Question

Without using the concept of inverse of a matrix, find the matrix `[[x       y],[z       u]]` such that
`[[5     -7],[-2         3]][[x        y],[z         u]]=[[-16       -6],[7                   2]]`

Answer 1

\[Given: \begin{bmatrix}5 & - 7 \\ - 2 & 3\end{bmatrix}\begin{bmatrix}x & y \\ z & u\end{bmatrix} = \begin{bmatrix}- 16 & - 6 \\ 7 & 2\end{bmatrix}\]
\[ \Rightarrow \begin{bmatrix}5x - 7z & 5y - 7u \\ - 2x + 3z & - 2y + 3u\end{bmatrix} = \begin{bmatrix}- 16 & - 6 \\ 7 & 2\end{bmatrix}\]
\[\]
The corresponding elements of two equal matrices  are equal .
\[ \therefore 5x - 7z = - 16 . . . \left( 1 \right) \]
\[5y - 7u = - 6 . . . \left( 2 \right) \]
\[\]
\[ - 2y + 3u = 2 \]
\[ \Rightarrow 3u = 2 + 2y\]
\[ \Rightarrow u = \frac{2 + 2y}{3} . . . \left( 3 \right) \]
\[ - 2x + 3z = 7 \]
\[ \Rightarrow 3z = 7 + 2x \]
\[ \Rightarrow z = \frac{7 + 2x}{3} . . . \left( 4 \right) \]
\[\] putting the value of z in eq.(1,we get )
\[5x - 7\left( \frac{7 + 2x}{3} \right) = - 16\]
\[ \Rightarrow 5x - \frac{49 + 14x}{3} = - 16\]
\[ \Rightarrow \frac{15x - 49 - 14x}{3} = - 16\]
\[ \Rightarrow x - 49 = - 48\]
\[ \Rightarrow x = - 48 + 49\]
\[ \therefore x = 1\]
\[\]
putting  the value of x in eq.(4),we get 
\[z = \frac{7 + 2\left( 1 \right)}{3}\]
\[ \Rightarrow z = \frac{9}{3} = 3\]
\[\]
putting  the value of  u in eq.(2), we get
\[5y - 7\left( \frac{2 + 2y}{3} \right) = - 6\]
\[ \Rightarrow 5y - \frac{14 + 14y}{3} = - 6\]
\[ \Rightarrow \frac{15y - 14 - 14y}{3} = - 6\]
\[ \Rightarrow y - 14 = - 18\]
\[ \Rightarrow y = - 18 + 14\]
\[ \therefore y = - 4\]
\[\]
putting the value of  y  in eq.(3),we get
\[u = \frac{2 + 2\left( - 4 \right)}{3}\]
\[ \Rightarrow u = - 2\]
\[\]
\[ \therefore \begin{bmatrix}x & y \\ z & u\end{bmatrix} = \begin{bmatrix}1 & - 4 \\ 3 & - 2\end{bmatrix}\]