Question

Find $x,y,a$ and $b$, if $\left[\begin{array}{ccc}2x-3y& a-b& 3\\ 1& x+4y& 3a+4b\end{array}\right]=\left[\begin{array}{ccc}1& -2& 3\\ 1& 6& 29\end{array}\right]$.

Answer 1

Since, matrices are equal then, their corresponding elements are also equal.

$\Rightarrow$ $2x-3y=1$ ------ ( 1 )

$\Rightarrow$ $a-b=-2$ ----- ( 2 )

$\Rightarrow$ $x+4y=6$ ------ ( 3 )

$\Rightarrow$ $3a+4b=29$ ------ ( 4 )

From equation ( 2 ),

$a-b=-2$

$a=-2+b$

Substituting value of $a$ in equation ( 4 ), we get

$\Rightarrow$ $3(-2+b)+4b=29$

$\Rightarrow$ $-6+3b+4b=29$

$\Rightarrow$ $7b=35$

$\therefore$ $b=5$

Substituting $b=5$ in equation ( 2 ), we get

$\Rightarrow$ $a-5=-2$

$\therefore$ $a=3$

From equation ( 3 ),

$x+4y=6$

$x=6-4y$

Substituting value of $x$ in equation ( 1 ) we get,

$\Rightarrow$ $2(6-4y)-3y=1$

$\Rightarrow$ $12-8y-3y=1$

$\Rightarrow$ $-11y=-11$

$\therefore$ $y=1$

Substituting $y=1$ in equation ( 3 ) we get,

$\Rightarrow$ $x+4\left(1\right)=6$

$\therefore$ $x=2$

$\therefore$ $x=2,y=1,a=3$ and $b=5$