Target Exam

CUET

Subject

-- Applied Mathematics - Section B2

Chapter

Algebra

Question:

If $A'=\begin{bmatrix}-2 & 3\\1 & 2\end{bmatrix}$ and $B=\begin{bmatrix}-1 & 0\\2 & 3\end{bmatrix}$

then $(3A+2B)'$ is :

Options:

$\begin{bmatrix}-2 & 5\\1 & 10\end{bmatrix}$

$\begin{bmatrix}8 & 13\\-2 & 12\end{bmatrix}$

$\begin{bmatrix}-8 & 13\\3 & 12\end{bmatrix}$

$\begin{bmatrix}7 & 10\\-3 & 12\end{bmatrix}$

Correct Answer:

$\begin{bmatrix}-8 & 13\\3 & 12\end{bmatrix}$

Explanation:

$A'=\begin{pmatrix}-2&3\\1&2\end{pmatrix}$

$A=(A')^T=\begin{pmatrix}-2&1\\3&2\end{pmatrix}$

$B=\begin{pmatrix}-1&0\\2&3\end{pmatrix}$

$3A=\begin{pmatrix}-6&3\\9&6\end{pmatrix}$

$2B=\begin{pmatrix}-2&0\\4&6\end{pmatrix}$

$3A+2B=\begin{pmatrix}-8&3\\13&12\end{pmatrix}$

$(3A+2B)'=\begin{pmatrix}-8&13\\3&12\end{pmatrix}$

The value of $(3A+2B)'$ is $\begin{pmatrix}-8&13\\3&12\end{pmatrix}$.