 & - 9 & - 5 \\ 2 & 23 & 13 \end{matrix}]$

$Adj A = {[\begin{matrix} 0 & 2 & 1 \\ - 1 & - 9 & - 5 \\ 2 & 23 & 13 \end{matrix}]}^{'} = [\begin{matrix} 0 & - 1 & 2 \\ 2 & - 9 & 23 \\ 1 & - 5 & 13 \end{matrix}]$

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

$A^{- 1} = \frac{1}{- 1} [\begin{matrix} 0 & - 1 & 2 \\ 2 & - 9 & 23 \\ 1 & - 5 & 13 \end{matrix}] . . . . . . . . . (i)$

Now, given equation are:

$2 x - 3 y + 5 z = 11, 3 x + 2 y - 4 z = - 5, x + y - 2 z = - 3$

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

$[\begin{matrix} 2 & - 3 & 5 \\ 3 & 2 & - 4 \\ 1 & 1 & - 2 \end{matrix}] [\begin{matrix} x \\ y \\ z \end{matrix}] = [\begin{matrix} 11 \\ - 5 \\ - 3 \end{matrix}] A X = B$

$\Rightarrow X = A^{- 1} B . . . . . . . . . (2)$

Put the value of $A^{- 1}$ from equation $(1)$ , we get

$X = A^{- 1} B$

$[\begin{matrix} x \\ y \\ z \end{matrix}] = (- 1) [\begin{matrix} 0 & - 1 & 2 \\ 2 & - 9 & 23 \\ 1 & - 5 & 13 \end{matrix}] [\begin{matrix} 11 \\ - 5 \\ - 3 \end{matrix}]$

$[\begin{matrix} x \\ y \\ z \end{matrix}] = - [\begin{matrix} 0 \times 11 + (- 1) \times (- 5) + 2 \times (- 3) \\ 2 \times 11 + (- 9) \times (- 5) + 23 \times (- 3) \\ 1 \times 11 + (- 5) \times (- 5) + 13 \times (- 3) \end{matrix}]$

$[\begin{matrix} x \\ y \\ z \end{matrix}] = - [\begin{matrix} 0 + 5 - 6 \\ 22 + 45 - 69 \\ 11 + 25 - 39 \end{matrix}]$

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

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

Using the property of equality of matrices, we get

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

M L Aggarwal Solutions for Chapter: Determinants, Exercise 5: EXERCISE 4.5

M L Aggarwal Mathematics Solutions for Exercise - M L Aggarwal Solutions for Chapter: Determinants, Exercise 5: EXERCISE 4.5

Questions from M L Aggarwal Solutions for Chapter: Determinants, Exercise 5: EXERCISE 4.5 with Hints & Solutions

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