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If $A=\begin{bmatrix}0 & 1\\1 & 0\end{bmatrix},$then $A^2$ is equal to

\begin{array}{1 1}(A)\;\begin{bmatrix}0 & 1\\1 & 0\end{bmatrix} & (B)\;\begin{bmatrix}1 & 0\\1 & 0\end{bmatrix}\\(C)\;\begin{bmatrix}0 & 1\\0 & 1\end{bmatrix} & (D)\;\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}\end{array}

1 Answer

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: $\begin{bmatrix}AB\end{bmatrix}_{i,j} = A_{i,1}B_{1,j} + A_{i,2}B_{2,j} + A_{i,3}B_{3,j} ... A_{i,n}B_{n,j}$
Given
$A=\begin{bmatrix}0 & 1\\1 & 0\end{bmatrix}$
To find
$A^2=A.A$
$\;\;\;=\begin{bmatrix}0 & 1\\1 & 0\end{bmatrix}\begin{bmatrix}0 & 1\\1 & 0\end{bmatrix}$
$\;\;\;=\begin{bmatrix}0+1 & 0+0\\0+0 & 1+0\end{bmatrix}$
$\;\;\;=\begin{bmatrix}1 & 0\\0& 1\end{bmatrix}$
Hence (D) is the right option.
answered Apr 1, 2013 by sharmaaparna1
 

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