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Home  >>  CBSE XII  >>  Math  >>  Matrices
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If possible,using elementary row transformations,find the inverse of the following matrices $\quad\begin{bmatrix}2 & 0 & -1\\5 & 1 & 0\\0& 1 & 3\end{bmatrix}$

Note: This is part 3 of a 3 part question, split as 3 separate questions here.
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Toolbox:
  • There are six operations (transformations) on a matrix,three of which are due to rows and three due to columns which are known as elementary operations or transformations.
  • Row/Column Switching: Interchange of any two rows or two columns, i.e, $R_i\leftrightarrow R_j$ or $\;C_i\leftrightarrow C_j$
  • Row/Column Multiplication: The multiplication of the elements of any row or column by a non zero number: i.e, i.e $R_i\rightarrow kR_i$ where $k\neq 0$ or $\;C_j\rightarrow kC_j$ where $k\neq 0$
  • Row/Column Addition:The addition to the element of any row or column ,the corresponding elements of any other row or column multiplied by any non zero number: i.e $R_i\rightarrow R_i+kR_j$ or $\;C_i\rightarrow C_i+kC_j$, where $i \neq j$.
  • If A is a matrix such that A$^{-1}$ exists, then to find A$^{-1}$ using elementary row operations, write A = IA and apply a sequence of row operation on A = IA till we get, I = BA. The matrix B will be the inverse of A. Similarly, if we wish to find A$^{-1}$ using column operations, then, write A = AI and apply a sequence of column operations on A = AI till we get, I = AB.
Step1:
In order to use row elementary transformation to find the inverse we write as\[A=I_3A\]
$\begin{bmatrix}2 & 0 & -1\\5 & 1 & 0\\0 & 1 & 3\end{bmatrix}=\begin{bmatrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{bmatrix}A$
Apply $R_1\leftrightarrow R_2$
$\begin{bmatrix}5 & 1 & 0\\2 & 0 & -1\\0 & 1 & 3\end{bmatrix}=\begin{bmatrix}0 & 1 & 0\\1 & 0 & 0\\0 & 0 & 1\end{bmatrix}A$
Step2:
Apply $R_1\rightarrow R_1-2R_2$
$\begin{bmatrix}1 & 1& 2\\2 & 0 & -1\\0 & 1 & 3\end{bmatrix}=\begin{bmatrix}-2 & 1 & 0\\1 & 0 & 0\\0 & 0 & 1\end{bmatrix}A$
Step3:
Apply $R_2\rightarrow R_2-2R_1$
$\begin{bmatrix}1 & 1& 2\\0 & -2 & -5\\0 & 1 & 3\end{bmatrix}=\begin{bmatrix}-2 & 1 & 0\\5 & -2 & 0\\0 & 0 & 1\end{bmatrix}A$
Step4:
Apply $R_2\rightarrow (-1)R_2$
$\begin{bmatrix}1 & 1& 2\\0 & 2 & 5\\0 & 1 & 3\end{bmatrix}=\begin{bmatrix}-2 & 1 & 0\\-5 & 2 & 0\\0 & 0 & 1\end{bmatrix}A$
Step5:
Apply $R_2\rightarrow R_2-R_3$
$\begin{bmatrix}1 & 1& 2\\0 & 1 & 2\\0 & 1 & 3\end{bmatrix}=\begin{bmatrix}-2 & 1 & 0\\-5 & 2 & -1\\0 & 0 & 1\end{bmatrix}A$
Step6:
Apply $R_1\rightarrow R_1-R_2$
$\begin{bmatrix}1 & 0& 0\\0 & 1 & 2\\0 & 1 & 3\end{bmatrix}=\begin{bmatrix}3 & -1 & 1\\-5 & 2 & -1\\0 & 0 & 1\end{bmatrix}A$
Step7:
Apply $R_3\rightarrow R_3-R_2$
$\begin{bmatrix}1 & 0& 0\\0 & 1 & 2\\0 & 0 & 1\end{bmatrix}=\begin{bmatrix}3 & -1 & 1\\-5 & 2 & -1\\5 & -2 & 2\end{bmatrix}A$
Step8:
Apply $R_2\rightarrow R_2-2R_3$
$\begin{bmatrix}1 & 0& 0\\0 & 1 & 0\\0 & 0 & 1\end{bmatrix}=\begin{bmatrix}3 & -1 & 1\\-15 & 6 & -5\\5 & -2 & 2\end{bmatrix}A$
$\Rightarrow A^{-1}=\begin{bmatrix}3 & -1 & 1\\-15 & 6 & -5\\5 & -2 & 2\end{bmatrix}$
answered Apr 1, 2013 by sharmaaparna1
 

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