The matrix A satisfying the equation [1301]A=[110−1] is
A
[14−10]
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B
[1−410]
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C
[1−20−1]
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D
none of these
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Solution
The correct option is D none of these [1301]A=[110−1] Let P=[1301],Q=[110−1] PA=Q ⇒A=P−1Q We have P=[1301] |P|=1 Now adjP=CT=[10−31]T ⇒adjP=[1−301] Hence, P−1=[1−301] So, A=[1−301][110−1] ⇒A=[140−1]