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If A =\begin{bmatrix} 1 &2 &-3 \\ 3 &2 &-2 \\ 2 &-1 &1 \end{bmatrix} , then find A^{-1} and use it to solve the following system of the equations :
x + 2y -3z = 6
3x + 2y - 2z = 3
2x - y + z = 2

 

 
 
 
 
 

Answers (1)

Given:

A =\begin{bmatrix} 1 &2 &-3 \\ 3 &2 &-2 \\ 2 &-1 &1 \end{bmatrix}

\\|A| = 1(2-2)-2(3+4)-3(-3-4) \\ |A|= -14+21 =7 \\

\\ \operatorname{Adj}(A)=\left[\begin{array}{rrr} 0 & -7 & -7 \\ 1 & 7 & 5 \\ 2 & -7 & -4 \end{array}\right]^T

\\ \operatorname{Adj}(A)=\left[\begin{array}{rrr} 0 & 1 & 2 \\ -7 & 7 & -7 \\ -7 & 5 & -4 \end{array}\right] \\\\ A^{-1}=\frac{1}{|A|} \text { Adj } A \\\\ =\frac{1}{7}\left[\begin{array}{rrr} 0 & 1 & 2 \\ -7 & 7 & -7 \\ -7 & 5 & -4 \end{array}\right] \\\\ =\left[\begin{array}{rrr} 0 & \frac{1}{7} & \frac{2}{7} \\ -1 & 1 & -1 \\ -1 & \frac{5}{7} & -\frac{4}{7} \end{array}\right]

x + 2y -3z = 6
3x + 2y - 2z = 3
2x - y + z = 2

The given system of linear equations can be written in matirx form as AX = B

\\\mathrm{X}=\left[\begin{array}{l} \mathrm{x} \\ \mathrm{y} \\ \mathrm{z} \end{array}\right] \\\\ \mathrm{B}=\left[\begin{array}{l} 6 \\ 3 \\ 2 \end{array}\right]\\

X=A^{-1} B

\\ \left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{rrr} 0 & \frac{1}{7} & \frac{2}{7} \\ -1 & 1 & -1 \\ -1 & \frac{5}{7} & -\frac{4}{7} \end{array}\right]\left[\begin{array}{l} 6 \\ 3 \\ 2 \end{array}\right] \\ \\ \\ \left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{r} 1 \\ -5 \\ -5 \end{array}\right]

Hence x = 1,  y=-5, z=-5
 

Posted by

Safeer PP

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