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Home  >>  CBSE XII  >>  Math  >>  Determinants
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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 $

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1 Answer

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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
 
|A|=ad-bc.
 
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 by sreemathi.v
edited Feb 28, 2013 by vijayalakshmi_ramakrishnans
 

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