
If $\left( \begin{matrix}
x & 0 \\
1 & y \\
\end{matrix} \right)+\left( \begin{matrix}
-2 & 1 \\
3 & 4 \\
\end{matrix} \right)=\left( \begin{matrix}
3 & 5 \\
6 & 3 \\
\end{matrix} \right)-\left( \begin{matrix}
2 & 4 \\
2 & 1 \\
\end{matrix} \right)$ then
A . $x=-3,y=-2$
B. $x=3,y=-2$
C. $x=3,y=2$
D. $x=-3,y=2$
Answer
163.5k+ views
Hint:In this question, we have given the matrices, and all they are of $2\times 2$ order and we have to find the value of x and y. to find the values, we have to add and subtract the matrices. First, we solve the left-hand side by adding the matrices, then we solve the right-hand side by subtracting the matrices. Then we equate both equations and simplify them to get the value of x and y.
Complete step by step Solution:
Given $\left( \begin{matrix}
x & 0 \\
1 & y \\
\end{matrix} \right)+\left( \begin{matrix}
-2 & 1 \\
3 & 4 \\
\end{matrix} \right)=\left( \begin{matrix}
3 & 5 \\
6 & 3 \\
\end{matrix} \right)-\left( \begin{matrix}
2 & 4 \\
2 & 1 \\
\end{matrix} \right)$
All the matrices are of $2\times 2$ order.
We have to find the value of x and y.
To find the value of x and y, we can add and subtract the given matrices.
First we add the matrices $\left( \begin{matrix}
x & 0 \\
1 & y \\
\end{matrix} \right)$ and $\left( \begin{matrix}
-2 & 1 \\
3 & 4 \\
\end{matrix} \right)$
Then $\left( \begin{matrix}
x & 0 \\
1 & y \\
\end{matrix} \right)$ + $\left( \begin{matrix}
-2 & 1 \\
3 & 4 \\
\end{matrix} \right)$ = $\left( \begin{matrix}
x+(-2) & 0+1 \\
1+3 & y+4 \\
\end{matrix} \right)$
Hence $\left( \begin{matrix}
x & 0 \\
1 & y \\
\end{matrix} \right)$ + $\left( \begin{matrix}
-2 & 1 \\
3 & 4 \\
\end{matrix} \right)$ = $\left( \begin{matrix}
x-2 & 1 \\
4 & y+4 \\
\end{matrix} \right)$
Now we subtract the matrices $\left( \begin{matrix}
3 & 5 \\
6 & 3 \\
\end{matrix} \right)$ and $\left( \begin{matrix}
2 & 4 \\
2 & 1 \\
\end{matrix} \right)$
$\left( \begin{matrix}
3 & 5 \\
6 & 3 \\
\end{matrix} \right)$ - $\left( \begin{matrix}
2 & 4 \\
2 & 1 \\
\end{matrix} \right)$ = $\left( \begin{matrix}
3-2 & 5-4 \\
6-2 & 3-1 \\
\end{matrix} \right)$
$\left( \begin{matrix}
3 & 5 \\
6 & 3 \\
\end{matrix} \right)$ - $\left( \begin{matrix}
2 & 4 \\
2 & 1 \\
\end{matrix} \right)$ = $\left( \begin{matrix}
1 & 1 \\
4 & 2 \\
\end{matrix} \right)$
Now we equate both the matrices and get
$\left( \begin{matrix}
x-2 & 1 \\
4 & y+4 \\
\end{matrix} \right)$ = $\left( \begin{matrix}
1 & 1 \\
4 & 2 \\
\end{matrix} \right)$
Now we compute and simplify the above terms, and we get
$x-2=1$ and
$y+4=2$
By simplifying the above equations, we get
$x=3,y=-2$
Hence the value of $x=3,y=-2$
Therefore, the correct option is (B).
Note: Keep in mind before adding and subtracting any matrices that they have an equal number of columns and rows to be added. As the given matrices are of order $2\times 2$, so we can add it simply. Similarly we can add a $2\times 3$ matrix with a $2\times 3$ matrix or $3\times 3$ matrix with $3\times 3$ matrix. However, we cannot add $2\times 3$ matrix with a $3\times 2$ matrix. Similarly, we cannot add $2\times 2$ matrix with a $3\times 3$ matrix. The order in which we add the matrix is not important because the addition of two matrices is commutative.
Complete step by step Solution:
Given $\left( \begin{matrix}
x & 0 \\
1 & y \\
\end{matrix} \right)+\left( \begin{matrix}
-2 & 1 \\
3 & 4 \\
\end{matrix} \right)=\left( \begin{matrix}
3 & 5 \\
6 & 3 \\
\end{matrix} \right)-\left( \begin{matrix}
2 & 4 \\
2 & 1 \\
\end{matrix} \right)$
All the matrices are of $2\times 2$ order.
We have to find the value of x and y.
To find the value of x and y, we can add and subtract the given matrices.
First we add the matrices $\left( \begin{matrix}
x & 0 \\
1 & y \\
\end{matrix} \right)$ and $\left( \begin{matrix}
-2 & 1 \\
3 & 4 \\
\end{matrix} \right)$
Then $\left( \begin{matrix}
x & 0 \\
1 & y \\
\end{matrix} \right)$ + $\left( \begin{matrix}
-2 & 1 \\
3 & 4 \\
\end{matrix} \right)$ = $\left( \begin{matrix}
x+(-2) & 0+1 \\
1+3 & y+4 \\
\end{matrix} \right)$
Hence $\left( \begin{matrix}
x & 0 \\
1 & y \\
\end{matrix} \right)$ + $\left( \begin{matrix}
-2 & 1 \\
3 & 4 \\
\end{matrix} \right)$ = $\left( \begin{matrix}
x-2 & 1 \\
4 & y+4 \\
\end{matrix} \right)$
Now we subtract the matrices $\left( \begin{matrix}
3 & 5 \\
6 & 3 \\
\end{matrix} \right)$ and $\left( \begin{matrix}
2 & 4 \\
2 & 1 \\
\end{matrix} \right)$
$\left( \begin{matrix}
3 & 5 \\
6 & 3 \\
\end{matrix} \right)$ - $\left( \begin{matrix}
2 & 4 \\
2 & 1 \\
\end{matrix} \right)$ = $\left( \begin{matrix}
3-2 & 5-4 \\
6-2 & 3-1 \\
\end{matrix} \right)$
$\left( \begin{matrix}
3 & 5 \\
6 & 3 \\
\end{matrix} \right)$ - $\left( \begin{matrix}
2 & 4 \\
2 & 1 \\
\end{matrix} \right)$ = $\left( \begin{matrix}
1 & 1 \\
4 & 2 \\
\end{matrix} \right)$
Now we equate both the matrices and get
$\left( \begin{matrix}
x-2 & 1 \\
4 & y+4 \\
\end{matrix} \right)$ = $\left( \begin{matrix}
1 & 1 \\
4 & 2 \\
\end{matrix} \right)$
Now we compute and simplify the above terms, and we get
$x-2=1$ and
$y+4=2$
By simplifying the above equations, we get
$x=3,y=-2$
Hence the value of $x=3,y=-2$
Therefore, the correct option is (B).
Note: Keep in mind before adding and subtracting any matrices that they have an equal number of columns and rows to be added. As the given matrices are of order $2\times 2$, so we can add it simply. Similarly we can add a $2\times 3$ matrix with a $2\times 3$ matrix or $3\times 3$ matrix with $3\times 3$ matrix. However, we cannot add $2\times 3$ matrix with a $3\times 2$ matrix. Similarly, we cannot add $2\times 2$ matrix with a $3\times 3$ matrix. The order in which we add the matrix is not important because the addition of two matrices is commutative.
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