Find a 2 × 2 matrix A such that

Solution:

Let $\mathrm{A}=\left[\begin{array}{ll}w & x \\ y & z\end{array}\right]$

Now,

$\left[\begin{array}{cc}w & x \\ y & z\end{array}\right]\left[\begin{array}{cc}1 & -2 \\ 1 & 4\end{array}\right]=6 I_{2}$

$\Rightarrow\left[\begin{array}{ll}w+x & -2 w+4 x \\ y+z & -2 y+4 z\end{array}\right]=6\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$

$\Rightarrow\left[\begin{array}{ll}w+x & -2 w+4 x \\ y+z & -2 y+4 z\end{array}\right]=\left[\begin{array}{ll}6 & 0 \\ 0 & 6\end{array}\right]$

The corresponding elements of two equal matrices are equal.

$\therefore w+x=6$      

$\Rightarrow w=6-x$      ...(1)

$-2 w+4 x=0$               ...(2)

Putting the value of $w$ in $e q .$ (2), we get

$-2(6-x)+4 x=0$

$\Rightarrow-12+2 x+4 x=0$

$\Rightarrow-12+6 x=0$

$\Rightarrow 6 x=12$

$\Rightarrow x=2$

Putting the value of $x$ in eq. $(1)$, we get

$w=6-2$

$\Rightarrow w=4$

Now,

$y+z=0$

$\Rightarrow y=-z$              ...(3)

$-2 y+4 z=6$                       ...(4)

Putting the value of $y$ in eq. (4), we get

$-2(-z)+4 z=6$

$\Rightarrow 2 z+4 z=6$

$\Rightarrow 6 z=6$

$\Rightarrow z=1$

Putting the value of $z$ in eq. (3), we get

$y=-1$

$\therefore A=\left[\begin{array}{cc}4 & 2 \\ -1 & 1\end{array}\right]$