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Home  >>  CBSE XII  >>  Math  >>  Matrices
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If $A=\begin{bmatrix}1 & 0 & -1\\2 & 1 & 3\\0 & 1 & 1\end{bmatrix},then\;verify\;that\;A^2+A=A(A+I),where\;I\;is\;3\times 3\;unit\;matrix.$

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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: $\begin{bmatrix}AB\end{bmatrix}_{i,j} = A_{i,1}B_{1,j} + A_{i,2}B_{2,j} + A_{i,3}B_{3,j} ... A_{i,n}B_{n,j}$
  • The sum / difference $A(+/-)B$ of two $m$-by-$n$ matrices $A$ and $B$ is calculated entrywise: $(A (+/-) B)_{i,j} = A_{i,j} +/- B_{i,j}$ where 1 ≤ i ≤ m and 1 ≤ j ≤ n.
Step1:
Given
$A=\begin{bmatrix}1 & 0 &-1\\2 & 1 &3\\0 & 1 & 1\end{bmatrix}$
LHS:-
$A^2+A$
$A^2=A.A=\begin{bmatrix}1 & 0 &-1\\2 & 1 &3\\0 & 1 & 1\end{bmatrix}\begin{bmatrix}1 & 0 &-1\\2 & 1 &3\\0 & 1 & 1\end{bmatrix}$
$A.A=\begin{bmatrix}1(1)+0(2)+-1(0) & 1(0)+0(1)+(-1)(1) & 1(-1)+0(3)+1(-1)\\2(1)+1(2)+3(0) &2(0)+1(1)+3(1) &2(-1)+1(3)+3(1)\\0(1)+1(2)+1(0)& 0(0)+1(1)+1(1)& 0(-1)+1(3)+1(1)\end{bmatrix}$
$\;\;\;=\begin{bmatrix}1+0+0 & 0+0-1& -1+0-1\\2+2+0 &0+1+3 &-2+3+3\\0+2+0 &0+1+1 & 0+3+1\end{bmatrix}$
$\;\;\;=\begin{bmatrix}1 & -1& -2\\4 &4 &4\\2 &2 & 4\end{bmatrix}$
$A^2+A=\begin{bmatrix}1 & -1 &2\\4 & 4 &4\\2 & 2 & 4\end{bmatrix}+\begin{bmatrix}1 & 0 &-1\\2 & 1 &3\\0 & 1 & 1\end{bmatrix}$
$\;\;\;\qquad=\begin{bmatrix}1+1 & -1+0& -2-1\\4+2 &4+1 &4+3\\2+0 &2+1 & 4+1\end{bmatrix}$
$\;\;\;\qquad=\begin{bmatrix}2 & -1& -3\\6 &5 &7\\2 &3 & 5\end{bmatrix}$
Step2:
RHS:-
$A(A+I)=\begin{bmatrix}1 & 0& -1\\2 & 1 & 3\\0 & 1 & 1\end{bmatrix}\bigg(\begin{bmatrix}1 & 0& -1\\2 & 1 & 3\\0 & 1 & 1\end{bmatrix}+\begin{bmatrix}1 & 0& 0\\0 & 1 & 0\\0 & 0 & 1\end{bmatrix}\bigg)$
$A+I=\begin{bmatrix}1 & 0& -1\\2 & 1 & 3\\0 & 1 & 1\end{bmatrix}+\begin{bmatrix}1 & 0& 0\\0 & 1 & 0\\0 & 0 & 1\end{bmatrix}$
$\;\;\;\qquad=\begin{bmatrix}1+1 & 0+0& -1+0\\2+0 &1+1 &3+0\\0+0 &1+0 & 1+1\end{bmatrix}$
$\;\;\;\qquad=\begin{bmatrix}2 & 0& -1\\2 &2 &3\\0 &1 & 2\end{bmatrix}$
$A(A+I)=\begin{bmatrix}1 & 0& -1\\2 &1 &3\\0 &1 & 1\end{bmatrix}\begin{bmatrix}2 & 0& -1\\2 &2 &3\\0 &1 & 2\end{bmatrix}$
$\;\;\;\;=\begin{bmatrix}1(2)+0(2)+-1(0) & 1(0)+0(2)+(1)(-1) & 1(-1)+0(3)+2(-1)\\2(2)+1(2)+3(0) &2(0)+1(2)+3(1) &2(-1)+1(3)+3(2)\\0(2)+1(2)+1(0)& 0(0)+1(2)+1(1)& 0(-1)+1(3)+1(2)\end{bmatrix}$
$\;\;\;=\begin{bmatrix}2+0-0 & 0+0-1& -1+0-2\\4+2+0 &0+2+3 &-2+3+6\\0+2+0 &0+2+1 & 0+3+2\end{bmatrix}$
$\;\;\;=\begin{bmatrix}2 & -1& -3\\6 &5 &7\\2 &3 & 5\end{bmatrix}$
$\Rightarrow LHS=RHS.$
$\Rightarrow A^2+A=A(A+1).$
answered Mar 23, 2013 by sharmaaparna1
 

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