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Home  >>  CBSE XII  >>  Math  >>  Matrices
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Does $A=\begin{bmatrix}5 & 3\\1 & 2\end{bmatrix}$ satisfy the equation $\;A^2-3A-7I=0$ and hence find $\;A^{-1}$

$\begin{array}{1 1} No, \;it\; does\; not; A^{-1}=-1/7\begin{bmatrix}-2 & -1 \\1 & 5\end{bmatrix} \\ Yes \;it \;does; A^{-1}=-1/7\begin{bmatrix}-2 & -3 \\1 & 5\end{bmatrix} \\ Yes\; it\; does; A^{-1}=-1/14\begin{bmatrix}-2 & -3 \\1 & 5\end{bmatrix} \\ No \;it\; does\; not; A^{-1}=-1/7\begin{bmatrix}-2 & -3 \\1 & 5\end{bmatrix} \end{array} $
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Toolbox:
  • 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.
  • 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.
  • 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:
LHS:-$A^2-3A+7I=0.$
$A^2=\begin{bmatrix}5 & 3\\-1 & -2\end{bmatrix}\begin{bmatrix}5 & 3\\-1 & -2\end{bmatrix}$
$\;\;\;=\begin{bmatrix}5(5)-3 & 15-6\\-5+2 & -3+4\end{bmatrix}\Rightarrow \begin{bmatrix}22 & 9\\-3 & 1\end{bmatrix}$
3A=$3\begin{bmatrix}5 & 3\\-1 & -2\end{bmatrix}=\begin{bmatrix}15 & 9\\-3 & -6\end{bmatrix}$
$A^2-3A-7I=\begin{bmatrix}22 & 9\\-3 & 1\end{bmatrix}+(-1)\begin{bmatrix}15 & 9\\-3 & -6\end{bmatrix}+(-1)7\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}$
$\Rightarrow \begin{bmatrix}22 & 9\\-3 & 1\end{bmatrix}+\begin{bmatrix}-15 &- 9\\3 & 6\end{bmatrix}+\begin{bmatrix}-7 & 0\\0 & -7\end{bmatrix}$
$\Rightarrow \begin{bmatrix}22-15-7 & 9-9+0\\-3+3+0 & 1+6-7\end{bmatrix}$
$\Rightarrow \begin{bmatrix}0 & 0\\0 & 0\end{bmatrix}$
Hence $A^2-3A-7I=0.$
Step2:
To find $A^{-1}$
Let $A^{-1}$ be B
$B=\begin{bmatrix}x & y\\P & Q\end{bmatrix}$
We have $A^2-3A-7I=0.$
$-7I=-A^2+3A$
$-7A.B=-A^2+3A$
$-7\begin{bmatrix}5 & 3\\-1 & -2\end{bmatrix}\begin{bmatrix}x & y\\P & Q\end{bmatrix}=\begin{bmatrix}-22 & -9\\3 & -1\end{bmatrix}+\begin{bmatrix}15& 9\\-3& -6\end{bmatrix}$
-7$\begin{bmatrix}5x+3p & 5y+3Q\\-x-2P & -y-2Q\end{bmatrix}=\begin{bmatrix}-7 & 0\\0 & -7\end{bmatrix}$
-7$\begin{bmatrix}5x+3p & 5y+3Q\\-x-2P & -y-2Q\end{bmatrix}=7\begin{bmatrix}-1 & 0\\0 & -1\end{bmatrix}$
Since the above two matrices are equal hence their corresponding elements should be equal.
-5x-3P=-1-----(1)
x+2P=0-----(2)
-5y-3Q=0-----(3)
y+2Q=-1------(4)
Consider (1) & (2) and add
-5x-3P=-1
5x+10P=0
_______________
7P=-1
P=-1/7.
substitute in (2)
x+2(-1/7)=0.
x=2/7
Consider (3) & (4) and add
-5y-3Q=0
5y+10Q=-5
__________________
7Q=-5
Q=-5/7
substitute in (4)
y+2(-5/7)=-1
y=-1+10/7=3/7
y=3/7.
B=$\begin{bmatrix}2/7 & 3/7 \\-1/7 & -5/7\end{bmatrix}$
$\Rightarrow B=-1/7\begin{bmatrix}-2 & -3 \\1 & 5\end{bmatrix}$
$A^{-1}=-1/7\begin{bmatrix}-2 & -3 \\1 & 5\end{bmatrix}$

 

answered Mar 19, 2013 by sharmaaparna1
edited Mar 21, 2013 by sharmaaparna1
 

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