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Solve the following.
If A = [1 -2 0; 2 -1 -1; 0 -2 1], then find A⁻¹. Using A⁻¹, solve the system of linear equations…
(a)[3]
If $A = \begin{bmatrix} 1 & -2 & 0 \\ 2 & -1 & -1 \\ 0 & -2 & 1 \end{bmatrix}$, then find $A^{-1}$.
(b)[1]
Using $A^{-1}$, solve the system of linear equations given below:
$x - 2y = -5$
$2x - y - z = -3$
$-2y + z = -4$
Answer
Answer (a)
Matrix: $A^{-1} = \begin{bmatrix} -3 & 2 & 2 \\ -2 & 1 & 1 \\ -4 & 2 & 3 \end{bmatrix}$
$|A|=1(-1-2)+2(2-0)=-3+4=1\neq0$, so $A^{-1}$ exists.
The cofactors give $\operatorname{adj}A=\begin{bmatrix}-3&2&2\\-2&1&1\\-4&2&3\end{bmatrix}$ and $A^{-1}=\frac{1}{|A|}\operatorname{adj}A$.
Final answer: $\begin{bmatrix}-3&2&2\\-2&1&1\\-4&2&3\end{bmatrix}$
Answer (b)
Solution: $x = 1$, $y = 3$, $z = 2$
The system is $AX=B$ with $A$ as above, $X=\begin{bmatrix}x\\y\\z\end{bmatrix}$, $B=\begin{bmatrix}-5\\-3\\-4\end{bmatrix}$. Then $X=A^{-1}B=\begin{bmatrix}15-6-8\\10-3-4\\20-6-12\end{bmatrix}=\begin{bmatrix}1\\3\\2\end{bmatrix}$.
Final answer: $x=1,\ y=3,\ z=2$