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

SOLUTION

Given equations are

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

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

By (i) – (ii) x 2, we get

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

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

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

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

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

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

By (i) x 2 –  (ii), we get

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

= ${\displaystyle \left[\begin{array}{ccc}12& -12& 0\\ -8& 4& 2\end{array}\right]-\left[\begin{array}{ccc}3& 2& 8\\ -2& 1& -7\end{array}\right]}$

= ${\displaystyle \left[\begin{array}{ccc}12-3& -12-2& 0-8\\ -8+2& 4 -1& 2+7\end{array}\right]}$

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

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

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