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

Hint: 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
\[\begin{align}
  & \left[ \begin{matrix}
   1 & x & 1 \\
\end{matrix} \right]\left[ \begin{matrix}
   1 & 2 & 3 \\
   4 & 5 & 6 \\
   3 & 2 & 5 \\
\end{matrix} \right]\left[ \begin{matrix}
   1 \\
   2 \\
   3 \\
\end{matrix} \right]=O...(1) \\
 & \Rightarrow XYZ=O \\
\end{align}\]
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)
 \[\begin{align}
  & \left[ \begin{matrix}
   1+4x+3 & 2+5x+2 & 3+6x+5 \\
\end{matrix} \right]\left[ \begin{matrix}
   1 \\
   2 \\
   3 \\
\end{matrix} \right]=O. \\
 & \Rightarrow \left[ \begin{matrix}
   4x+4 & 5x+4 & 6x+8 \\
\end{matrix} \right]\left[ \begin{matrix}
   1 \\
   2 \\
   3 \\
\end{matrix} \right]=O \\
\end{align}\]

 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

\[\begin{align}
  & .\left[ \begin{matrix}
   4x+4 & 5x+4 & 6x+8 \\
\end{matrix} \right]\left[ \begin{matrix}
   1 \\
   2 \\
   3 \\
\end{matrix} \right]=O \\
 & \Rightarrow \left[ 4x+4+2\left( 5x+4 \right)+3\left( 6x+8 \right) \right]=O \\
\end{align}\]
We use the condition of equality between matrices,
\[\begin{align}
  & 4x+4+2\left( 5x+4 \right)+3\left( 6x+8 \right)=\left[ 0 \right] \\
 & \Rightarrow 32x+36=0 \\
 & \Rightarrow x=\dfrac{-36}{32}=\dfrac{-9}{8} \\
\end{align}\]

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 $1\times n.$