Question

If $m\left[\begin{array}{cc}-3& 4\end{array}\right]+n\left[\begin{array}{cc}4& -3\end{array}\right]=\left[\begin{array}{cc}10& -11\end{array}\right]$ , then $3m+7n$ is equal to

• A. $3$
• B. $5$
• C. $10$
• D. $1$

Answer 1

The correct option is D

$1$

Explanation for the correct answer:

Solve the given expression

Given, $m\left[-\begin{array}{cc}3& 4\end{array}\right]+n\left[\begin{array}{cc}4& -3\end{array}\right]=\left[\begin{array}{cc}10& -11\end{array}\right]$

$$\begin{gathered} \Rightarrow \left[\begin{array}{cc}-3m& 4m\end{array}\right]+\left[\begin{array}{cc}4n& -3n\end{array}\right]=\left[\begin{array}{cc}10& -11\end{array}\right] \\ \Rightarrow \left[\begin{array}{cc}-3m+4n& 4m-3n\end{array}\right]=\left[\begin{array}{cc}10& -11\end{array}\right] \end{gathered}$$

So, $-3m+4n=10$ and $4m-3n=-11$

Solving the two equations by replacing $m=\frac{4n-10}{3}$ in $4m-3n=-11$, we get

$$\begin{gathered} \Rightarrow 4\left(\frac{4n-10}{3}\right)-3n=-11 \\ \Rightarrow 16n-40-9n=-33 \\ \Rightarrow 7n=7 \\ \Rightarrow n=1 \end{gathered}$$

Now putting $n=1$ in $m=\frac{4n-10}{3}$, we get,

$$\begin{gathered} \Rightarrow m=\frac{4\left(1\right)-10}{3} \\ \Rightarrow m=-2 \end{gathered}$$

Putting these values in $3m+7n$, we get, $3(-2)+7\left(1\right)=1$

Therefore, option (D) is the correct answer.