Find $A^{-1}$, if $A=\left[\begin{array}{rrr}1 & 2 & 5 \\ 1 & -1 & -1 \\ 2 & 3 & -1\end{array}\right]$. Hence solve the following system of linear equations:
$x+2 y+5 z=10, x-y-z=-2,2 x+3 y-z=-11$
Here,
$A=\left[\begin{array}{ccc}1 & 2 & 5 \\ 1 & -1 & -1 \\ 2 & 3 & -1\end{array}\right]$
$|A|=\left|\begin{array}{ccc}1 & 2 & 5 \\ 1 & -1 & -1 \\ 2 & 3 & -1\end{array}\right|$
$=1(1+3)-2(-1+2)+5(3+2)$
$=4-2+25$
$=27$
Let $\mathrm{C}_{i j}$ be the co factors of the elements $\mathrm{a}_{i j}$ in $\mathrm{A}\left[a_{i j}\right]$. Then,
$C_{11}=(-1)^{1+1}\left|\begin{array}{cc}-1 & -1 \\ 3 & -1\end{array}\right|=4, \quad C_{12}=(-1)^{1+2}\left|\begin{array}{cc}1 & -1 \\ 2 & -1\end{array}\right|=-1, \quad C_{13}=(-1)^{1+3}\left|\begin{array}{cc}1 & -1 \\ -2 & 3\end{array}\right|=5$
$C_{21}=(-1)^{2+1}\left|\begin{array}{cc}2 & 5 \\ 3 & -1\end{array}\right|=17, \quad C_{22}=(-1)^{2+2}\left|\begin{array}{cc}1 & 5 \\ 2 & -1\end{array}\right|=-11, \quad C_{23}=(-1)^{2+3}\left|\begin{array}{ll}1 & 2 \\ 2 & 3\end{array}\right|=1$
$C_{31}=(-1)^{3+1}\left|\begin{array}{cc}2 & 5 \\ -1 & -1\end{array}\right|=3, \quad C_{32}=(-1)^{3+2}\left|\begin{array}{cc}1 & 5 \\ 1 & -1\end{array}\right|=6, \quad C_{33}=(-1)^{3+3}\left|\begin{array}{cc}1 & 2 \\ 1 & -1\end{array}\right|=-3$
$\operatorname{adj} A=\left[\begin{array}{ccc}4 & -1 & 5 \\ 17 & -11 & 1 \\ 3 & 6 & -3\end{array}\right]^{\mathrm{T}}$
$=\left[\begin{array}{ccc}4 & 17 & 3 \\ -1 & -11 & 6 \\ 5 & 1 & -3\end{array}\right]$
$\Rightarrow A^{-1}=\frac{1}{|A|}$ adj $A$
$=\frac{1}{27}\left[\begin{array}{ccc}4 & 17 & 3 \\ -1 & -11 & 6 \\ 5 & 1 & -3\end{array}\right]$
The given system of equations can be written in matrix form as follows:
$\left[\begin{array}{ccc}1 & 2 & 5 \\ 1 & -1 & -1 \\ 2 & 3 & -1\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{c}10 \\ -2 \\ -11\end{array}\right]$
$X=A^{-1} B$
$\Rightarrow\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\frac{1}{27}\left[\begin{array}{ccc}4 & 17 & 3 \\ -1 & -11 & 6 \\ 5 & 1 & -3\end{array}\right]\left[\begin{array}{c}10 \\ -2 \\ -11\end{array}\right]$
$\Rightarrow\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\frac{1}{27}\left[\begin{array}{c}40-34-33 \\ -10+22-66 \\ 50-2+33\end{array}\right]$
$\Rightarrow\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\frac{1}{27}\left[\begin{array}{c}-27 \\ -54 \\ 81\end{array}\right]$
$\therefore x=\frac{-27}{27}, y=\frac{-54}{27}$ and $z=\frac{81}{27}$
$\Rightarrow x=-1, y=-2$ and $z=3$