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  • If A is an invertible matrix of order 2, then det (A^-1) is equal to (A) det (A) (B) 1/det (A) (C) 1 (D) 0

Q : 18       If A is an invertible matrix of order 2, then det  \small (A^-^1) is equal to

                (A) \small det(A)       (B)  \small \frac{1}{det (A)}       (C)  \small 1       (D) \small 0

Answers (1)

best_answer

Given that the matrix is invertible hence A^{-1} exists and A^{-1} = \frac{1}{|A|}adjA

Let us assume a matrix of the order of 2;

A = \begin{bmatrix} a &b \\ c &d \end{bmatrix}.

Then |A| = ad-bc.

adjA = \begin{bmatrix} d &-b \\ -c & a \end{bmatrix}    and  |adjA| = ad-bc

Now,

A^{-1} = \frac{1}{|A|}adjA   

Taking determinant both sides;

|A^{-1}| = |\frac{1}{|A|}adjA| = \begin{bmatrix} \frac{d}{|A|} &\frac{-b}{|A|} \\ \\ \frac{-c}{|A|} & \frac{a}{|A|} \end{bmatrix}

\therefore|A^{-1}| = \begin{vmatrix} \frac{d}{|A|} &\frac{-b}{|A|} \\ \\ \frac{-c}{|A|} & \frac{a}{|A|} \end{vmatrix} = \frac{1}{|A|^2}\begin{vmatrix} d &-b \\ -c& a \end{vmatrix} = \frac{1}{|A|^2}(ad-bc) =\frac{1}{|A|^2}.|A| = \frac{1}{|A|}

Therefore we get;

|A^{-1}| = \frac{1}{|A|}

Hence the correct answer is B.

 

Posted by

Divya Prakash Singh

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