- Home
- Standard 12
- Mathematics
If $\mathrm{A}=\left[\begin{array}{rrr}2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2\end{array}\right],$ find $\mathrm{A}^{-1}$. Using $\mathrm{A}^{-1}$ solve the system of equations $2x - 3y + 5z = 11$ ; $3x + 2y - 4z = - 5$ ; $x + y - 2z = - 3$
$x=1, y=2,z=3$
$x=1, y=2,z=3$
$x=1, y=2,z=3$
$x=1, y=2,z=3$
Solution
$A=\left[\begin{array}{ccc}2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2\end{array}\right]$
$\therefore|A|=2(-4+4)+3(-6+4)+5(3-2)=0-6+5=-1 \neq 0$
Now,
$A_{11}=0, A_{12}=2, A_{13}=1$
$A_{11}=-1, A_{22}=-9, A_{23}=-5$
$A_{11}=2, A_{32}=23, A_{33}=13$
$\therefore A^{-1}=\frac{1}{|A|}(\operatorname{adj} A)=-\left[\begin{array}{ccc}0 & -1 & 2 \\ 2 & -9 & 23 \\ 1 & -5 & 13\end{array}\right]=\left[\begin{array}{ccc}0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13\end{array}\right]……(1)$
Now, the given system of equations can be written in the form of $A X=B$, where
$A=\left[\begin{array}{ccc}2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2\end{array}\right], X=\left[\begin{array}{l}x \\ y \\ z\end{array}\right]$ and $B=\left[\begin{array}{c}11 \\ -5 \\ -3\end{array}\right]$
The solution of the system of equations is given by $X=A^{-1} B$
$\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{ccc}0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13\end{array}\right]\left[\begin{array}{c}11 \\ -5 \\ -3\end{array}\right] \quad\left[\begin{array}{ll}U \sin g(1)\end{array}\right]$
$=\left[\begin{array}{c}0-5+6 \\ -22-45+69 \\ -11-25+39\end{array}\right]=\left[\begin{array}{l}1 \\ 2 \\ 3\end{array}\right]$
Hence, $x=1, y=2,$ and $z=3$
