Question

If $$2\text{A + 3B = }\left[\begin{array}{ccc}2& -1& 4\\ 3& 2& 5\end{array}\right]$$ and $$\text{A + 2B = }\left[\begin{array}{ccc}5& 0& 3\\ 1& 6& 2\end{array}\right]$$ then find B-
A)$\left[\begin{array}{ccc}8& -1& 2\\ -1& 10& -1\end{array}\right]$ B)$\left[\begin{array}{ccc}8& 1& 2\\ -1& 10& -1\end{array}\right]$ C) $\left[\begin{array}{ccc}8& 1& -2\\ -1& 10& -1\end{array}\right]$ D) $\left[\begin{array}{ccc}8& 1& 2\\ 1& 10& 1\end{array}\right]$

Answer 1

Hint: To find the value of B, first multiply $$\text{A + 2B = }\left[\begin{array}{ccc}5& 0& 3\\ 1& 6& 2\end{array}\right]$$ by 2 on both sides then subtract the given $$2\text{A + 3B = }\left[\begin{array}{ccc}2& -1& 4\\ 3& 2& 5\end{array}\right]$$ from the obtained multiplication result. You will get the value of B.

Complete step by step answer:

 Given, $$2\text{A + 3B = }\left[\begin{array}{ccc}2& -1& 4\\ 3& 2& 5\end{array}\right]$$ ---- (i)
And $$\text{A + 2B = }\left[\begin{array}{ccc}5& 0& 3\\ 1& 6& 2\end{array}\right]$$----- (ii)
Here, the matrices given are of order $\left(2\times 3\right)$ .We have to find B. First we multiply eq. (ii) by 2 on both sides,
$$\begin{gathered} \Rightarrow \text{2}\left(\text{A + 2B}\right)\text{ = 2}\left[\begin{array}{ccc}5& 0& 3\\ 1& 6& 2\end{array}\right] \\ \Rightarrow 2\text{A + 4B = }\left[\begin{array}{ccc}10& 0& 6\\ 2& 12& 4\end{array}\right] \end{gathered}$$

Now on subtracting eq. (i) from the obtained result, we have
$$\begin{gathered} \Rightarrow 2\text{A + 4B - }\left(2\text{A + 3B}\right)=\left[\begin{array}{ccc}10& 0& 6\\ 2& 12& 4\end{array}\right]-\left[\begin{array}{ccc}2& -1& 4\\ 3& 2& 5\end{array}\right] \\ \Rightarrow \text{2A + 4B - 2A - 3B = }\left[\begin{array}{ccc}10-2& 0-\left(-1\right)& 6-4\\ 2-3& 12-2& 4-5\end{array}\right] \end{gathered}$$
On solving the above expression, we get
$\Rightarrow \text{B = }\left[\begin{array}{ccc}8& 1& 2\\ -1& 10& -1\end{array}\right]$
Hence, the correct answer is ‘B’.

Note: Here, the order of matrices is $\left(2\times 3\right)$ because order=number of rows × number of columns. Also, since the order of both matrices [the obtained one and eq.(i)] is the same, so we can subtract the elements in the same position easily and obtain the matrix of the same order.