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Home  >>  CBSE XII  >>  Math  >>  Matrices
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Find x,y,z if $A=\begin{bmatrix}0 & 2y & z\\x & y & -z\\x & -y & z\end{bmatrix}\;satisfies\;A'=A^{-1}$.

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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.
  • If the order of 2 matrices are equal, their corresponding elements are equal, i.e, if $A_{ij}=B_{ij}$, then any element $a_{ij}$ in matrix A is equal to corresponding element $b_{ij}$ in matrix B.
Step1:
In order to use row elementary transformation to find inverse we can write
$A=I_3A.$
Multiply $A^T$ on both side,
$A^T.A=A^T(I_3.A)$
$\Rightarrow A^T.A=A^T.I_3A$
Given $A=\begin{bmatrix}0 & 2y & z\\x & y & -z\\x & -y & z\end{bmatrix}$
$A'=\begin{bmatrix}0 & x & x\\2y & y & -y\\z & -z& z\end{bmatrix}$
Step2:
$\begin{bmatrix}0 & x & x\\2y & y & -y\\z & -z & z\end{bmatrix}\begin{bmatrix}0 & 2y & z\\x & y & -z\\x & -y & z\end{bmatrix}=\begin{bmatrix}0 & x & x\\2y & y & -y\\z & -z & z\end{bmatrix}\begin{bmatrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0& 1\end{bmatrix}A$
 
$\;\;\;=\begin{bmatrix}0+x^2+x^2 & 0+xy-xy& 0-xz+xz\\0+xy-xy & 4y^2+y^2+y^2 & 2yz-zy-zy\\0-xz+xz &2zy -zy-zy & z^2+z^2+z^2\end{bmatrix}=\begin{bmatrix}0 & 0+x+0 & 0+0+x\\2y+0+0 & y+0+0 & 0+0-y\\z+0+0 &0-z+0 & 0+0+z\end{bmatrix}A$
 
$\;\;\;=\begin{bmatrix}2x^2 & 0 & 0\\0 & 6y^2 & 0\\0 & 0 & 3z^2\end{bmatrix}=\begin{bmatrix}0 & x & x\\2y & y & -y\\z & -z & z\end{bmatrix}A$
Step3:
Apply $R_1=\frac{1}{2x^2}R_1$
$\begin{bmatrix}1 & 0 & 0\\0 & 6y^2 &0\\0 & 0 & 3z^2\end{bmatrix}=\begin{bmatrix}0 &1/2x & 1/2x\\2y & y & -y\\z & -z & z\end{bmatrix}A$
Step4:
Apply $R_2=\frac{1}{6y^2}R_2$
$\begin{bmatrix}1 & 0 & 0\\0 & 1 &0\\0 & 0 & 3z^2\end{bmatrix}=\begin{bmatrix}0 &1/2x & 1/2x\\1/3y & 1/6y & -1/6y\\z & -z & z\end{bmatrix}A$
Step5:
Apply $R_3=\frac{1}{3z^2}R_3$
$\begin{bmatrix}1 & 0 & 0\\0 & 1 &0\\0 & 0 & 1\end{bmatrix}=\begin{bmatrix}0 &1/2x & 1/2x\\1/3y & 1/6y & -1/6y\\1/3z & -1/3z & 1/3z\end{bmatrix}A$
$\Rightarrow A^{-1}=\begin{bmatrix}0 & 1/2x & 1/2x\\1/3y & 1/6y & -1/6y\\1/3z & -1/3z &1/3z\end{bmatrix}$
Step6:
Given $A'=A^{-1}$
$\Rightarrow \begin{bmatrix}0 & x & x\\2y & y & -y\\z & -z & z\end{bmatrix}=\begin{bmatrix}0 & 1/2x &1/2x\\1/3y & 1/6y & -1/6y\\1/3z & -1/3z & 1/3z\end{bmatrix}$
Given the two matrices are equal hence their corresponding elements should be equal.
x=1/2x.
$2x^2=1$
$x^2=1/2$
$x=\pm\frac{1}{\sqrt 2}$
$2y=\frac{1}{3y}$
$6y^2=1$
$y^2=1/6$
$y=\pm \frac{1}{\sqrt 6}$
z=1/3z
3$z^2=1.$
$z^2=1/3$
$z=\pm\sqrt {\frac{1}{3}}$

 

answered Apr 1, 2013 by sharmaaparna1
 

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