Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Determinants

Question:

If $X+Y=\left[\begin{array}{ll}2 & 1 \\ 3 & 5\end{array}\right]$ and $X-Y=\left[\begin{array}{ll}0 & 3 \\ 7 & 9\end{array}\right]$, then $Y$ is equal to

Options:

$\left[\begin{array}{cc}-1 & 1 \\ -2 & -2\end{array}\right]$

$\left[\begin{array}{cc}-1 & 1 \\ 2 & 2\end{array}\right]$

$\left[\begin{array}{cc}1 & -1 \\ -2 & -2\end{array}\right]$

$\left[\begin{array}{cc}1 & -1 \\ 2 & 2\end{array}\right]$

Correct Answer:

$\left[\begin{array}{cc}1 & -1 \\ -2 & -2\end{array}\right]$

Explanation:

$X+Y=\left[\begin{array}{ll}2 & 1 \\ 3 & 5\end{array}\right]$            .....(1)

$X-Y=\left[\begin{array}{ll}0 & 3 \\ 7 & 9\end{array}\right]$            .....(2)

eq. (1) - eq. (2)

⇒  2Y = $\left[\begin{array}{cc}2 & -2 \\ -4 & -4\end{array}\right]$   so  $Y=\left[\begin{array}{cc}1 & -1 \\ -2 & -2\end{array}\right]$

→  Option C