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Home  >>  CBSE XII  >>  Math  >>  Matrices
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If $ A = \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix} $ and $ I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} $ find $k$ so that $A^2 = kA - 2I$

$\begin{array}{1 1} k=1\\ k=2 \\ k=-1 \\ k=0 \end{array} $

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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.
  • We can then match the corresponding elements and solve the resulting equations to find the values of the unknown variables.
Given:$A=\begin{bmatrix}3 & -2\\4 & -2\end{bmatrix}$
$A^2=A\times A=\begin{bmatrix}3 & -2\\4 & -2\end{bmatrix}\begin{bmatrix}3 & -2\\4 & -2\end{bmatrix}$
$\Rightarrow \begin{bmatrix}9-8 & -6+4\\12-8 & -8+4\end{bmatrix}$
$\Rightarrow \begin{bmatrix}1 & -2\\4 & -4\end{bmatrix}$
$A^2=kA-2I.$
$ \begin{bmatrix}1 & -2\\4 & -4\end{bmatrix}=k\begin{bmatrix}3 & -2\\4 & -2\end{bmatrix}-2\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}$
$ \begin{bmatrix}1 & -2\\4 & -4\end{bmatrix}=\begin{bmatrix}3k & -2k\\4k & -2k\end{bmatrix}+(-1)\begin{bmatrix}2 & 0\\0 & 2\end{bmatrix}$
$\;\;\;\;=\begin{bmatrix}3k & -2k\\4k & -2k\end{bmatrix}+\begin{bmatrix}-2 & 0\\0 & -2\end{bmatrix}$
$\;\;\;\;=\begin{bmatrix}3k-2 & -2k+0\\4k+0 & -2k-2\end{bmatrix}$
The above given two matrices are equal hence there corresponding elements are equal.
3K-2=1-----(1)
-2k+0=-2----(2)
4k+0=4-----(3)
-2k-2=-4-----(4)
From eq(1)
3k-2=1
3k=1+2
3k=3
k=1
From eq(2)
-2k=-2
k=1
From eq(3)
4k=4
k=1.
From eq(4)
-2k-2=-4
-2k=-4+2
-2k=-2
k=1
Hence k=1.
answered Mar 4, 2013 by sharmaaparna1
 

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