Question

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

Answer 1

Since the corresponding elements of two equal matrices are equal,

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

$$\begin{gathered} \Rightarrow 2x-3y=1...\left(1\right) \\ x+4y=6 \\ \Rightarrow x=6-4y...\left(2\right) \\ \mathrm{Putting}\mathrm{the}\mathrm{value}\mathrm{of}x\mathrm{in}\mathrm{eq}.\left(1\right),\mathrm{we}\mathrm{get} \\ 2\left(6-4y\right)-3y=1 \\ \Rightarrow 12-8y-3y=1 \\ \Rightarrow 12-11y=1 \\ \Rightarrow -11y=-11 \\ \Rightarrow y=\frac{-11}{-11}=1 \\ \mathrm{Putting}\mathrm{the}\mathrm{value}\mathrm{of}y\mathrm{in}\mathrm{eq}.\left(2\right),\mathrm{we}\mathrm{get} \\ x=6-4\left(1\right) \\ \Rightarrow x=6-4 \\ \Rightarrow x=2 \\ \mathrm{Now}, \\ a-b=-2 \\ \Rightarrow a=-2+b...\left(3\right) \\ 3a+4b=29...\left(4\right) \\ \mathrm{Putting}\mathrm{the}\mathrm{value}\mathrm{of}a\mathrm{in}\mathrm{eq}.\left(4\right),\mathrm{we}\mathrm{get} \\ 3\left(-2+b\right)+4b=29 \\ \Rightarrow -6+3b+4b=29 \\ \Rightarrow -6+7b=29 \\ \Rightarrow 7b=29+6 \\ \Rightarrow 7b=35 \\ \Rightarrow b=\frac{35}{7}=5 \\ \mathrm{Putting}\mathrm{the}\mathrm{value}\mathrm{of}b\mathrm{in}\mathrm{eq}.\left(3\right),\mathrm{we}\mathrm{get} \\ a=-2+5 \\ \Rightarrow a=3 \\ \therefore a=3,b=5,x=2\mathrm{and}y=1 \end{gathered}$$