If ⎡⎢⎣x+3z+42y−7−6a−10b−3−210⎤⎥⎦=⎡⎢⎣063y−2−6−32c+22b+4−210⎤⎥⎦, then the value of a is:
A
−1
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B
−2
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C
1
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D
2
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Solution
The correct option is B−2 As the given matrices are equal , therefore, their corresponding elements must be equal. Comparing the element a22 of both matrices, we get a−1=−3, ⇒a=−2