# Solve the following equations

Question:

If $A=\left[\begin{array}{rrr}2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2\end{array}\right]$, find $A^{-1}$ and hence solve the system of linear equations

$2 x-3 y+5 z=11,3 x+2 y-4 z=-5, x+y+2 z=-3$

Solution:

Here,

$A=\left[\begin{array}{ccc}2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2\end{array}\right]$

$|A|=\left|\begin{array}{ccc}2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2\end{array}\right|$

$=2(-4+4)+3(-6+4)+5(3-2)$

$=0-6+5$

$=-1$

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,

$C_{11}=(-1)^{1+1}\left|\begin{array}{ll}2 & -4 \\ 1 & -2\end{array}\right|=0, \quad C_{12}=(-1)^{1+2}\left|\begin{array}{ll}3 & -4 \\ 1 & -2\end{array}\right|=2, \quad C_{13}=(-1)^{1+3}\left|\begin{array}{ll}3 & 2 \\ 1 & 1\end{array}\right|=1$

$C_{21}=(-1)^{2+1}\left|\begin{array}{cc}-3 & 5 \\ 1 & -2\end{array}\right|=-1, \quad C_{22}=(-1)^{2+2}\left|\begin{array}{cc}2 & 5 \\ 1 & -2\end{array}\right|=-9, \quad C_{23}=(-1)^{2+3}\left|\begin{array}{cc}2 & -3 \\ 1 & 1\end{array}\right|=-5$

$C_{31}=(-1)^{3+1}\left|\begin{array}{cc}-3 & 5 \\ 2 & -4\end{array}\right|=2, \quad C_{32}=(-1)^{3+2}\left|\begin{array}{cc}2 & 5 \\ 3 & -4\end{array}\right|=23, \quad C_{33}=(-1)^{3+3}\left|\begin{array}{cc}2 & -3 \\ 3 & 2\end{array}\right|=13$

adj $A=\left[\begin{array}{ccc}0 & 2 & 1 \\ -1 & -9 & -5 \\ 2 & 23 & 13\end{array}\right]^{\mathrm{T}}$

$=\left[\begin{array}{ccc}0 & -1 & 2 \\ 2 & -9 & 23 \\ 1 & -5 & 13\end{array}\right]$

$\Rightarrow A^{-1}=\frac{1}{|A|} a d j A$

$=\frac{1}{-1}\left[\begin{array}{ccc}0 & -1 & 2 \\ 2 & -9 & 23 \\ 1 & -5 & 13\end{array}\right]$

The given system of equations can be written in matrix form as follows:

$\left[\begin{array}{ccc}2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{c}11 \\ -5 \\ -3\end{array}\right]$

$X=A^{-1} B$

$\Rightarrow\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\frac{1}{-1}\left[\begin{array}{ccc}0 & -1 & 2 \\ 2 & -9 & 23 \\ 1 & -5 & 13\end{array}\right]\left[\begin{array}{l}11 \\ -5 \\ -3\end{array}\right]$

$\Rightarrow\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\frac{1}{-1}\left[\begin{array}{c}0+5-6 \\ 22+45-69 \\ 11+25-39\end{array}\right]$

$\Rightarrow\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\frac{1}{-1}\left[\begin{array}{l}-1 \\ -2 \\ -3\end{array}\right]$

$\Rightarrow x=\frac{-1}{-1}, y=\frac{-2}{-1}$ and $z=\frac{-3}{-1}$

$\therefore x=1, y=2$ and $z=3$