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Please solve RD Sharma class 12 chapter Algebra of matrices exercise 4.3 question 25 maths textbook solution

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Answer:

x=-1 or x=-2

Hint: Matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.

Given: \left[\begin{array}{lll}x & 4 & 1\end{array}\right]\left[\begin{array}{ccc}2 & 1 & 2 \\ 1 & 0 & 2 \\ 0 & 2 & -4\end{array}\right]\left[\begin{array}{c}x \\ 4 \\ -1\end{array}\right]=0

Firstly, we multiply first two matrices

\Rightarrow[2 x+4+0 \quad x+0+2 \quad 2 x+8-4]\left[\begin{array}{c}x \\ 4 \\ -1\end{array}\right]=0 \\\\ \Rightarrow\left[\begin{array}{lll}2 x+4 & x+2 & 2 x+4\end{array}\right]\left[\begin{array}{c}x \\ 4 \\ -1\end{array}\right]=0\\\\ \Rightarrow[(2 x+4) x+4(x+2)+(-1)(2 x+4)]=0

\Rightarrow 2 x^{2}+4 x+4 x+8-2 x-4=0 \\\\ \Rightarrow 2 x^{2}+6 x+4=0 \\\\ \Rightarrow 2 x^{2}+2 x+4 x+4=0 \\\\ \Rightarrow 2 x(x+1)+4(x+1)=0 \\\\ \Rightarrow(x+1)(2 x+4)=0 \\ \\

Either x+1=0      or         \quad 2 x+4=0

\Rightarrow x=-1 \quad \text { or } \quad 2 x=-4 \Rightarrow x=-2

Hence, x=-1 \text { or }-2

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