If m begin{bmatrix} -3 & 4 end{bmatrix} + n b | vedclass

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(D) Given that,$m \begin{bmatrix} -3 & 4 \end{bmatrix} + n \begin{bmatrix} 4 & -3 \end{bmatrix} = \begin{bmatrix} 10 & -11 \end{bmatrix}$ Multiplying

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(D) Given that,$m \begin{bmatrix} -3 & 4 \end{bmatrix} + n \begin{bmatrix} 4 & -3 \end{bmatrix} = \begin{bmatrix} 10 & -11 \end{bmatrix}$ Multiplying the scalars $m$ and $n$ into the matrices,we get: $\begin{bmatrix} -3m & 4m \end{bmatrix} + \begin{bmatrix} 4n & -3n \end{bmatrix} = \begin{bmatrix} 10 & -11 \end{bmatrix}$ Adding the matrices: $\begin{bmatrix} -3m + 4n & 4m - 3n \end{bmatrix} = \begin{bmatrix} 10 & -11 \end{bmatrix}$ By equating the corresponding elements,we get the following system of linear equations: $-3m + 4n = 10 \quad \dots (i)$ $4m - 3n = -11 \quad \dots (ii)$ To solve this system,multiply equation $(i)$ by $3$ and equation $(ii)$ by $4$: $-9m + 12n = 30 \quad \dots (iii)$ $16m - 12n = -44 \quad \dots (iv)$ Adding equations $(iii)$ and $(iv)$: $7m = -14 \Rightarrow m = -2$ Substituting $m = -2$ into equation $(i)$: $-3(-2) + 4n = 10 \Rightarrow 6 + 4n = 10 \Rightarrow 4n = 4 \Rightarrow n = 1$ Now,calculate $3m + 7n$: $3m + 7n = 3(-2) + 7(1) = -6 + 7 = 1$

If $m \begin{bmatrix} -3 & 4 \end{bmatrix} + n \begin{bmatrix} 4 & -3 \end{bmatrix} = \begin{bmatrix} 10 & -11 \end{bmatrix}$,then $3m + 7n$ is equal to

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