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Home  >>  CBSE XII  >>  Math  >>  Determinants
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Find the inverse of the matrix (if it exists): \[ \begin{bmatrix} 1&2&3 \\ 0&2&4 \\ 0&0&5 \\ \end{bmatrix} \]

$\begin{array}{1 1} \frac{-1}{10}\begin{bmatrix}10 & -10 & 2\\0 & 5 & -4\\0 & 0 & 2\end{bmatrix} \\\frac{1}{10}\begin{bmatrix}10 & -10 & 2\\0 & 5 & -4\\0 & 0 & 2\end{bmatrix} \\ 10\begin{bmatrix}10 & -10 & 2\\0 & 5 & -4\\0 & 0 & 2\end{bmatrix} \\ -10\begin{bmatrix}10 & -10 & 2\\0 & 5 & -4\\0 & 0 & 2\end{bmatrix} \end{array} $

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Toolbox:
  • (i)A matrix is said to be singular if $|A|$ =0.
  • (ii)A square matrix is said to be invertible only if it is a non-singular matrix.
  • (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.
Let A=$\begin{bmatrix}1 & 2& 3\\0 & 2& 4\\0 & 0 & 5\end{bmatrix}$
Expanding along $R_1$ we get,
$|A|=1(2\times 5-4\times 0)-2(0\times 5-4\times 0)+3(0\times 0-2\times 0)$
$\;\;\;\;=10-0-0=10$
Since |A|$\neq 0$ inverse exists,
Now let us find the adjoint of the A.$A_{ij}=(-1)^{i+j}M_{ij}$
Where $A_{ij}$ is the cofactor and $M_{ij}$ is the minor.
$M_{11}=\begin{bmatrix}2 & 4\\0 & 5\end{bmatrix}=10-0=10$
$M_{12}=\begin{bmatrix}0 & 4\\0 & 5\end{bmatrix}=0-0=0$
$M_{13}=\begin{bmatrix}0 & 2\\0 & 0\end{bmatrix}=0-0=0$
$M_{21}=\begin{bmatrix}2 & 3\\0 & 5\end{bmatrix}=10-0=10$
$M_{22}=\begin{bmatrix}1 & 3\\0 & 5\end{bmatrix}=5-0=5$
$M_{23}=\begin{bmatrix}1 & 2\\0 & 0\end{bmatrix}=0-0=0$
$M_{31}=\begin{bmatrix}2 & 3\\2 & 4\end{bmatrix}=8-6=2$
$M_{32}=\begin{bmatrix}1 & 3\\0 & 4\end{bmatrix}=4-0=4$
$M_{33}=\begin{bmatrix}1 & 2\\0 & 2\end{bmatrix}=2-0=2$
$A_{11}=(-1)^{1+1}.1=10.$
$A_{12}=(-1)^{1+2}.0=0.$
$A_{13}=(-1)^{1+3}.0=0.$
$A_{21}=(-1)^{2+1}.10=-10.$
$A_{22}=(-1)^{2+2}.5=5.$
$A_{23}=(-1)^{2+3}.0=0.$
$A_{31}=(-1)^{3+1}.2=2.$
$A_{32}=(-1)^{3+2}.4=-4.$
$A_{33}=(-1)^{3+3}.2=2.$
Now the (adj A)=$\begin{bmatrix}a_{11} & a_{21}&a_{31}\\a_{12} &a_{22} &a_{32}\\a_{13} &a_{23} & a_{33}\end{bmatrix}$
$\;\;\;\;\qquad\quad=\begin{bmatrix}10 & -10 & 2\\0 & 5 & -4\\0 & 0 & 2\end{bmatrix}$
The inverse of the matrix is $A^{-1}=\frac{1}{|A|}adj \;A$
|A|=10.
Therefore inverse =$\frac{1}{10}\begin{bmatrix}10 & -10 & 2\\0 & 5 & -4\\0 & 0 & 2\end{bmatrix}$
answered Mar 6, 2013 by vijayalakshmi_ramakrishnans
 

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