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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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If \( A = \begin{bmatrix} 4 & 3 \\ 2 & 5 \end{bmatrix}\), find $x$ and $y$ such that $ A^2-xA+yI=0. $

$\begin{array}{1 1} x = 9,y = 4 \\ x = 7,y = 14 \\ x = 9,y = -14 \\ x = 9,y = 14 \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: $\begin{bmatrix}AB\end{bmatrix}_{i,j} = A_{i,1}B_{1,j} + A_{i,2}B_{2,j} + A_{i,3}B_{3,j} ... A_{i,n}B_{n,j}$
  • 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.
  • We can then match the corresponding elements and solve the resulting equations to find the values of the unknown variables.
Step1:
Given:
$A=\begin{bmatrix}4 & 3\\2 & 5\end{bmatrix}$
$A^2=A.A=\begin{bmatrix}4 & 3\\2 & 5\end{bmatrix}\begin{bmatrix}4 & 3\\2 & 5\end{bmatrix}$
$\Rightarrow \begin{bmatrix}4(4)+3(2) & 4(3)+3(5)\\2(4)+5(2) & 2(3)+5(5)\end{bmatrix}$
$\Rightarrow \begin{bmatrix}16+6 & 12+15\\8+10 & 6+25\end{bmatrix}$
$\Rightarrow \begin{bmatrix}22 & 27\\18& 31\end{bmatrix}$
Step2:
Substitute the value of A
$A^2-xA+yI=0$
$ \begin{bmatrix} 22& 27\\18 & 31\end{bmatrix}-x\begin{bmatrix}4 & 3\\2 & 5\end{bmatrix}+y\begin{bmatrix}1 &0\\0 &1\end{bmatrix}=0$
$ \begin{bmatrix} 22& 27\\18 & 31\end{bmatrix}-\begin{bmatrix}4x & 3x\\2x & 5x\end{bmatrix}+\begin{bmatrix}y &0\\0 &y\end{bmatrix}=0$
$\begin{bmatrix}22-4x+y &27-3x\\18-2x+0 & 31-5x+y\end{bmatrix}=0$
$22-4x+y=0$-----(1)
$27-3x=0$-------(2)
$18-2x=0$-------(3)
$31-5x+y=0$-------(4)
From equation (2) we have
27-3x=0
-3x=-27
x=9.
Step3:
Substitute the value of x in equation (1)
22-4x+y=0.
22-4(9)+y=0.
22-36+y=0.
-14+y=0.
14=y.
Therefore x=9,y=14.
answered Apr 9, 2013 by sharmaaparna1
 

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