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Home  >>  CBSE XII  >>  Math  >>  Determinants
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If $A = \begin{bmatrix} 2 &-3 & 5\\ 3& 2 & -4\\ 1& 1& -2 \end{bmatrix}$, $\;$ find $A^{-1}.\;$ Using $ A^{-1}$ solve the system of equations: \[2x-3y+5z=11\] \[3x+2y-4z = -5\] \[x+y-2z = -3\]

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Toolbox:
  • A matrix is said to be singular if |A|= 0.
  • A matrix is said to be invertible if |A|$\neq 0$.
  • If A is a non-singular matrix,AX=B,then $X=A^{-1}B.$
  • Using this we can solve the system of equation which has unique solution.
We know $A^{-1}=\frac{1}{|A|}(adj \;A)$
Let us now find if 'A' is a singular or a nonsingular matrix.
The value of determinant A,|A| can be found by expanding along $R_1$
$|A|=2(2\times -2 -1\times -4)-(-3)(3\times -2-1\times -4)+5(3\times 1-2\times 1)$
$\;\;\;=2(-4+4)+3(-6+4)+5(3-2)$
$\;\;\;=0-6+5=-1\neq 0$
Hence it is a non-singular matrix.
Therefore $A^{-1}$ exists
Let us find the (adj A) by finding the minors and cofactors.
$M_{11}=\begin{vmatrix}2 & -4\\1 & -2\end{vmatrix}=-4+4=0$
$M_{12}=\begin{vmatrix}3 &-4\\1 & -2\end{vmatrix}=-6+4=-2$
$M_{13}=\begin{vmatrix}3 &2\\1 & 1\end{vmatrix}=3-2=1$
$M_{21}=\begin{vmatrix}-3 &5\\1 & -2\end{vmatrix}=6-5=1$
$M_{22}=\begin{vmatrix}2 &5\\1 & -2\end{vmatrix}=-4-5=-9$
$M_{23}=\begin{vmatrix}2 &-3\\1 & 1\end{vmatrix}=2+3=5$
$M_{31}=\begin{vmatrix}-3 &5\\2 & -4\end{vmatrix}=12-10=2$
$M_{32}=\begin{vmatrix}2 &5\\3 & -4\end{vmatrix}=-8-15=-23$
$M_{33}=\begin{vmatrix}2 &-3\\3 & 2\end{vmatrix}=4+9=13$
$A_{11}=(-1)^{1+1}.0=0$
$A_{12}=(-1)^{1+2}.-2=2$
$A_{13}=(-1)^{1+3}.1=1$
$A_{21}=(-1)^{2+1}.1=-1$
$A_{22}=(-1)^{2+2}.-9=-9$
$A_{23}=(-1)^{2+3}.5=-5$
$A_{31}=(-1)^{3+1}.2=2$
$A_{32}=(-1)^{3+2}.-23=23$
$A_{33}=(-1)^{3+3}.13=13$
Therefore adj A=$\begin{bmatrix}A_{11} & A_{21} & A_{31}\\A_{12} & A_{22} & A_{32}\\A_{13} & A_{23} & A_{33}\end{bmatrix}=\begin{bmatrix}0 & -1 & 2\\-2 & -9 & 23\\1 & -5 & 13\end{bmatrix}$
$A^{-1}=\frac{1}{|A|}(adj \;A)$ We know |A|=-1
$A^{-1}=\frac{1}{-1}\begin{bmatrix}0 & -1 & 2\\2 & -9 & 23\\1 & -5 & 13\end{bmatrix}=\begin{bmatrix}0 & 1 & -2\\-2 & 9 & -23\\-1 & 5 & -13\end{bmatrix}$
Using this let us solve the system of given equation\[2x-3y+5z=11\] \[3x+2y-4z = -5\] \[x+y-2z = -3\]
This can be written in the form AX=B.
$\begin{bmatrix}2 & -3 & -5\\3 & 2 & -4\\1 & 1 & -2\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}11\\-5\\-3\end{bmatrix}$
Where $A=\begin{bmatrix}2 & -3 & 5\\3 & 2 & -4\\1 & 1 & -2\end{bmatrix}X=\begin{bmatrix}x\\y\\z\end{bmatrix}B=\begin{bmatrix}11\\-5\\-3\end{bmatrix}$
We know AX=B,then $X=A^{-1}B.$
Therefore $\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0 & 1 & -2\\-2 & 9 & -23\\-1 & 5 & -13\end{bmatrix}\begin{bmatrix}11\\-5\\-3\end{bmatrix}$
Matrix multiplication can be done by multiplying the rows of matrix A with the column of matrix B.
Therefore $\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0-5+6\\-22-45+69\\-11-25+39\end{bmatrix}=\begin{bmatrix}1\\2\\3\end{bmatrix}$
Hence x=1,y=2 and z=3.
answered Mar 5, 2013 by vijayalakshmi_ramakrishnans
 

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