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)

GivenA = \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}

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