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Home  >>  CBSE XII  >>  Math  >>  Determinants
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\[ \text{Let A = } \begin{bmatrix} 3 & 7 \\ 2 & 5 \end{bmatrix} \text{and B = } \begin{bmatrix} 6 & 8 \\ 7 & 9 \end{bmatrix}. \text{Verify that } (AB^{-1}) = B^{-1}A^{-1}\]

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Toolbox:
  • (i)A matrix is said to be singular if $|A|$ =0.
  • (ii)A matrix is said to be invertible only if $|A|\neq 0$.
  • (iii) $A^{-1}=\frac{1}{|A|}adj \;A$
  • (iv)The adjoint of a square matrix A=[$a_{ij}]_{n\times n}$ is defined as the transpose of the matrix$ [A_{ij}]n\times n$
  • where $A_{ij}$ is the cofactor of the element $[a_{ij}].$
  • Adjoint is denoted by adj A.
Given$A=\begin{bmatrix}3 & 7\\2 & 5\end{bmatrix}$
 
The value of determinant of A is
 
$|A|=3\times 5-7\times 2$
 
$\;\;\;=15-14=1.$
 
The adj A of a square matrix of order 2 can also be obtained by interchanging $a_{11}$ and $a_{11}$ and by changing signs of $a_{12}$ and $a_{21}$
 
(i.e)$adj \;A=\begin{bmatrix}5 & -7\\2 & 3\end{bmatrix}$
 
Now $A^{-1}=\frac{1}{|A|}(adj A)$,we know |A|=1.
 
$A^{-1}=1\begin{bmatrix}5 & -7\\2 & 3\end{bmatrix}$
 
Similarly $B=\begin{bmatrix}6 & 8\\7 & 9\end{bmatrix}$
 
The value of the determinant B is
 
$|B|=6\times 9-8\times 7=54-56=-2.$
 
We can obtain the adjoint of B as we did before,
 
$(adj B)=\begin{bmatrix}9 & -8\\-7 & 6\end{bmatrix}$
 
Hence $B^{-1}=\frac{1}{|B|}(adj B),$we know |B|=-2.
 
$B^{-1}=\frac{1}{-2}\begin{bmatrix}9 & -8\\-7 & 6\end{bmatrix}$
 
$\;\;\;=\begin{bmatrix}\frac{-9}{2} & 4\\\frac{7}{2} & 3\end{bmatrix}$
 
Now let us obtain $B^{-1}A^{-1}$
 
Matrix multiplication can be done by multiplying the rows and corresponding columns.
 
$B^{-1}A^{-1}=\begin{bmatrix}\frac{-9}{2} & -4\\\frac{7}{2} & 3\end{bmatrix}\begin{bmatrix}5 & -7\\-2 & 3\end{bmatrix}$
 
$\;\;\;=\begin{bmatrix}\frac{-45}{2}-8 & \frac{63}{2}\\\frac{35}{2}+6 & \frac{-49}{2}-9\end{bmatrix}=\begin{bmatrix}\frac{-61}{2} & 87/2\\\frac{47}{2} & \frac{-67}{2}\end{bmatrix}$
 
Hence $B^{-1}A^{-1}=\begin{bmatrix}-61/2 & 87/2\\47/2 & -67/2\end{bmatrix}$-------(1)
 
Let us obtain $(AB)=\begin{bmatrix}3 & 7\\2 & 5\end{bmatrix}\begin{bmatrix}6 & 8\\7 & 9\end{bmatrix}$
 
We can multiply these two matrices as we did before
 
$\;\;\;=\begin{bmatrix}18+49 & 24+63\\12+35 & 16+45\end{bmatrix}$
 
$\;\;\;=\begin{bmatrix}67 & 87\\47 & 61\end{bmatrix}$
 
Now |AB| can be obtained as follows :
 
$|AB|=67\times 61-87\times 47$
 
$\;\;\;=4087-4089=-2.$
 
Hence |AB|=-2.
 
We can obtain adj(AB) as before:
 
$adj(AB)=\begin{bmatrix}61 & -87\\-47 & 67\end{bmatrix}$
 
Therefore $(AB)^{-1}=\frac{1}{|AB|}adj \;AB$
 
$\;\;\;=\frac{1}{-2}\begin{bmatrix}61 & -87\\-47 & 67\end{bmatrix}$
 
$\;\;\;=\begin{bmatrix}-61/2 & 87/2\\47/2 & -67/2\end{bmatrix}$----(2)
 
Hence from equ(1) and equ(2) we have
 
$B^{-1}A^{-1}=(AB)^{-1}$
 
Hence the result is verified.

 

answered Feb 26, 2013 by sreemathi.v
edited Feb 28, 2013 by sreemathi.v
 

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