Find the value of x, y and z if
\[\left\{ 5\left[ \begin{matrix}
   1 \\
   0 \\
   1 \\
\end{matrix}\text{ }\begin{matrix}
   0 \\
   1 \\
   1 \\
\end{matrix} \right]-3\left[ \begin{matrix}
   1 \\
   -2 \\
   3 \\
\end{matrix}\text{ }\begin{matrix}
   2 \\
   3 \\
   1 \\
\end{matrix} \right] \right\}\left[ \begin{matrix}
   2 \\
   1 \\
\end{matrix} \right]=\left[ \begin{matrix}
   x+1 \\
   y-1 \\
   2z \\
\end{matrix} \right]\]

Hint: To find value of x, y and z we will compare matrices at LHS and RHS of above expression. To do so, first we will arrange the LHS by making it a single matrix. 5 and 3 can be multiplied inside using $k\left[ \begin{matrix}
   a & b \\
   c & d \\
\end{matrix} \right]=\left[ \begin{matrix}
   ka & kb \\
   kc & kd \\
\end{matrix} \right]$ and then both matrices can be added or subtracted term wise to obtain a single matrix. Then, we will use matrix multiplication to finally form a last matrix at LHS then comparing to RHS given is result.

Complete step by step answer:
We will first open the matrix, we have the LHS of given expression as
\[\left\{ 5\left[ \begin{matrix}
   1 \\
   0 \\
   1 \\
\end{matrix}\text{ }\begin{matrix}
   0 \\
   1 \\
   1 \\
\end{matrix} \right]-3\left[ \begin{matrix}
   1 \\
   -2 \\
   3 \\
\end{matrix}\text{ }\begin{matrix}
   2 \\
   3 \\
   1 \\
\end{matrix} \right] \right\}\left[ \begin{matrix}
   2 \\
   1 \\
\end{matrix} \right]\]
Now, matrix multiplication by scalar is done as given below:
\[k\left[ \begin{matrix}
   a & b \\
   c & d \\
\end{matrix} \right]=\left[ \begin{matrix}
   ka & kb \\
   kc & kd \\
\end{matrix} \right]\]
Where $\left[ \begin{matrix}
   a & b \\
   c & d \\
\end{matrix} \right]$ is a matrix and k is a scalar.
That is, scalar is multiplied to each term of matrix.
Applying this in above matrix where 5 is scalar of matrix $\left[ \begin{matrix}
   1 \\
   0 \\
   1 \\
\end{matrix}\text{ }\begin{matrix}
   0 \\
   1 \\
   1 \\
\end{matrix} \right]$ and +3 is scalar of matrix $\left[ \begin{matrix}
   1 \\
   -2 \\
   3 \\
\end{matrix}\text{ }\begin{matrix}
   2 \\
   3 \\
   1 \\
\end{matrix} \right]$ we get:
\[\begin{align}
  & \text{LHS}=\left\{ 5\left[ \begin{matrix}
   1 \\
   0 \\
   1 \\
\end{matrix}\text{ }\begin{matrix}
   0 \\
   1 \\
   1 \\
\end{matrix} \right]-\left( 3 \right)\left[ \begin{matrix}
   1 \\
   -2 \\
   3 \\
\end{matrix}\text{ }\begin{matrix}
   2 \\
   3 \\
   1 \\
\end{matrix} \right] \right\}\left[ \begin{matrix}
   2 \\
   1 \\
\end{matrix} \right] \\
 & \text{LHS}=\left\{ \left[ \begin{matrix}
   5 \\
   0 \\
   5 \\
\end{matrix}\text{ }\begin{matrix}
   0 \\
   5 \\
   5 \\
\end{matrix} \right]-\left[ \begin{matrix}
   3 \\
   -6 \\
   9 \\
\end{matrix}\text{ }\begin{matrix}
   6 \\
   9 \\
   1 \\
\end{matrix} \right] \right\}\left[ \begin{matrix}
   2 \\
   1 \\
\end{matrix} \right] \\
\end{align}\]
Now, addition or subtraction is done term wise as given below:
\[\left[ \begin{matrix}
   a & b \\
   c & d \\
\end{matrix} \right]+\left[ \begin{matrix}
   e & f \\
   g & h \\
\end{matrix} \right]=\left[ \begin{matrix}
   a+e & b+f \\
   c+g & d+h \\
\end{matrix} \right]\]
Where $\left[ \begin{matrix}
   a & b \\
   c & d \\
\end{matrix} \right]\text{ and }\left[ \begin{matrix}
   e & f \\
   g & h \\
\end{matrix} \right]$ are matrices of $2\times 2$ order.
Applying this in above we get:
\[\begin{align}
  & \text{LHS}=\left\{ \left[ \begin{matrix}
   5-3 \\
   0+6 \\
   5-9 \\
\end{matrix}\text{ }\begin{matrix}
   0-6 \\
   5-9 \\
   5-1 \\
\end{matrix} \right] \right\}\left[ \begin{matrix}
   2 \\
   1 \\
\end{matrix} \right] \\
 & \text{LHS}=\left\{ \left[ \begin{matrix}
   2 \\
   6 \\
   -4 \\
\end{matrix}\text{ }\begin{matrix}
   -6 \\
   -4 \\
   4 \\
\end{matrix} \right] \right\}\left[ \begin{matrix}
   2 \\
   1 \\
\end{matrix} \right] \\
\end{align}\]
So, finally we are left with multiplication of above obtained matrix.
Matrix multiplication is done as below:
\[\left[ \begin{matrix}
   a & b \\
   c & d \\
\end{matrix} \right]\cdot \left[ \begin{matrix}
   e & f \\
   g & h \\
\end{matrix} \right]=\left[ \begin{matrix}
   ae+bg & af+bh \\
   ce+dg & cf+dh \\
\end{matrix} \right]\]
That is, multiplying the term wise of a row with corresponding column of the second matrix and adding them.
Here $\left[ \begin{matrix}
   a & b \\
   c & d \\
\end{matrix} \right]\text{ and }\left[ \begin{matrix}
   e & f \\
   g & h \\
\end{matrix} \right]$ are matrices of $2\times 2$ order.
Applying this matrix multiplication method to our question, we get:
\[\begin{align}
  & \text{LHS}=\left[ \begin{matrix}
   2 \\
   6 \\
   -4 \\
\end{matrix}\text{ }\begin{matrix}
   -6 \\
   -4 \\
   4 \\
\end{matrix} \right]\left[ \begin{matrix}
   2 \\
   1 \\
\end{matrix} \right] \\
 & \text{LHS}=\left[ \begin{matrix}
   2\times 2+(-6)\times 1 \\
   6\times 2+\left( -4 \right)\times 1 \\
   -4\times 2+4\times 1 \\
\end{matrix} \right] \\
 & \text{LHS}=\left[ \begin{matrix}
   -2 \\
   8 \\
   -4 \\
\end{matrix} \right] \\
\end{align}\]
Comparing this obtained LHS by our given RHS we get:
\[\left[ \begin{matrix}
   -2 \\
   8 \\
   -4 \\
\end{matrix} \right]=\left[ \begin{matrix}
   x+1 \\
   y-1 \\
   2z \\
\end{matrix} \right]=\left[ \begin{matrix}
   -2 \\
   8 \\
   -4 \\
\end{matrix} \right]\]
Comparing each term we get:
\[\begin{align}
  & -2=x+1,8=y-1\text{ and }-4=2z \\
 & \Rightarrow x+1=-2 \\
 & x=-2-1 \\
 & x=-3 \\
 & \Rightarrow y-1=8 \\
 & y=8+1 \\
 & y=9 \\
 & \Rightarrow 2z=-4 \\
 & z=\dfrac{-4}{2}=-2 \\
\end{align}\]

Hence, value of x = -3, y = 9 and z = -2

Note: The key point to note here is that, the above given are matrices not determinants. Matrices are represented as $\left[ {} \right]$ and determinants as $\left| {} \right|$ If any scalar term is multiplied then it is multiplied to each and every term of a matrix, but if it is a determinant then that scalar is only multiplied to one row or column and not all. Here, we have a matrix, so we have multiplied 5 and 3 to each and every element inside.