Question

matrix Find x such that [1 x 1] [1 3 2 2 5 1 15 3 2 ] [ 1 2 x] = 0
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Answer 1

$\large\underline{\sf{Solution-}}$

Given matrix equation is

$\rm :\longmapsto\:\begin{gathered}\sf \left[\begin{array}{ccc}1&x&1\end{array}\right]\end{gathered}\begin{gathered}\sf \left[\begin{array}{ccc}1&3&2\\2&5&1\\ 15&3&2\end{array}\right]\end{gathered}\begin{gathered}\sf\left[\begin{array}{c} 1\\2\\x&\end{array}\right]\end{gathered} = 0$

$\rm :\longmapsto\:\begin{gathered}\sf \left[\begin{array}{ccc}1 + 2x + 15&3 + 5x + 3&2 + x + 2\end{array}\right]\end{gathered}\begin{gathered}\sf\left[\begin{array}{c} 1\\2\\x&\end{array}\right]\end{gathered} = 0$

$\rm :\longmapsto\:\begin{gathered}\sf \left[\begin{array}{ccc}16 + 2x&6 + 5x&4 + x\end{array}\right]\end{gathered}\begin{gathered}\sf\left[\begin{array}{c} 1\\2\\x&\end{array}\right]\end{gathered} = 0$

$\rm :\longmapsto\:[16 + 2x + 2(6 + 5x) + x(4 + x)] = 0$

$\rm :\longmapsto\:[16 + 2x + 12 + 10x + 4x + {x}^{2} ] = 0$

$\rm :\longmapsto\:[ {x}^{2} + 16x + 28] = [0]$

So, on comparing, we get

$\rm :\longmapsto\: {x}^{2} + 16x + 28 = 0$

$\rm :\longmapsto\: {x}^{2} + 14x + 2x+ 28 = 0$

$\rm :\longmapsto\:x(x + 14) + 2(x + 14) = 0$

$\rm :\longmapsto\:(x + 14)(x + 2) = 0$

$\rm :\longmapsto\:x + 14 = 0 \: \: \: or \: \: \: x + 2= 0$

$\bf\implies \:x = - 14 \: \: \: or \: \: \: x = - 2$

Additional Information :-

1. Matrix multiplication of two matrices A and B is defined if number of columns of pre multiplier is equals to number of rows of post multiplier.

2. Matrix multiplication may or may not be Commutative.

3. Matrix multiplication is Associative.

4. Matrix multiplication is Distributive, i.e. A(B + C) = AB + AC