If A = begin{bmatrix} 0 & 2 3 & -4 end{bmatr | vedclass

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(C) Given,$kA = \begin{bmatrix} 0 & 3a \ 2b & 24 \end{bmatrix}$.Substituting $A$,we get $k \begin{bmatrix} 0 & 2 \ 3 & -4 \end{bmatrix} = \begin{bma

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$ -6, -4, -9$

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(C) Given,$kA = \begin{bmatrix} 0 & 3a \ 2b & 24 \end{bmatrix}$.Substituting $A$,we get $k \begin{bmatrix} 0 & 2 \ 3 & -4 \end{bmatrix} = \begin{bmatrix} 0 & 3a \ 2b & 24 \end{bmatrix}$.Multiplying $k$ inside the matrix: $\begin{bmatrix} 0 & 2k \ 3k & -4k \end{bmatrix} = \begin{bmatrix} 0 & 3a \ 2b & 24 \end{bmatrix}$.Comparing the corresponding elements:$1) -4k = 24 \implies k = -6$.$2) 2k = 3a \implies 2(-6) = 3a \implies -12 = 3a \implies a = -4$.$3) 3k = 2b \implies 3(-6) = 2b \implies -18 = 2b \implies b = -9$.Thus,the values are $k = -6, a = -4, b = -9$.

If $A = \begin{bmatrix} 0 & 2 \\ 3 & -4 \end{bmatrix}$ and $kA = \begin{bmatrix} 0 & 3a \\ 2b & 24 \end{bmatrix}$,then the values of $k, a, b$ are respectively

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