If $A=\left(\begin{array}{ccc}2 & 1 & 1 \\ 1 & -2 & -1 \\ 0 & 3 & -5\end{array}\right)$, find $A^{-1}$ Using $A^{-1}$, solve the following system of linear equations: $2 x+y+z=1$ $x-2 y-z=\frac{3}{2}$ $3 y-5 z=9$ HINT: Here $A=\left(\begin{array}{ccc}2 & 1 & 1 \\ 1 & -2 & -1 \\ 0 & 3 & -5\end{array}\right)$ $\mathrm{X}=\left(\begin{array}{l} x \\ y \\ z \end{array}\right) \text { and } \mathrm{B}=\left(\begin{array}{l} 1 \\ 3 / 2 \\ 9 \end{array}\right)$

If $A=\left(\begin{array}{ccc}2 & 1 & 1 \\ 1 & -2 & -1 \\ 0 & 3 & -5\end{array}\right)$ , find $A^{-1}$ Using $A^{-1}$ , solve the following system of linear equations:
$2 x+y+z=1$
$x-2 y-z=\frac{3}{2}$
$3 y-5 z=9$ HINT: Here $A=\left(\begin{array}{ccc}2 & 1 & 1 \\ 1 & -2 & -1 \\ 0 & 3 & -5\end{array}\right)$ $\mathrm{X}=\left(\begin{array}{l} x \\ y \\ z \end{array}\right) \text { and } \mathrm{B}=\left(\begin{array}{l} 1 \\ 3 / 2 \\ 9 \end{array}\right)$

Solution:

It is given,
$\begin{array}{l} A=\left[\begin{array}{ccc} 2 & 1 & 1 \\ 1 & -2 & -1 \\ 0 & 3 & -5 \end{array}\right] \\ A^{-1}=\frac{1}{|A|} \operatorname{adj}(A) \end{array}$
Determinant of matrix $A$ is
$\begin{array}{l} |A|=\left|\begin{array}{ccc} 2 & 1 & 1 \\ 1 & -2 & -1 \\ 0 & 3 & -5 \end{array}\right| \\ =2(-2 x-5-(-1) \times 3)-(1 \times-5-(-1) \times 0)+(1 \times 3-(-2) \times 0) \\ =2(10+3)-(-5)+(3) \\ =26+5+3 \\ =34 \\ |A| \neq 0 \end{array}$
$\therefore \mathrm{A}^{-1}$ is possible.
$\begin{array}{l} A^{T}=\left[\begin{array}{ccc} 2 & 1 & 0 \\ 1 & -2 & 3 \\ 1 & -1 & -5 \end{array}\right] \\ \operatorname{Adj}(A)=\left[\begin{array}{ccc} 13 & 8 & 1 \\ 5 & -10 & 3 \\ 3 & -6 & -5 \end{array}\right] \\ A^{-1}=\frac{1}{|A|} \operatorname{adj}(A) \\ A^{-1}=\frac{1}{34}\left[\begin{array}{ccc} 13 & 8 & 1 \\ 5 & -10 & 3 \\ 3 & -6 & -5 \end{array}\right] \end{array}$
The given set of lines are : $-$
$\begin{array}{l} 2 x+y+z=1 \\ x-2 y-z=\frac{3}{2} \\ 3 y-5 z=9 \end{array}$
Now converting the following equations in matrix form,
$\begin{array}{l} \mathrm{AX}=\mathrm{B} \\ \text { Where } \mathrm{A}=\left[\begin{array}{ccc} 2 & 1 & 1 \\ 1 & -2 & -1 \\ 0 & 3 & -5 \end{array}\right], \mathrm{X}=\left[\begin{array}{l} x \\ y \\ z \end{array}\right], \mathrm{B}=\left[\begin{array}{l} 1 \\ 3 \\ \frac{1}{2} \\ 9 \end{array}\right] \end{array}$
Pre – multiplying by $A^{-1}$
$\begin{array}{l} A^{-1} A X=A^{-1} B \\ I X=A^{-1} B \\ X=A^{-1} B \\ {\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\frac{1}{34}\left[\begin{array}{ccc} 13 & 8 & 1 \\ 5 & -10 & 3 \\ 3 & -6 & -5 \end{array}\right]\left[\begin{array}{l} 1 \\ 3 \\ 2 \\ 9 \end{array}\right]} \\ {\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\frac{1}{34}\left[\begin{array}{c} 1 \times 13+\frac{3}{2} \times 8+9 \times 1 \\ 1 \times 5+\frac{3}{2} \times-10+9 \times 3 \\ 1 \times 3+\frac{3}{2} \times-6+9 \times-5 \end{array}\right]} \\ {\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\frac{1}{34}\left[\begin{array}{c} 13+12+9 \\ 5-15+27 \\ 3-9-45 \end{array}\right]} \\ {\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\frac{1}{34}\left[\begin{array}{c} 34 \\ 17 \\ -51 \end{array}\right]=\left[\begin{array}{c} 1 \\ \frac{1}{2} \\ -\frac{3}{2} \end{array}\right]} \\ \therefore x=1, y=\frac{1}{2}, z=-\frac{3}{2} \end{array}$