Without using the concept of inverse of a matrix, find the matrix

Question:

Without using the concept of inverse of a matrix, find the matrix $\left[\begin{array}{ll}x & y \\ z & u\end{array}\right]$ such that

$\left[\begin{array}{rr}5 & -7 \\ -2 & 3\end{array}\right]\left[\begin{array}{ll}x & y \\ z & u\end{array}\right]=\left[\begin{array}{rr}-16 & -6 \\ 7 & 2\end{array}\right]$

Solution:

Given : $\left[\begin{array}{cc}5 & -7 \\ -2 & 3\end{array}\right]\left[\begin{array}{ll}x & y \\ z & u\end{array}\right]=\left[\begin{array}{cc}-16 & -6 \\ 7 & 2\end{array}\right]$

$\Rightarrow\left[\begin{array}{cc}5 x-7 z & 5 y-7 u \\ -2 x+3 z & -2 y+3 u\end{array}\right]=\left[\begin{array}{cc}-16 & -6 \\ 7 & 2\end{array}\right]$

The corresponding elements of two equal matrices are equal.

$\therefore 5 x-7 z=-16$    ...(1)

$5 y-7 u=-6$                     ...(2)

$-2 y+3 u=2$

$\Rightarrow 3 u=2+2 y$

$\Rightarrow u=\frac{2+2 y}{3}$    ...(3)

$-2 x+3 z=7$

$\Rightarrow 3 z=7+2 x$

$\Rightarrow z=\frac{7+2 x}{3}$                 ...(4)

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

$5 x-7\left(\frac{7+2 x}{3}\right)=-16$

$\Rightarrow 5 x-\frac{49+14 x}{3}=-16$

$\Rightarrow \frac{15 x-49-14 x}{3}=-16$

$\Rightarrow x-49=-48$

$\Rightarrow x=-48+49$

$\therefore x=1$

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

$5 y-7\left(\frac{2+2 y}{3}\right)=-6$

$\Rightarrow 5 y-\frac{14+14 y}{3}=-6$

$\Rightarrow \frac{15 y-14-14 y}{3}=-6$

$\Rightarrow y-14=-18$

$\Rightarrow y=-18+14$

$\therefore y=-4$

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

$u=\frac{2+2(-4)}{3}$

$\Rightarrow u=-2$

$\therefore \quad\left[\begin{array}{ll}x & y \\ z & u\end{array}\right]=\left[\begin{array}{ll}1 & -4 \\ 3 & -2\end{array}\right]$

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