Question

What is the value of x for which $\left[ {1{\text{ 2 1}}} \right]\left( {\begin{array}{*{20}{c}}
  1&2&0 \\
  2&0&1 \\
  1&0&2
\end{array}} \right)\left[ \begin{gathered}
  0 \\
  2 \\
  x \\
\end{gathered} \right] = 0$?

Answer 1

Hint- Here, we will proceed by firstly checking whether the multiplication of the matrices can occur or not by using the condition for the multiplication of two matrices and then using the algebra of multiplication of matrices to solve, the given equation in order to obtain value of x.

Complete step-by-step answer:
The given equation is $\left[ {1{\text{ 2 1}}} \right]\left( {\begin{array}{*{20}{c}}
  1&2&0 \\
  2&0&1 \\
  1&0&2
\end{array}} \right)\left[ \begin{gathered}
  0 \\
  2 \\
  x \\
\end{gathered} \right] = 0{\text{ }} \to {\text{(1)}}$
As we know that for the multiplication AB of any two matrices A and B having orders $a \times b$ and $c \times d$ respectively to exist, the number of columns of the first matrix should be equal to the number of rows of the second matrix i.e., b=c.
Consider matrix A = $\left[ {1{\text{ 2 1}}} \right]$ which has an order of $1 \times 3$ and matrix B = $\left( {\begin{array}{*{20}{c}}
  1&2&0 \\
  2&0&1 \\
  1&0&2
\end{array}} \right)$ which has an order of $3 \times 3$ and matrix C = $\left[ \begin{gathered}
  0 \\
  2 \\
  x \\
\end{gathered} \right]$ which has an order of $1 \times 3$.
Now, the given equation (1) becomes
$ \Rightarrow {\text{ABC}} = 0 \to {\text{(2)}}$
The multiplication matrix AB obtained by multiplying two matrices A and B together exists because the number of columns in matrix A (i.e., 3) is equal to the number of rows in matrix B (i.e., 3) and is given by
$
  {\text{AB}} = \left[ {1{\text{ 2 1}}} \right]\left( {\begin{array}{*{20}{c}}
  1&2&0 \\
  2&0&1 \\
  1&0&2
\end{array}} \right) \\
   \Rightarrow {\text{AB}} = \left[ {\left( {1 \times 1} \right) + \left( {2 \times 2} \right) + \left( {0 \times 1} \right){\text{ }}\left( {1 \times 2} \right) + \left( {2 \times 0} \right) + \left( {1 \times 1} \right){\text{ }}\left( {1 \times 1} \right) + \left( {2 \times 0} \right) + \left( {1 \times 2} \right)} \right] \\
   \Rightarrow {\text{AB}} = \left[ {{\text{5 3 3}}} \right] \\
 $
Also, the multiplication matrix ABC obtained by multiplying two matrices AB and C together exists because the number of columns in matrix AB (i.e., 3) is equal to the number of rows in matrix C (i.e., 3).
Now, equation (2) becomes
\[
   \Rightarrow \left[ {{\text{5 3 3}}} \right]{\text{C}} = 0 \\
   \Rightarrow \left[ {{\text{5 3 3}}} \right]\left[ \begin{gathered}
  0 \\
  2 \\
  x \\
\end{gathered} \right] = 0 \\
   \Rightarrow 5 \times 0 + 3 \times 2 + 3 \times x = 0 \\
   \Rightarrow 3x = - 6 \\
   \Rightarrow x = - 2 \\
 \]
Hence, x=-2 is the required value of x which will satisfy the given equation.

Note- The order of any matrix is denoted by $a \times b$ where a is the number of rows in the matrix and b is the number of columns in the matrix. Also, when two matrices A and B (having orders $a \times b$ and $c \times d$ respectively) are multiplied with each other the resultant matrix obtained AB will have order of $a \times d$ where b = c.