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Q.
Solve the following system of equations $x - y + z = 4$, $x -2 y + 2z = 9$ and $2x + y + 3z = 1$.
Determinants
Solution:
Given system of equations is $x - y + z = 4$, $x - 2y + 2z =9$ and $2x + y + 3z = 1$, which is written in the matrix form as
$\left[\begin{matrix}1&-1&1\\ 1&-2&2\\ 2&1&3\end{matrix}\right]\left[\begin{matrix}x\\ y\\ z\end{matrix}\right]=\left[\begin{matrix}4\\ 9\\ 1\end{matrix}\right]$
or $AX = B$
where, $A =\left[\begin{matrix}1&-1&1\\ 1&-2&2\\ 2&1&3\end{matrix}\right], B=\left[\begin{matrix}4\\ 9\\ 1\end{matrix}\right] and X=\left[\begin{matrix}x\\ y\\ z\end{matrix}\right]$
Now, $|A| = \left|\begin{matrix}1&-1&1\\ 1&-2&2\\ 2&1&3\end{matrix}\right|=1\left(-6-2\right)+1\left(3-4\right)+1\left(1+4\right)=-8-1+5=-4 \ne0$
$\therefore \quad$ A is invertible
So, there exists a unique solution $X = A^{-1}B$.
Now, cofactor matrix of $A =\left[\begin{matrix}-8&1&5\\ 4&1&-3\\ 0&-1&-1\end{matrix}\right]$
$\therefore \quad adj A =\left[\begin{matrix}-8&1&5\\ 4&1&-3\\ 0&-1&-1\end{matrix}\right]^{T}=\left[\begin{matrix}-8&4&0\\ 1&1&-1\\ 5&-3&-1\end{matrix}\right]$
So, $A^{-1}=\frac{1}{\left|A\right|} \left(adj A\right)=\frac{-1}{4} \left[\begin{matrix}-8&4&0\\ 1&1&-1\\ 5&-3&-1\end{matrix}\right]$
Now, $X=A^{-1}B=\frac{-1}{4} \left[\begin{matrix}-8&4&0\\ 1&1&-1\\ 5&-3&-1\end{matrix}\right]\left[\begin{matrix}4\\ 9\\ 1\end{matrix}\right]$
$\Rightarrow \quad\left[\begin{matrix}x\\ y\\ z\end{matrix}\right]=\frac{-1}{4} \left[\begin{matrix}-32+36+0\\ 4+9-1\\ 20-27-1\end{matrix}\right]=\frac{-1}{4}\left[\begin{matrix}4\\ 12\\ -8\end{matrix}\right]=\left[\begin{matrix}-1\\ -3\\ 2\end{matrix}\right]$
On comparing, we get $x = -1$, $y = -3$ and $z = 2$