Question

Find the value of x, such that x is a real number, if $\left[ \begin{matrix} x & -5 & -1 \\\ \end{matrix} \right]\left[ \begin{matrix} 1 & 0 & 2 \\\ 0 & 2 & 1 \\\ 2 & 0 & 3 \\\ \end{matrix} \right]\left[ \begin{matrix} x \\\ 4 \\\ 1 \\\ \end{matrix} \right]=0$ .

Answer 1

Hint:In this question, we will use the multiplication rule of the matrix to evaluate the expression on the left hand side and use equality of the matrix to find value of x.

Complete step-by-step answer:
In a given question, we will first multiply matrices given on the left hand side of the equation.
Let us first multiply matrix $\left[ \begin{matrix} 1 & 0 & 2 \\\ 0 & 2 & 1 \\\ 2 & 0 & 3 \\\ \end{matrix} \right]$ with $\left[ \begin{matrix} x \\\ 4 \\\ 1 \\\ \end{matrix} \right]$
For this, we will multiply elements of the first row of first matrix with corresponding elements of the first column of second matrix and add them up. We will then repeat the same process with the second row and put value in the second element of the product matrix. And lastly with third rows and put value in the third element of product matrix.
Hence, our product for these two matrices will be $\left[ \begin{matrix} 1 & 0 & 2 \\\ 0 & 2 & 1 \\\ 2 & 0 & 3 \\\ \end{matrix} \right]\left[ \begin{matrix} x \\\ 4 \\\ 1 \\\ \end{matrix} \right]=0$

& =\left[ \begin{matrix} 1\times x+0\times 4+2\times 1 \\\ 0\times x+2\times 4+1\times 1 \\\ 2\times x+0\times 4+3\times 1 \\\ \end{matrix} \right] \\\ & =\left[ \begin{matrix} x+0+2 \\\ 0+8+1 \\\ 2x+0+3 \\\ \end{matrix} \right] \\\ & =\left[ \begin{matrix} x+2 \\\ 9 \\\ 2x+3 \\\ \end{matrix} \right] \\\ \end{aligned}$$ Now, we will multiply the matrix $\left[ \begin{matrix} x & -5 & -1 \\\ \end{matrix} \right]$ with matrix given above, we get, $$\begin{aligned} & \left[ \begin{matrix} x & -5 & -1 \\\ \end{matrix} \right]\left[ \begin{matrix} x+2 \\\ 9 \\\ 2x+3 \\\ \end{matrix} \right] \\\ & =\left[ x\left( x+2 \right)+\left( -5 \right)\times 9+\left( -1 \right)\times \left( 2x+3 \right) \right] \\\ \end{aligned}$$ Applying distributive law, we get, $$\begin{aligned} & {{x}^{2}}+2x-45-2x-3 \\\ & ={{x}^{2}}-49 \\\ \end{aligned}$$ Now, comparing this value, which is value of left hand side of given equation with value of right hand side of the equation, which is zero, we get, ${{x}^{2}}-49=0$ Add 49 on both sides of the equation. We get, ${{x}^{2}}=49$ . Taking root on both sides of the equation, we get, $\begin{aligned} & \sqrt{{{x}^{2}}}=\pm \sqrt{49} \\\ & \Rightarrow x=\pm 7 \\\ \end{aligned}$ $\Rightarrow x=7$ or $-7$ Hence, the value of x in the given equation is $\pm 7$ . Note: In this type of question, before starting the multiplication of matrices, check the order of matrices and see if both matrices are multiplied or not.Two matrices can be multiplied if the number of columns in the first matrix is equal to the number of rows in the second one,then matrix multiplication is possible.