Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section A

Chapter

Matrices

Question:

If $X+Y=\begin{bmatrix}7 & 0\\2 & 5\end{bmatrix}$ and $X-Y=\begin{bmatrix}3 & 0 \\0 & 3\end{bmatrix}$. Then X and Y are :

Options:

$X=\begin{bmatrix}5 & 0\\1 & 4\end{bmatrix}$ and $Y=\begin{bmatrix}2 & 0\\1 & 1\end{bmatrix}$

$X=\begin{bmatrix}0 & 5\\1 & 4\end{bmatrix}$ and $Y=\begin{bmatrix}0 & 2\\1 & 1\end{bmatrix}$

$X=\begin{bmatrix}1 & 4\\0 & 5\end{bmatrix}$ and $Y=\begin{bmatrix}0 & 2\\1 & 1\end{bmatrix}$

$X=\begin{bmatrix}1 & 4\\5 & 0\end{bmatrix}$ and $Y=\begin{bmatrix}0 & 2\\1 & 1\end{bmatrix}$

Correct Answer:

$X=\begin{bmatrix}5 & 0\\1 & 4\end{bmatrix}$ and $Y=\begin{bmatrix}2 & 0\\1 & 1\end{bmatrix}$

Explanation:

The correct answer is Option (1) → $X=\begin{bmatrix}5 & 0\\1 & 4\end{bmatrix}$ and $Y=\begin{bmatrix}2 & 0\\1 & 1\end{bmatrix}$

$X+Y=\begin{bmatrix}7 & 0\\2 & 5\end{bmatrix}$ ....(1)

$X-Y=\begin{bmatrix}3 & 0 \\0 & 3\end{bmatrix}$  ....(2)

$X+Y+X-Y=\begin{bmatrix}10 & 0\\2 & 8\end{bmatrix}$

$2X=\begin{bmatrix}10 & 0\\2 & 8\end{bmatrix}⇒X=\begin{bmatrix}5& 0\\1 & 4\end{bmatrix}$

so from (1)

$Y=\begin{bmatrix}7& 0\\2 & 5\end{bmatrix}-\begin{bmatrix}5& 0\\1 & 4\end{bmatrix}$

$Y=\begin{bmatrix}2 & 0\\1 & 1\end{bmatrix}$