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Home  >>  CBSE XII  >>  Math  >>  Determinants
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$If \; A^{-1}=\begin{bmatrix} 3 &-1 & 1\\ -15&6 &-5 \\ 5& -2&2 \end{bmatrix} \; and \; B = \begin{bmatrix} 1 & 2 & -2\\ -1 & 3 & 0\\ 0 & -2 & 1 \end{bmatrix}, \; find \; (AB)^{-1} $

$\begin{array}{1 1} (AB)^{-1}=\begin{bmatrix}9 & -2 & 5\\-2 & -1 & 0\\1 & 0 & 2\end{bmatrix} \\(AB)^{-1}=\begin{bmatrix}9 & -3 & 5\\-2 & 1 & 0\\1 & 0 & 2\end{bmatrix} \\ (AB)^{-1}=\begin{bmatrix}-9 & -3 & 5\\-2 & 1 & 0\\-1 & 0 & 2\end{bmatrix} \\ (AB)^{-1}=\begin{bmatrix}9 & -3 & 5\\-2 & 1 & 1\\1 & 1 & 2\end{bmatrix} \end{array} $

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Toolbox:
  • A matrix is said to be singular if|A|=0.
  • A matrix is said to be invertible if $|A|\neq 0.$
  • $A^{-1}=\frac{1}{|A|}(adj\;A)$
  • (adj A)=$\begin{bmatrix}A_{11} & A_{21} & A_{31}\\A_{12} & A_{22} & A_{32}\\A_{13} & A_{23} & A_{33}\end{bmatrix}$
  • Where $A_{11},A_{12}....A_{33}$ are cofactors.
Given $A^{-1}=\begin{bmatrix} 3 &-1 & 1\\ -15&6 &-5 \\ 5& -2&2 \end{bmatrix} \; and \; B = \begin{bmatrix} 1 & 2 & -2\\ -1 & 3 & 0\\ 0 & -2 & 1 \end{bmatrix}$
We know that $(AB)^{-1}=B^{-1}A^{-1}$
Let us find $B^{-1}$
$B= \begin{bmatrix} 1 & 2 & -2\\ -1 & 3 & 0\\ 0 & -2 & 1 \end{bmatrix}$
Let us determine the determinant of B by expanding along $R_1$
$|B|=1(3-0)-2(-1-0)-2(2-0)$
$\;\;\;=3+2-4=1\neq 0.$
Hence B is invertible.
To find the adjoint of B,let us find the cofactors of respective elements.
$M_{11}=\begin{vmatrix}3 & 0\\-2 & 1\end{vmatrix}$=3-0=3.
$M_{12}=\begin{vmatrix}-1 & 0\\0 & 1\end{vmatrix}$=-1-0= - 1.
$M_{13}=\begin{vmatrix}-1 & 3\\0 & -2\end{vmatrix}$=2-0= 2.
$M_{21}=\begin{vmatrix}2 & -2\\-2 & 1\end{vmatrix}$=2-4= - 2.
$M_{22}=\begin{vmatrix}1 & -2\\0 & 1\end{vmatrix}$=1-0=1.
$M_{31}=\begin{vmatrix}2 & -2\\3 & 0\end{vmatrix}$=0+6=6.
$M_{32}=\begin{vmatrix}1 & -2\\-1 & 0\end{vmatrix}$=0-2= - 2.
$M_{33}=\begin{vmatrix}1 & 2\\-1 & 3\end{vmatrix}$=3+2=5.
$A_{11}=(-1)^{1+1}.3=3$
$A_{12}=(-1)^{1+2}.(-1)=1$
$A_{13}=(-1)^{1+3}23=2$
$A_{21}=(-1)^{2+1}.(-2)=2$
$A_{22}=(-1)^{2+2}.1=1$
$A_{31}=(-1)^{3+1}.6=6$
$A_{32}=(-1)^{3+2}.(-2)=2$
$A_{33}=(-1)^{3+3}.5=5$
Hence (adj B)=$\begin{bmatrix}A_{11} & A_{21} & A_{31}\\A_{12} & A_{22} & A_{32}\\A_{13} & A_{23} & A_{33}\end{bmatrix}$
$\;\;\;\;\;\;\;\;\;\;=\begin{bmatrix}3 & 2 & 6\\1 & 1 & 2\\2 & 2 & 5\end{bmatrix}$
$B^{-1}=\frac{1}{|B|}.(adj B)$,|B|=1.
$B^{-1}=\frac{1}{1}\begin{bmatrix}3 & 2 & 6\\1 & 1 & 2\\2 & 2 & 5\end{bmatrix}=\begin{bmatrix}3 & 2 & 6\\1 & 1 & 2\\2 & 2 & 5\end{bmatrix}$
$(AB)^{-1}=B^{-1}A^{-1}$
$A^{-1}=\begin{bmatrix} 3 &-1 & 1\\ -15&6 &-5 \\ 5& -2&2 \end{bmatrix}$
$B^{-1}A^{-1}=\begin{bmatrix}3 & 2 & 6\\1 & 1 & 2\\2 & 2 & 5\end{bmatrix}\begin{bmatrix} 3 &-1 & 1\\ -15&6 &-5 \\ 5& -2&2 \end{bmatrix}$
Matrix multiplication can be done by multiplying the rows of $B^{-1}$ with columns of $A^{-1}.$
$B^{-1}A^{-1}=\begin{bmatrix}9-30+30 & -3+12-12 & 3-10+12\\3-15+10 & -1+6-4 & -1-5+4\\6-30+25 & -2+12-10 & 2-10+10\end{bmatrix}$
$\;\;\;\;\;\;\;=\begin{bmatrix}9 & -3 & 5\\-2 & 1 & 0\\1 & 0 & 2\end{bmatrix}$
$(AB)^{-1}=\begin{bmatrix}9 & -3 & 5\\-2 & 1 & 0\\1 & 0 & 2\end{bmatrix}$
answered Mar 5, 2013 by vijayalakshmi_ramakrishnans
 

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