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Home  >>  CBSE XII  >>  Math  >>  Matrices
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If $A=\begin{bmatrix}\alpha & \beta\\\gamma & -\alpha\end{bmatrix}$ is such that $A^2=I,$ then which of the following is true:

\[\begin{array}{1 1}(A)\;1+{\alpha}^2+\beta\gamma =0& (B)\;1-{\alpha}^2+\beta\gamma=0\\(C)\;1-{\alpha}^2-\beta\gamma=0 & (D)\;1+{\alpha}^2-\beta\gamma=0\end{array}\]
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1 Answer

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Toolbox:
  • If A is an m-by-n matrix and B is an n-by-p matrix, then their matrix product AB is the m-by-p matrix whose entries are given by dot product of the corresponding row of A and the corresponding column of B: $\begin{bmatrix}AB\end{bmatrix}_{i,j} = A_{i,1}B_{1,j} + A_{i,2}B_{2,j} + A_{i,3}B_{3,j} ... A_{i,n}B_{n,j}$
Step1:
Given
$A=\begin{bmatrix}\alpha & \beta\\\gamma & -\alpha\end{bmatrix}$
$A^2=I$
$A^2=A.A=\begin{bmatrix}\alpha & \beta\\\gamma & -\alpha\end{bmatrix}\begin{bmatrix}\alpha & \beta\\\gamma & -\alpha\end{bmatrix}$
$\;\;\;\;=\begin{bmatrix}{\alpha}^2+\beta\gamma & \alpha\beta-\alpha\beta\\\alpha\gamma -\alpha\gamma&\beta\gamma -{\alpha}^2\end{bmatrix}$
$\;\;\;\;=\begin{bmatrix}{\alpha}^2+\beta\gamma & \alpha\beta-\alpha\beta\\\alpha\gamma -\alpha\gamma&\beta\gamma +{\alpha}^2\end{bmatrix}=I$
Step2:
$\begin{bmatrix}{\alpha}^2+\beta\gamma & \alpha\beta-\alpha\beta\\\alpha\gamma -\alpha\gamma&\beta\gamma -{\alpha}^2\end{bmatrix}=\begin{bmatrix}1 &0\\0 & 1\end{bmatrix}$
${\alpha}^2+\beta\gamma=1$
$\beta \gamma+{\alpha}^2=1$
$\Rightarrow 1-{\alpha}^2-\beta \gamma=0$
$1-\beta\gamma-{\alpha}^2=0$
$\Rightarrow$ (C) is the required answer.

 

answered Mar 19, 2013 by sharmaaparna1
edited Mar 19, 2013 by sharmaaparna1
 

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