Question
If $\left[\begin{array}{cc}2 a+b & 3 a-b \\ c+2 d & 2 c-d\end{array}\right]=\left[\begin{array}{cc}2 & 3 \\ 4 & -1\end{array}\right]$, find $\mathrm{a}, \mathrm{b}, c$ and $d$

Answer

$\left[\begin{array}{cc}2 a+b & 3 a-b \\ c+2 d & 2 c-d\end{array}\right]=\left[\begin{array}{cc}2 & 3 \\ 4 & -1\end{array}\right]$

∴ By equality of matrices, we get 2a + b = 2 ….(i) 3a – b = 3 ….(ii) c + 2d = 4 ….(iii) 2c – d = -1 ….(iv) Adding (i) and (ii), we get 5a = 5 ∴ a = 1 Substituting a = 1 in (i), we get 2(1) + b = 2 ∴ b = 0 By (iii) + (iv) x 2, we get 5c = 2

$\therefore c=\frac{2}{5}$

Substituting $c=\frac{2}{5}$ in (iii), we get

$\begin{aligned} & \frac{2}{5}+2 d=4 \\ & \therefore 2 d=4-\frac{2}{5} \\ & \therefore 2 d=\frac{18}{5} \\ & \therefore d=\frac{9}{5}\end{aligned}$

[Note: Answer given in the textbook is $d=\frac{3}{5}$.

However, as per our calculation it is $d=\frac{9}{5}$.]

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