## Pages

### Find matrices A and B, if 2A – B = [6-60-421] and A – 2B = [328-21-7].

Exercise 2.2 | Q 12 | Page 47

#### QUESTION

Find matrices A and B, if 2A – B = $\left[\begin{array}{ccc}6& -6& 0\\ -4& 2& 1\end{array}\right]$ and A – 2B = $\left[\begin{array}{ccc}3& 2& 8\\ -2& 1& -7\end{array}\right]$.

#### SOLUTION

Given equations are

2A – B = $\left[\begin{array}{ccc}6& -6& 0\\ -4& 2& 1\end{array}\right]$           ...(i)

and A – 2B = $\left[\begin{array}{ccc}3& 2& 8\\ -2& 1& -7\end{array}\right]$    ...(ii)

By (i) – (ii) x 2, we get

3B = $\left[\begin{array}{ccc}6& -6& 0\\ -4& 2& 1\end{array}\right]-2\left[\begin{array}{ccc}3& 2& 8\\ -2& 1& -7\end{array}\right]$

$\left[\begin{array}{ccc}6& -6& 0\\ -4& 2& 1\end{array}\right]-\left[\begin{array}{ccc}6& 4& 16\\ -4& 2& -14\end{array}\right]$

∴ 3B = $\left[\begin{array}{ccc}0& -10& -16\\ 0& 0& 15\end{array}\right]$

∴ B = $\frac{1}{3}\left[\begin{array}{ccc}0& -10& -16\\ 0& 0& 15\end{array}\right]$

∴ B = $\left[\begin{array}{ccc}0& \frac{-10}{3}& \frac{-16}{3}\\ 0& 0& 5\end{array}\right]$

By (i) x 2 –  (ii), we get

3A = $2\left[\begin{array}{ccc}6& -6& 0\\ -4& 2& 1\end{array}\right]-\left[\begin{array}{ccc}3& 2& 8\\ -2& 1& -7\end{array}\right]$

∴ 3A = $\left[\begin{array}{ccc}9& -14& -8\\ -6& 3& 9\end{array}\right]$

∴ A = $\frac{1}{3}\left[\begin{array}{ccc}9& 14& -8\\ -6& 3& 9\end{array}\right]$

∴ A = $\left[\begin{array}{ccc}3& \frac{-14}{3}& \frac{-8}{3}\\ -2& 1& 3\end{array}\right]$.