# If A is an invertible matrix of order 2, then det$(A^{-1})\;$ is equal to: $(A)\; det\;(A) \quad (B)\; \frac{1}{det\;(A)} \quad (C)\; 1 \quad (D)\; 0$

Toolbox:
• (i) The adj A of a matrix of order 2 can be obtained by interchanging $a_{11}$ and $a _{22}$ and changing the symbols of $a_{12}$ and $a_{21}$.
• (ii) $A^{-1}=\frac{1}{|A|}adj \;A$
Since it is given A is an invertible matrix.

$|A|\neq0$

Therefore $A^{-1}=\frac{1}{|A|}adj \;A$

Also it is given matrix A is of order 2,

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

Then the value of the determinant is

and the adjoint can be obtained by interchanging $a_{11}$ and $a_{22}$ and changing the symbols of $a_{12}$ and $a_{21}$

Hence adj A=$\begin{bmatrix}d & -b\\-c & a\end{bmatrix}$

Therefore $A^{-1}=\frac{1}{|A|}adj \;A$

$\;\;\;\qquad\qquad=\frac{1}{|A|}\begin{bmatrix}d & -b\\-c & a\end{bmatrix}$

$\;\;\;\qquad\qquad=\begin{bmatrix}\frac{d}{|A|} & \frac{-b}{|A|}\\\frac{-c}{|A|} & \frac{a}{|A|}\end{bmatrix}$

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

Taking $\frac{1}{|A|}$ as a common factor from $R_1$ and $R_2$.

$|A^{-1}|=\frac{1}{|A|^2}\begin{vmatrix}d & -b\\-c & a\end{vmatrix}$

Expanding the determinant we get,

$|A^{-1}|=\frac{1}{|A|^2}(ad-bc)$

But we know (ad-bc)=|A|.

Therefore $|A^{-1}|=\frac{1}{|A|^2}|A|$

$\qquad\qquad\qquad=\frac{1}{|A|}$

det($A^{-1})=\frac{1}{det(A)}$

Hence the correct answer is B.

answered Feb 26, 2013
edited Feb 28, 2013