Find x, y, a and b if
(i) $\left[\begin{array}{ccc}2 x-3 y & a-b & 3 \\ 1 & x+4 y & 3 a+4 b\end{array}\right]=\left[\begin{array}{ccc}1 & -2 & 3 \\ 1 & 6 & 29\end{array}\right]$
Since the corresponding elements of two equal matrices are equal,
$\left[\begin{array}{ccc}2 x-3 y & a-b & 3 \\ 1 & x+4 y & 3 a+4 b\end{array}\right]=\left[\begin{array}{ccc}1 & -2 & 3 \\ 1 & 6 & 29\end{array}\right]$
$\Rightarrow 2 x-3 y=1$ ....(1)
$x+4 y=6$
$\Rightarrow x=6-4 y$ .....(2)
Putting the value of $x$ in eq. (1), we get
$2(6-4 y)-3 y=1$
$\Rightarrow 12-8 y-3 y=1$
$\Rightarrow 12-11 y=1$
$\Rightarrow-11 y=-11$
$\Rightarrow y=\frac{-11}{-11}=1$
Putting the value of $y$ in eq. (2), we get
$x=6-4(1)$
$\Rightarrow x=6-4$
$\Rightarrow x=2$
Now,
$a-b=-2$
$\Rightarrow a=-2+b$ ....(3)
$3 a+4 b=29$ .....(4)
Putting the value of $a$ in eq. (4), we get
$3(-2+b)+4 b=29$
$\Rightarrow-6+3 b+4 b=29$
$\Rightarrow-6+7 b=29$
$\Rightarrow 7 b=29+6$
$\Rightarrow 7 b=35$
$\Rightarrow b=\frac{35}{7}=5$
Putting the value of $b$ in eq. (3), we get
$a=-2+5$
$\Rightarrow a=3$
$\therefore a=3, b=5, x=2$ and $y=1$
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