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Home  >>  CBSE XII  >>  Math  >>  Matrices
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If $A=\begin{bmatrix}3 & -5\\-4 & 2\end{bmatrix},$ then find $A^2-5A-14I.$Hence,obtain $A^3$.

$\begin{array}{1 1} A^3=\begin{bmatrix}187 &195\\156& 148\end{bmatrix} \\ A^3=\begin{bmatrix}187 &-195\\-156& 148\end{bmatrix} \\ A^3=\begin{bmatrix}7 &-195\\-156& 148\end{bmatrix} \\ A^3=\begin{bmatrix}187 & 0 \\-156& 148\end{bmatrix}\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: $\begin{bmatrix}AB\end{bmatrix}_{i,j} = A_{i,1}B_{1,j} + A_{i,2}B_{2,j} + A_{i,3}B_{3,j} ... A_{i,n}B_{n,j}$
  • 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.
  • An identity matrix or unit matrix of size n is the n × n square matrix with ones on the main diagonal and zeros elsewhere. An identity matrix of order 2, $I_{2}= \begin{bmatrix} 1 &0 \\ 0&1 \end{bmatrix}$
Step1:
Given
$A= \begin{bmatrix}3 & -5\\-4 & 2\end{bmatrix}$
$A^2=A.A\Rightarrow \begin{bmatrix}3 & -5\\-4 & 2\end{bmatrix}\begin{bmatrix}3 & -5\\-4 & 2\end{bmatrix}$
$\Rightarrow \begin{bmatrix}3(3)+(-5)(-4) & 3(-5)+(-5)(2)\\-4(3)+2(-4) & -4(-5)+2(2)\end{bmatrix}$
$\Rightarrow \begin{bmatrix}9+20 & -15-10\\-12-8 & 20+4\end{bmatrix}$
$\Rightarrow \begin{bmatrix}29 & -25\\-20 & 24\end{bmatrix}$
Step2:
$A^2-5A-14I$
$\Rightarrow \begin{bmatrix}29 & -25\\-20 & 24\end{bmatrix}-5\begin{bmatrix}3 & -5\\-4 & 2\end{bmatrix}-14\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}$
$\Rightarrow \begin{bmatrix}29 & -25\\-20 & 24\end{bmatrix}+\begin{bmatrix}15 & 25\\20 & -10\end{bmatrix}+\begin{bmatrix}-14 & 0\\0 & -14\end{bmatrix}$
$\Rightarrow \begin{bmatrix}29-15-14 & -25+25+0\\-20+20+0 & 24-10-14\end{bmatrix}$
$\Rightarrow \begin{bmatrix}29-29 & 0\\0 & 24-24\end{bmatrix}$
$\Rightarrow \begin{bmatrix}0 & 0\\0 & 0\end{bmatrix}$
Hence proved
Step3:
$A^3\Rightarrow A^2.A$
$\Rightarrow \begin{bmatrix}29 & -25\\-20 & 24\end{bmatrix}\begin{bmatrix}3 & -5\\-4 & 2\end{bmatrix}$
$\Rightarrow \begin{bmatrix}29(3)+(-25)(-4) &29(-5)+ (-25)(2)\\-20 (3)+24(-4)& -20(-5)+(24)(2)\end{bmatrix}$
$\Rightarrow \begin{bmatrix}87+100 &-145-50\\-60-96& 100+48\end{bmatrix}$
$\Rightarrow \begin{bmatrix}187 &-195\\-156& 148\end{bmatrix}$
answered Mar 25, 2013 by sharmaaparna1
 

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