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Home  >>  CBSE XII  >>  Math  >>  Determinants
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If $A=\begin{bmatrix}1 & 2 & 0\\-2 & -1 & -2\\0 & -1 & 1\end{bmatrix},find\;A^{-1}.$\[Using\;A^{-1}, solve\; the\; system\; of\; linear\; equations\; x-2y=10,2x-y-z=8,-2y+Z=7.\]

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Toolbox:
  • If |A|$\neq 0$,then it is a non-singular matrix.
  • $A^{-1}=\frac{1}{|A|}adj \;A$
  • AX=B$\Rightarrow X=A^{-1}B.$
Given:x-2y=10.
$\qquad$ 2x-y-z=8.
$\qquad$ -2y+z=7.
This can be written of the form AX=B.
$\begin{bmatrix}1 & -2 & 0\\2 & -1 & -1\\0 & -2 & 1\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}10\\8\\7\end{bmatrix}$
Where $A=\begin{bmatrix}1 & -2 & 0\\2 & -1 & -1\\0 & -2 & 1\end{bmatrix},X=\begin{bmatrix}x\\y \\z\end{bmatrix}$ and $B=\begin{bmatrix}10\\8\\7\end{bmatrix}$
To solve the above system of equations,let us find $A^{-1}$.
Let us first find if A is singular or non-singular matrix.
To find if its singular or non-singular,let us evaluate the value of the determinant.
|A|=1(-1-2)-(-2)(2-0)+0.
$\;\;\;$=-3+4=1$\neq 0$.
Hence it is a non-singular matrix.
Next let us find the adj A.
To find adj A,Let us find the cofactors of the respective elements
$A_{11}=(-1)^{1+1}\begin{vmatrix}-1 & -1\\-2 & 1\end{vmatrix}$=-1-2=-3.
$A_{12}=(-1)^{1+2}\begin{vmatrix}2 & -1\\0 & 1\end{vmatrix}$=-(2-0)=-2.
$A_{13}=(-1)^{1+3}\begin{vmatrix}2 & -1\\0 & -2\end{vmatrix}$=-4-0=-4.
$A_{21}=(-1)^{2+1}\begin{vmatrix}-2 & 0\\-2 & 1\end{vmatrix}$=-(-2-0)=2.
$A_{22}=(-1)^{2+2}\begin{vmatrix}1 & 0\\0 & 1\end{vmatrix}$=1-0=1.
$A_{23}=(-1)^{2+3}\begin{vmatrix}1 & -2\\0 & -2\end{vmatrix}$=-(-2)=2.
$A_{31}=(-1)^{3+1}\begin{vmatrix}-2 & 0\\-1 &- 1\end{vmatrix}$=2-0=2.
$A_{32}=(-1)^{3+2}\begin{vmatrix}1 & 0\\2 & -1\end{vmatrix}$=-(-1-0)=1.
$A_{33}=(-1)^{3+3}\begin{vmatrix}1 & -2\\2 & -1\end{vmatrix}$=-1+4=3.
$Adj\;A=\begin{bmatrix}A_{11} & A_{21} & A_{31}\\A_{12} & A_{22} & A_{32}\\A_{13} & A_{23} & A_{33}\end{bmatrix}$
$\qquad=\begin{bmatrix}-3 & 2 & 2\\-2 & 1 & 1\\-4 & 2 & 3\end{bmatrix}$
$A^{-1}=\frac{1}{|A|}adj\;A$ we know |A|=1.
Therefore $A^{-1}=1\begin{bmatrix}-3 & 2 & 2\\-2 & 1 & 1\\-4 & 2 & 3\end{bmatrix}$
We know X=$A^{-1}B.$
$\begin{bmatrix}-3 & 2 & 2\\-2 & 1 & 1\\-4 & 2 & 3\end{bmatrix}\begin{bmatrix}10 \\8\\7\end{bmatrix}=\begin{bmatrix}x\\y\\z\end{bmatrix}$
$\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}-3 & 2 & 2\\-2 & 1 & 1\\-4 & 2 & 3\end{bmatrix}\begin{bmatrix}10\\8\\7\end{bmatrix}=\begin{bmatrix}-30+16+14\\-20+8+7\\-40+16+21\end{bmatrix}$
$\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0\\-5\\-3\end{bmatrix}$
Hence x=0,y=-5 and z=-3.
answered Mar 20, 2013 by sreemathi.v
 

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