Question

If the multiplication of matrices $\left( \begin{matrix}
   2 & 3 \\
   5 & 7 \\
\end{matrix} \right).\left( \begin{matrix}
   1 & -3 \\
   -2 & 4 \\
\end{matrix} \right)$ gives a matrix $\left( \begin{matrix}
   -4 & 6 \\
   -9 & x \\
\end{matrix} \right)$, then find the value of x?

Answer 1

Hint: We recall the procedure of multiplication of matrices before starting to solve the problem. We multiply the given matrices $\left( \begin{matrix}
   2 & 3 \\
   5 & 7 \\
\end{matrix} \right).\left( \begin{matrix}
   1 & -3 \\
   -2 & 4 \\
\end{matrix} \right)$ to proceed through the problem. Once, we find the multiplication of matrices we equate the resultant matrix with $\left( \begin{matrix}
   -4 & 6 \\
   -9 & x \\
\end{matrix} \right)$ by using the equality of matrices theorem to get the required value of ‘x’.

Complete step by step answer:
Given that we have a multiplication of two matrices $\left( \begin{matrix}
   2 & 3 \\
   5 & 7 \\
\end{matrix} \right).\left( \begin{matrix}
   1 & -3 \\
   -2 & 4 \\
\end{matrix} \right)$ which gives the resultant matrix $\left( \begin{matrix}
   -4 & 6 \\
   -9 & x \\
\end{matrix} \right)$. We need to find the value of ‘x’ that was present in the resultant matrix.
We have got the value of $\left( \begin{matrix}
   2 & 3 \\
   5 & 7 \\
\end{matrix} \right).\left( \begin{matrix}
   1 & -3 \\
   -2 & 4 \\
\end{matrix} \right)=\left( \begin{matrix}
   -4 & 6 \\
   -9 & x \\
\end{matrix} \right)$ ---(1).
We know that multiplication of two matrices $\left( \begin{matrix}
   a & b \\
   c & d \\
\end{matrix} \right)$, $\left( \begin{matrix}
   p & q \\
   r & s \\
\end{matrix} \right)$ is defined as $\left( \begin{matrix}
   a & b \\
   c & d \\
\end{matrix} \right)\times \left( \begin{matrix}
   p & q \\
   r & s \\
\end{matrix} \right)=\left( \begin{matrix}
   \left( a\times p \right)+\left( b\times r \right) & \left( a\times q \right)+\left( b\times s \right) \\
   \left( c\times p \right)+\left( d\times r \right) & \left( c\times q \right)+\left( d\times s \right) \\
\end{matrix} \right)$. We use this multiplication rule in the equation (1).
We have got the value of $\left( \begin{matrix}
   \left( 2\times 1 \right)+\left( 3\times -2 \right) & \left( 2\times -3 \right)+\left( 3\times 4 \right) \\
   \left( 5\times 1 \right)+\left( 7\times -2 \right) & \left( 5\times -3 \right)+\left( 7\times 4 \right) \\
\end{matrix} \right)=\left( \begin{matrix}
   -4 & 6 \\
   -9 & x \\
\end{matrix} \right)$.
We have got the value of $\left( \begin{matrix}
   2-6 & -6+12 \\
   5-14 & -15+28 \\
\end{matrix} \right)=\left( \begin{matrix}
   -4 & 6 \\
   -9 & x \\
\end{matrix} \right)$.
We have got the value of $\left( \begin{matrix}
   -4 & 6 \\
   -9 & 13 \\
\end{matrix} \right)=\left( \begin{matrix}
   -4 & 6 \\
   -9 & x \\
\end{matrix} \right)$ ---(2).
We know that according to the equality of the matrices if two matrices are said to be equal then each and every corresponding element in the two matrices are equal. i.e., if two matrices $\left( \begin{matrix}
   a & b \\
   c & d \\
\end{matrix} \right)$, $\left( \begin{matrix}
   p & q \\
   r & s \\
\end{matrix} \right)$ are said to be equal, then a = p, b = q, c = r and d = s. We use this result in equation (2).
So, we need to find the value of ‘x’ and we can see that all other elements are equal in both matrices.
So, we have got the value of x = 13.
We have found the value of x as 13.

∴ The value of x is 13.

Note: We should not make multiplication in matrices like we do addition and subtraction in matrices. We should not equate elements that are correspondingly the same on both matrices that we are equating. Similarly, we can expect problems that involve addition, subtraction in place of multiplication.