Question

Find the value of $x$ if $ \left[ \begin{matrix}
1 & x & 1 \\
\end{matrix} \right]\left[ \begin{matrix}
1 & 2 & 3 \\
4 & 5 & 6 \\
3 & 2 & 5 \\
\end{matrix} \right]\left[ \begin{matrix}
1 \\
2 \\
3 \\
\end{matrix} \right]=O $$$$$

Answer 1

Name the three multiplied matrices on the left hand side as $XYZ$. First multiply $X\times Y$ and then $XY\times Z$. Use the condition of equality between matrices to obtain a linear equation in $x$. Solve it to find the value of $x$.

Complete step by step answer:
The given equation with matrices involving an unknown $x$ is

& \left[ \begin{matrix} 1 & x & 1 \\\ \end{matrix} \right]\left[ \begin{matrix} 1 & 2 & 3 \\\ 4 & 5 & 6 \\\ 3 & 2 & 5 \\\ \end{matrix} \right]\left[ \begin{matrix} 1 \\\ 2 \\\ 3 \\\ \end{matrix} \right]=O...(1) \\\ & \Rightarrow XYZ=O \\\ \end{aligned}$$ Where $X=\left[ \begin{matrix} 1 & x & 1 \\\ \end{matrix} \right],Y=\left[ \begin{matrix} 1 & 2 & 3 \\\ 4 & 5 & 6 \\\ 3 & 2 & 5 \\\ \end{matrix} \right],Z=\left[ \begin{matrix} 1 \\\ 2 \\\ 3 \\\ \end{matrix} \right]$ and $O$ is zero matrix with all the entries as zeros. $$$$ We know from the multiplication of matrices that two matrices $A,B$ can only be multiplied when they have compatible order. The order of a matrix is defined as $m\times n$ where $m$ is the number of rows and $n$ is the number of column. So let $A,B$ be a matrix order $m\times n$and $A,B$ be matrix of order $q\times r$. The order of $A$ and $B$ are compatible if number columns of first matrix is equal to the number of rows of second matrix in the order of multiplication. In symbols $n=q$. The order of the product matrix is $m\times r$ $$$$ Let be $A=\mathop{\left[ {{a}_{ij}} \right]}_{i=1,j=1}^{i=m,j=n}$ be the first matrix with entries ${{a}_{ij}}$ and $B=\mathop{\left[ {{b}_{ij}} \right]}_{i=1,j=1}^{i=q,j=r}$ be the second matrix with entries ${{b}_{ij}}$. We denote their product as $C=AB$. Then the entries of $C$ are given by $$\mathop{\left[ {{c}_{ij}} \right]}_{i=1,j=1}^{i=m,j=r}={{a}_{i1}}{{b}_{1j}}+{{a}_{i2}}{{b}_{2j}}+...+{{a}_{in}}{{b}_{nj}}$$ Where $i$ is the row and $j$ is the column of the entry.$$$$ The matrices $A$ and $B$ are equal if and only if they are of same order and ${{a}_{ij}}={{b}_{ij}}$. $$$$ As the matrix $X$is of the order $1\times 3$ and the matrix $Y$ is of the order $3\times 3$. Then $XY$ will have the order $1\times 3$. Let us multiply the first two matrices row by column in the equation (1) $$\begin{aligned} & \left[ \begin{matrix} 1+4x+3 & 2+5x+2 & 3+6x+5 \\\ \end{matrix} \right]\left[ \begin{matrix} 1 \\\ 2 \\\ 3 \\\ \end{matrix} \right]=O. \\\ & \Rightarrow \left[ \begin{matrix} 4x+4 & 5x+4 & 6x+8 \\\ \end{matrix} \right]\left[ \begin{matrix} 1 \\\ 2 \\\ 3 \\\ \end{matrix} \right]=O \\\ \end{aligned}$$ Now shall multiply $XY$ and $Z$. The order of $XY$ is is $1\times 3$ and the order of $Z$ is $3\times 1$.So the order of $XYZ$ will be $1\times 1$. Let us multiply they row by column $$\begin{aligned} & .\left[ \begin{matrix} 4x+4 & 5x+4 & 6x+8 \\\ \end{matrix} \right]\left[ \begin{matrix} 1 \\\ 2 \\\ 3 \\\ \end{matrix} \right]=O \\\ & \Rightarrow \left[ 4x+4+2\left( 5x+4 \right)+3\left( 6x+8 \right) \right]=O \\\ \end{aligned}$$ We use the condition of equality between matrices, $$\begin{aligned} & 4x+4+2\left( 5x+4 \right)+3\left( 6x+8 \right)=\left[ 0 \right] \\\ & \Rightarrow 32x+36=0 \\\ & \Rightarrow x=\dfrac{-36}{32}=\dfrac{-9}{8} \\\ \end{aligned}$$ **So the value of $x$ is $\dfrac{-9}{8}$. $$$$** **Note:** It is to be noted that matrix multiplication is not a commutative or associative operation that means matrix can only be multiplied when they are in particular order. The matrix $X$ is a row matrix or row vector. The matrix $Y$ is a column matrix or column vector. A vector is a matrix with dimension $n\times 1$ or $1\times n.$