If [2a+b3a-bc+2d2c-d]=[234-1], find a, b, c and d.

Exercise 2.2 | Q 14 | Page 47

QUESTION

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

SOLUTION

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

∴ 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

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

${\displaystyle \frac{2}{5}+2d}$ = 4

∴ 2d = ${\displaystyle 4-\frac{2}{5}}$

∴ 2d = ${\displaystyle \frac{18}{5}}$

∴ d = ${\displaystyle \frac{9}{5}}$.