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If A = \(\begin{bmatrix} 0 & 2 \\[0.3em] 3& -4 \\[0.3em] \end{bmatrix}\) and kA = \(\begin{bmatrix} 0 & 3a \\[0.3em] 2b & 24 \\[0.3em] \end{bmatrix}\), then the values of k, a, b, are respectivelyA. –6, –12, –18 B. –6, 4, 9 C. –6, –4, –9 D. –6, 12, 18 |
Answer» A = \(\begin{bmatrix} 0 & 2 \\[0.3em] 3& -4 \\[0.3em] \end{bmatrix}\) and kA = \(\begin{bmatrix} 0 & 3a \\[0.3em] 2b & 24 \\[0.3em] \end{bmatrix}\) = \(\begin{bmatrix} 0 & 2k \\[0.3em] 3k & -4k \\[0.3em] \end{bmatrix}\) Comparing the equations, -4k = 24 k = -6 3k = 2b 3(-6) = 2b 2b = -18 b= - 9 3a = 2k 3a = 2(-6) 3a = -12 a = -4 Values are, k = -6, a = -4 & b = -9 Option (C) is the answer. |
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