Question

If $\begin{bmatrix}x+y & 2 \\ 1& x-y\end{bmatrix}=\begin{bmatrix}4& 2 \\ 1& 2\end{bmatrix}$, then the values of $x$ and $y$ are:}

• A. $x=3, y=1$
• B. $x=1, y=3$
• C. $x=2, y=3$
• D. $x=1, y=1$

Hint:

Equal matrices = Equal positions. Just solve the resulting simple algebra.

Answer 1

The Correct Option is A

To solve the given problem, we need to equate the corresponding elements of the two matrices provided in the question:

\(\begin{bmatrix} x+y & 2 \\ 1 & x-y \end{bmatrix} = \begin{bmatrix} 4 & 2 \\ 1 & 2 \end{bmatrix}\)

Let's equate the corresponding elements from both matrices:

1. For the element at the first row and first column: \(x + y = 4\)
2. For the element at the first row and second column: \(2 = 2\) (This is already satisfied and doesn't give us any equation.)
3. For the element at the second row and first column: \(1 = 1\) (This is already satisfied and doesn't give us any additional information.)
4. For the element at the second row and second column: \(x - y = 2\)

We now have the following system of equations:

1. \(x + y = 4\)
2. \(x - y = 2\)

To find the values of \(x\) and \(y\), we can solve this system using the method of elimination or substitution. Let's use elimination here:

Add the two equations:

1. \((x + y) + (x - y) = 4 + 2\)

This simplifies to:

1. \(2x = 6\)

Divide both sides by 2:

1. \(x = 3\)

Now substitute \(x = 3\) back into one of the original equations. Let's use \(x + y = 4\):

1. \(3 + y = 4\)

Solve for \(y\):

1. \(y = 4 - 3\)
2. \(y = 1\)

Therefore, the solution is \(x = 3\) and \(y = 1\).

Thus, the correct option is:

$x=3, y=1$