Question

Solve by matrix method $x+2y+3z=2,2x+3y+z=-1,x-y-z=-2$

Answer 1

The given equations is of the form $AX=B$

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

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

$=1\left[\begin{array}{cc}3& 1\\ -1& -1\end{array}\right]-2\left[\begin{array}{cc}2& 1\\ 1& -1\end{array}\right]+\left[\begin{array}{cc}2& 3\\ 1& -1\end{array}\right]$

$=1(-3+1)-2(-2-1)+3(-2-3)$

$=1(-2)-2(-3)+3(-5)$

$=-2+6-15$

$=-11\ne 0$

suits singular inverse exists :

Ad joint of A

$\Rightarrow A_{11}={(-1)}_{1+1}\left[\begin{array}{cc}3& 1\\ -1& -1\end{array}\right]$

$=(-1)(-3+1)$

$=(-1)(-2)$

$=2$

$\Rightarrow A_{12}={(-1)}_{1+2}\left[\begin{array}{cc}2& 1\\ 1& -1\end{array}\right]$

$=(-1)(-2-1)$

$=-3$

$\Rightarrow A_{13}={(-1)}_{1+3}\left[\begin{array}{cc}2& 3\\ 1& -1\end{array}\right]$

$={(-1)}_4(-2+3)$

$=(-1)(-5)$

$=-5$

$\Rightarrow A_{21}={(-1)}_{2+1}\left[\begin{array}{cc}2& 3\\ -1& -1\end{array}\right]$

$={(-1)}_3(-2+3)$

$=1$

$\Rightarrow A_{22}={(-1)}_{2+2}\left[\begin{array}{cc}1& 3\\ 1& -1\end{array}\right]$

$={(-1)}_4(-1-3)$

$=-4$

$\Rightarrow A_{23}={(-1)}_{2+3}\left[\begin{array}{cc}1& 2\\ 1& -1\end{array}\right]$

$=(-1-2)$

$=-3$

$\Rightarrow A_{31}={(-1)}_{3+1}\left[\begin{array}{cc}2& 3\\ 3& 1\end{array}\right]$

$=(2-9)$

$=-7$

$\Rightarrow A_{32}={(-1)}_{3+2}\left[\begin{array}{cc}1& 3\\ 2& 1\end{array}\right]$

$=(1-6)$

$=-5$

$\Rightarrow A_{33}={(-1)}_{3+3}\left[\begin{array}{cc}1& 2\\ 2& 3\end{array}\right]$

$=(-3-4)$

$=-1$

Here the ad joint of A is

$\left[\begin{array}{ccc}{A}_{11}& A_{21}& A_{31}\\ A_{12}& A_{22}& A_{32}\\ A_{13}& A_{23}& A_{33}\end{array}\right]$

$=\left[\begin{array}{ccc}2& 1& -7\\ -3& 4& -5\\ -5& -3& -1\end{array}\right]$

$\Rightarrow A^{-1}=\frac{-1}{11}\left[\begin{array}{ccc}2& 1& -7\\ -3& 4& -5\\ -5& -3& -1\end{array}\right]$

$\Rightarrow A^{-1}B=X$

$X=A^{-1}B$

$\Rightarrow \left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\frac{-1}{11}\left[\begin{array}{ccc}2& 1& -7\\ -3& 4& -5\\ -5& -3& -1\end{array}\right]\left[\begin{array}{c}2\\ -1\\ -2\end{array}\right]$

$=\frac{-1}{11}\left[\begin{array}{c}4-1+14\\ -6-4+10\\ -10+3+2\end{array}\right]$

$=\frac{-1}{11}\left[\begin{array}{c}17\\ 0\\ -5\end{array}\right]$

$=\left[\begin{array}{c}-17/11\\ 0\\ 5/11\end{array}\right]$

$\Rightarrow x=\frac{-17}{11}$ $y=0$ $z=\frac{5}{11}$

Hence, solved.