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If $A = \begin{bmatrix} 1 & 1 & 1\\ 1 & 0 & 2\\ 3 & 1 & 1 \end{bmatrix}$ , find $A ^{-1}$ , Hence solve the system of equations

$x + y + z = 6, \ x + 2z = 7, \ 3x + y + z =12$

Answers (1)

Given: $A = \begin{bmatrix} 1 & 1 & 1\\ 1 & 0 & 2\\ 3 & 1 & 1 \end{bmatrix}$

$|A| = 1(0-2) - 1 (1 - 6) + 1(1-0)$

$|A| = -2 + 5 + 1$

$|A| = 4 \neq 0$

$\therefore A^{-1}$ exists.

Now, cofactors are,

$\begin{matrix} a_{11} = -2 & a_{21} = 0 & a_{31} = 2\\ a_{12} = 5 & a_{22} = -2 & a_{32} = -1 \\ a_{13} = 1 & a_{23} =2 &a_{33} = -1 \end{matrix}$

$\text{adj}A = \begin{bmatrix} -2 & 5 & 1\\ 0 & -2 & 2\\ 2 & -1 &-1 \end{bmatrix}^T$

$\text{adj}A =\begin{bmatrix} -2 & 0 &2 \\ 5 & -2 & -1\\ 1& 2 & -1 \end{bmatrix}$

So, $A^{-1} = \frac{1}{|A|}(\text{adj}A)$

$A^{-1} =\frac{1}{4}\begin{bmatrix} -2 & 0 &2 \\ 5 & -2 & -1\\ 1& 2 & -1 \end{bmatrix}\qquad -(i)$

The given equations are:

$x + y + z = 6, \ x + 2z = 7, \ 3x + y + z =12$

Now,

$A = \begin{bmatrix} 1 & 1 & 1\\ 1 & 0 & 2\\ 3 & 1 & 1 \end{bmatrix}\quad X = \begin{bmatrix} x\\ y\\ z \end{bmatrix}\quad B = \begin{bmatrix} 6\\7 \\12 \end{bmatrix}$

$X = A^{-1}B$

$X =\frac{1}{4}\begin{bmatrix} -2 & 0 &2 \\ 5 & -2 & -1\\ 1& 2 & -1 \end{bmatrix}\begin{bmatrix} 6\\7 \\12 \end{bmatrix}$ Using (i)

$X =\frac{1}{4}\begin{bmatrix} -12 + 0+24 \\ 30 -1 4 + 12 \\ 6 + 14 -12 \end{bmatrix}$

$X =\frac{1}{4}\begin{bmatrix} 12 \\ 4 \\ 8 \end{bmatrix}$

$X =\begin{bmatrix} 3 \\ 1 \\ 2 \end{bmatrix}\qquad\therefore\quad \begin{matrix} x = 3\\y =1 \\ z = 2 \end{matrix}$

Posted by

Ravindra Pindel

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