MDoubt No 000003 - UPSC IAS Mathematics Optional

UPSC [CSE 2019 P1]
Question 1(d) : If $$A=\begin{bmatrix} 1 & 2 & 1\\ 1 & -4 & 1\\ 3 & 0 & -3 \end{bmatrix} \text{and } B=\begin{bmatrix} 2 & 1 & 1\\ 1 & -1 & 0\\ 2 & 1 & -1 \end{bmatrix}$$ then show that $AB=6I_3$. Use this result to solve the following system of equations : $$\begin{array}{r} 2x+y+z=5\\ x-y=0\\ 2x+y-z=1\\ \end{array}$$
Solution : Given $A=\begin{bmatrix} 1 & 2 & 1\\ 1 & -4 & 1\\ 3 & 0 & -3 \end{bmatrix} \text{and } B=\begin{bmatrix} 2 & 1 & 1\\ 1 & -1 & 0\\ 2 & 1 & -1 \end{bmatrix}$
$\implies AB=\begin{bmatrix} 1 & 2 & 1\\ 1 & -4 & 1\\ 3 & 0 & -3 \end{bmatrix}\begin{bmatrix} 2 & 1 & 1\\ 1 & -1 & 0\\ 2 & 1 & -1 \end{bmatrix}$
$\implies AB=\begin{bmatrix} 6 & 0 & 0\\ 0 & 6 & 0\\ 0 & 0 & 6 \end{bmatrix}=6\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix}=6I_3$
Hence Proved.
So, $AB=6I_3$
$\implies ABB^{-1}=6I_3B^{-1}$
$\implies A=6B^{-1}$
$\implies B^{-1}=\dfrac{1}{6}A$
Now, $$\begin{array}{r} 2x+y+z=5\\ x-y=0\\ 2x+y-z=1\\ \end{array}$$ This can be written as $\begin{bmatrix} 2 & 1 & 1\\ 1 & -1 & 0\\ 2 & 1 & -1 \end{bmatrix}\begin{bmatrix} x\\ y\\ z \end{bmatrix}=\begin{bmatrix} 5\\ 0\\ 1 \end{bmatrix}$.
$\implies BX=C$, where $B=\begin{bmatrix} 2 & 1 & 1\\ 1 & -1 & 0\\ 2 & 1 & -1 \end{bmatrix}, X=\begin{bmatrix} x\\ y\\ z \end{bmatrix}$ and $C=\begin{bmatrix} 5\\ 0\\ 1 \end{bmatrix}$
$\implies B^{-1}BX=B^{-1}C$
$\implies X=B^{-1}C$
$\implies X=\dfrac{1}{6}AC$
$\implies X=\dfrac{1}{6}\begin{bmatrix} 1 & 2 & 1\\ 1 & -4 & 1\\ 3 & 0 & -3 \end{bmatrix}\begin{bmatrix} 5\\ 0\\ 1 \end{bmatrix}$
$\implies X=\dfrac{1}{6}\begin{bmatrix} 6\\ 6\\ 12 \end{bmatrix}=\begin{bmatrix} 1\\ 1\\ 2 \end{bmatrix}$
$\therefore x=1, y=1, z=2$.