1.

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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