Question

Find the value of $x+y$ from the following equation:
$2\left( \begin{matrix}
   1 & 3 \\
   0 & x \\
\end{matrix} \right)+\left( \begin{matrix}
   y & 0 \\
   1 & 2 \\
\end{matrix} \right)=\left( \begin{matrix}
   5 & 6 \\
   1 & 8 \\
\end{matrix} \right)$

Answer 1

Hint: In the given matrix equation, solve the left hand side of the equation. To solve the left hand side of the equation, we have to multiply the first matrix by 2. Matrix multiplication is done by multiplying each element of the matrix by 2. Then add the elements of this resulting matrix with the elements of the second matrix present on the L.H.S. Then equate each element of the matrix on the left hand side of the equation to the corresponding elements on the right hand side of the equation. Here “corresponding elements” mean the element which is at first row and first column of the left hand side is equal to the elements which are at first row and first column of the matrix on the right hand side.

Complete step by step answer:
The matrix equation given in the above problem is:
$2\left( \begin{matrix}
   1 & 3 \\
   0 & x \\
\end{matrix} \right)+\left( \begin{matrix}
   y & 0 \\
   1 & 2 \\
\end{matrix} \right)=\left( \begin{matrix}
   5 & 6 \\
   1 & 8 \\
\end{matrix} \right)$
In this equation, we have to find the value of $x+y$.
First of all, let us solve the matrix expression written on the left hand side of the above equation.
$2\left( \begin{matrix}
   1 & 3 \\
   0 & x \\
\end{matrix} \right)+\left( \begin{matrix}
   y & 0 \\
   1 & 2 \\
\end{matrix} \right)$
Multiplying 2 to the first matrix is done by multiplying each element of the matrix by 2.
$\begin{align}
  & \left( \begin{matrix}
   1\left( 2 \right) & 3\left( 2 \right) \\
   0\left( 2 \right) & x\left( 2 \right) \\
\end{matrix} \right)+\left( \begin{matrix}
   y & 0 \\
   1 & 2 \\
\end{matrix} \right) \\
 & =\left( \begin{matrix}
   2 & 6 \\
   0 & 2x \\
\end{matrix} \right)+\left( \begin{matrix}
   y & 0 \\
   1 & 2 \\
\end{matrix} \right) \\
\end{align}$
Now, adding the above two matrices by adding each element of the first matrix to the corresponding elements of the second matrix.
$\begin{align}
  & \left( \begin{matrix}
   2+y & 6+0 \\
   0+1 & 2x+2 \\
\end{matrix} \right) \\
 & =\left( \begin{matrix}
   2+y & 6 \\
   1 & 2x+2 \\
\end{matrix} \right) \\
\end{align}$
Now, equating the above matrix to the R.H.S of the matrix equation given in the problem and we get,
$\left( \begin{matrix}
   2+y & 6 \\
   1 & 2x+2 \\
\end{matrix} \right)=\left( \begin{matrix}
   5 & 6 \\
   1 & 8 \\
\end{matrix} \right)$
As the above two matrices are equal to each other so the elements of the first matrix are equal to their corresponding elements.
$\begin{align}
  & 2+y=5; \\
 & 2x+2=8 \\
\end{align}$
Solving above two equations we get,
$2+y=5$
Subtracting 2 on both the sides we get,
$y=5-2=3$
Hence, we have got the value of y as 3.
$2x+2=8$
Subtracting 2 on both the sides we get,
$\begin{align}
  & 2x=8-2=6 \\
 & \Rightarrow 2x=6 \\
 & \Rightarrow x=\dfrac{6}{2}=3 \\
\end{align}$
Hence, we have got the value of x as 3.
Now, we are asked to find the value of $x+y$ so adding 3 with 3 we get 6.

Hence, we have the value of $x+y$ as 6.

Note: You can verify the values of x and y that we have got above by substituting in the matrix equation and see whether these values of x and y are satisfying the matrix equation or not.
We have got the value of x and y as 3 and 3 respectively.
The given matrix equation is:
$2\left( \begin{matrix}
   1 & 3 \\
   0 & x \\
\end{matrix} \right)+\left( \begin{matrix}
   y & 0 \\
   1 & 2 \\
\end{matrix} \right)=\left( \begin{matrix}
   5 & 6 \\
   1 & 8 \\
\end{matrix} \right)$
Substituting x and y as 3 in the above equation we get,
$\begin{align}
  & 2\left( \begin{matrix}
   1 & 3 \\
   0 & 3 \\
\end{matrix} \right)+\left( \begin{matrix}
   3 & 0 \\
   1 & 2 \\
\end{matrix} \right)=\left( \begin{matrix}
   5 & 6 \\
   1 & 8 \\
\end{matrix} \right) \\
 & \Rightarrow \left( \begin{matrix}
   1\left( 2 \right) & 3\left( 2 \right) \\
   0\left( 2 \right) & 3\left( 2 \right) \\
\end{matrix} \right)+\left( \begin{matrix}
   3 & 0 \\
   1 & 2 \\
\end{matrix} \right)=\left( \begin{matrix}
   5 & 6 \\
   1 & 8 \\
\end{matrix} \right) \\
 & \Rightarrow \left( \begin{matrix}
   2 & 6 \\
   0 & 6 \\
\end{matrix} \right)+\left( \begin{matrix}
   3 & 0 \\
   1 & 2 \\
\end{matrix} \right)=\left( \begin{matrix}
   5 & 6 \\
   1 & 8 \\
\end{matrix} \right) \\
 & \Rightarrow \left( \begin{matrix}
   2+3 & 6+0 \\
   0+1 & 6+2 \\
\end{matrix} \right)=\left( \begin{matrix}
   5 & 6 \\
   1 & 8 \\
\end{matrix} \right) \\
 & \Rightarrow \left( \begin{matrix}
   5 & 6 \\
   1 & 8 \\
\end{matrix} \right)=\left( \begin{matrix}
   5 & 6 \\
   1 & 8 \\
\end{matrix} \right) \\
\end{align}$
As you can see that L.H.S is equal to R.H.S so the values of x and y that we have got is satisfying the matrix equation and hence, are correct.