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Home  >>  CBSE XII  >>  Math  >>  Matrices
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if $ A = \begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix} ,$ prove that $ A^3 - 6A^2 + 7A + 2I = 0 $

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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 $c$.
  • 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.
  • $I=\begin{bmatrix}1 & 0& 0\\0 & 1 & 0\\0& 0 & 1\end{bmatrix}$
We need to prove that $LHS: A^3−6A^2+7A+2I= RHS: 0$.
$A^2=A\times A=\begin{bmatrix}1 & 0 & 2\\0 & 2 & 1\\2 & 0& 3\end{bmatrix}\begin{bmatrix}1 & 0 & 2\\0 & 2 & 1\\2 & 0& 3\end{bmatrix}$
$A^2 = \begin{bmatrix}1+0+4 & 0+0+0 & 2+0+6\\0+0+2 & 0+4+0 & 0+2+3\\2+0+6 & 0+0+0& 4+0+9\end{bmatrix}$
$A^2 = \begin{bmatrix}5 & 0 & 8\\2 & 4 & 5\\8 & 0& 13\end{bmatrix}$
$-6A^2=(-)6 \begin{bmatrix}5 & 0 & 8\\2 & 4 & 5\\8 & 0& 13\end{bmatrix} = \begin{bmatrix}-30 & 0 & -48\\-12 & -24 &-30 \\-48 & 0& -78\end{bmatrix}$
$A^3=A^2\times A.$
$A^3=A^2\times A\Rightarrow \begin{bmatrix}5 & 0 & 8\\2 & 4 & 5\\8 & 0& 13\end{bmatrix}\begin{bmatrix}1 & 0 & 2\\0 & 2 & 1\\2 & 0& 3\end{bmatrix}$
$A^3 = \begin{bmatrix}5+0+16 & 0+0+0 & 10+0+24\\2+0+10 & 0+8+0 & 4+4+15\\8+0+26 & 0+0+0& 16+0+39\end{bmatrix}$
$A^3 = \begin{bmatrix}21 & 0 & 34\\12 & 8 & 23\\34 & 0& 55\end{bmatrix}$
$7A = 7\begin{bmatrix}1 & 0 & 2\\0 & 2 & 1\\2 & 0& 3\end{bmatrix} = \begin{bmatrix}7 & 0 & 14\\0 & 14 & 7\\14 & 0& 21\end{bmatrix}$
$2I = 2\begin{bmatrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0& 1\end{bmatrix} = \begin{bmatrix}2 & 0 & 0\\0 & 2 & 0\\0 & 0& 2\end{bmatrix}$
Therefore, evaluating $LHS: A^3−6A^2+7A+2I$:
$LHS = \begin{bmatrix}21 & 0 & 34\\12 & 8 & 23\\34 & 0& 55\end{bmatrix}+\begin{bmatrix}-30 & 0 & -48\\-12 & -24 & -30\\-48 & 0& -78\end{bmatrix}+\begin{bmatrix}7 & 0 & 14\\0 & 14 & 7\\14 & 0& 21\end{bmatrix}+\begin{bmatrix}2 & 0 & 0\\0 & 2 & 0\\0 & 0& 2\end{bmatrix}$
$LHS = \begin{bmatrix}21-30+7+2 & 0+0+0+0 & 34-48+14+0\\12-12+0+0 & 8-24+14+2 & 23-30+7+0\\34-48+14+0 & 0+0+0+0& 55-78+21+2\end{bmatrix}$
$LHS = \begin{bmatrix}0 & 0 & 0\\0 & 0 & 0\\0& 0& 0\end{bmatrix}=0_{3\times 3} = RHS.$
answered Feb 13, 2013 by sreemathi.v
edited Feb 27, 2013 by balaji.thirumalai
 

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