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Home  >>  CBSE XII  >>  Math  >>  Matrices
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Find $x$, if $ \begin{bmatrix} x & -5 & -1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix} \begin{bmatrix} x \\ 4 \\ 1 \end{bmatrix} = 0 $

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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: $\begin{bmatrix}AB\end{bmatrix}_{i,j} = A_{i,1}B_{1,j} + A_{i,2}B_{2,j} + A_{i,3}B_{3,j} ... A_{i,n}B_{n,j}$
Step 1: LHS:
$\begin{bmatrix} x & -5 & -1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix} \begin{bmatrix} x \\ 4 \\ 1 \end{bmatrix} $
$\;\;\;=\begin{bmatrix} x & -5 & -1 \end{bmatrix} \begin{bmatrix} x+0+2 \\ 0+8+1\\ 2 x+0+3 \end{bmatrix} $
$\;\;\;=\begin{bmatrix} x & -5 & -1 \end{bmatrix} \begin{bmatrix} x+2 \\ 9\\ 2 x+3 \end{bmatrix} $
$\;\;\;=x(x+2)+(-5\times 9)+(-1)(2x+3)$
$\;\;\;=x^2+2x-45-2x-3$
$\;\;\;=x^2-48$
Step 2: Now equate the above equation of the LHS to that of the RHS
$x^2-48=0$
$x^2=48$
$x=\sqrt {48}$
$x=\sqrt {16\times 3}$
$x=4\sqrt {3}$
$x=\pm 4\sqrt 3$.
answered Mar 16, 2013 by sharmaaparna1
edited Mar 19, 2013 by sreemathi.v
 

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