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Home  >>  CBSE XII  >>  Math  >>  Matrices
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Find $ A^2 -5A + 6I $ , if $ A = \begin{bmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{bmatrix} $

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Toolbox:
  • Multiplication of two matrices is defined only if the number of columns of the left matrix is the same as the number of rows of the right matrix.
  • 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 scalar multiplication $cA$ of a matrix $A$ and a number $c$ (also called a scalar in the parlance of abstract algebra) is given by multiplying every entry of $A$ by $c$.
  • 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.
  • 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.
  • $I=\begin{bmatrix}1 & 0& 0\\0 & 1 & 0\\0& 0 & 1\end{bmatrix}$
Consider $A^2-5A+6I$.
$A^2=A\times A\Rightarrow \begin{bmatrix}2 & 0& 1\\2 & 1 & 3\\1& -1 & 0\end{bmatrix}\begin{bmatrix}2 & 0& 1\\2 & 1 & 3\\1& -1 & 0\end{bmatrix}$
 
$\Rightarrow A^2 = \begin{bmatrix}2\times 2+0\times 2+1\times 1 & 2\times 0+0\times 1+1\times (-1)& 2\times 1+0\times 3+1\times 01\\2\times 2+1\times 2+3\times 1 & 2\times 0+1\times 1+3\times (-1) & 2\times 1+1\times 3+3\times 0\\1\times 2+(-1)\times 2+0\times 1& 1\times 0+(-1)\times 1+0\times (-1) & 1\times 1+(-1)\times 3+0\times 0\end{bmatrix}$
$A^2 = \begin{bmatrix}5 & -1& 2\\9 & -2 & 5\\0& -1 & -2\end{bmatrix}$
$-5A = (-1)(5A) = (-1)(5)\begin{bmatrix}2 & 0& 1\\2 & 1 & 3\\1& -1 & 0\end{bmatrix}$
$-5A = \begin{bmatrix}-10 & 0& -5\\-10 & -5 & -15\\-5& 5 & 0\end{bmatrix}$
Since we know that $I=\begin{bmatrix}1 & 0& 0\\0 & 1 & 0\\0& 0 & 1\end{bmatrix}$, $6I = 6\begin{bmatrix}1 & 0& 0\\0 & 1 & 0\\0& 0 & 1\end{bmatrix}$
$6I = \begin{bmatrix}6 & 0& 0\\0 & 6 & 0\\0& 0 & 6\end{bmatrix}$
Therefore, $A^2-5A+6I=\begin{bmatrix}5 & -1& 2\\9 & -2 & 5\\0& -1 & -2\end{bmatrix}+\begin{bmatrix}-10 & 0& -5\\-10 & -5 & -15\\-5& 5 & 0\end{bmatrix}+\begin{bmatrix}6 & 0& 0\\0 & 6 & 0\\0& 0 & 6\end{bmatrix}$:
$\Rightarrow A^2-5A+6I = \begin{bmatrix}5-10+6 & -1+0+0& 2-5+0\\9-10+0 & -2-5+6 & 5-15+0\\0-5+0& -1+5+0 & -2+0+6\end{bmatrix}$
$\Rightarrow A^2-5A+6I = \begin{bmatrix}1 & -1& 3\\-1 & -1 & -10\\-5& 4 & 4\end{bmatrix}$

 

answered Feb 13, 2013 by sreemathi.v
edited Feb 27, 2013 by balaji.thirumalai
 

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