Target Exam

CUET

Subject

-- Applied Mathematics - Section B2

Chapter

Algebra

Question:

If $M =\begin{bmatrix}2\\-1\\3\end{bmatrix}$ and $N =\begin{bmatrix}7&1&-4\end{bmatrix}$, then $(MN)^T$ will be equal to:

Options:

$\begin{bmatrix}7&1&-4\\2&-1&3\\9&0&-1\end{bmatrix}$

$\begin{bmatrix}14&2&-8\\-7&-1&4\\21&3&-12\end{bmatrix}$

$\begin{bmatrix}14&-7&21\\2&-1&3\\1&3&-12\end{bmatrix}$

$\begin{bmatrix}14&-7&21\\2&-1&3\\-8&4&-12\end{bmatrix}$

Correct Answer:

$\begin{bmatrix}14&-7&21\\2&-1&3\\-8&4&-12\end{bmatrix}$

Explanation:

The correct answer is Option (4) → $\begin{bmatrix}14&-7&21\\2&-1&3\\-8&4&-12\end{bmatrix}$

$M=\begin{bmatrix}2\\-1\\3\end{bmatrix},\; N=\begin{bmatrix}7&1&-4\end{bmatrix}$

$MN=\begin{bmatrix}2\\-1\\3\end{bmatrix}\begin{bmatrix}7&1&-4\end{bmatrix} =\begin{bmatrix} 2\times7 & 2\times1 & 2\times(-4)\\ -1\times7 & -1\times1 & -1\times(-4)\\ 3\times7 & 3\times1 & 3\times(-4) \end{bmatrix}$

$MN=\begin{bmatrix} 14&2&-8\\ -7&-1&4\\ 21&3&-12 \end{bmatrix}$

$(MN)^T=\begin{bmatrix} 14&-7&21\\ 2&-1&3\\ -8&4&-12 \end{bmatrix}$

$(MN)^T=\begin{bmatrix} 14&-7&21\\ 2&-1&3\\ -8&4&-12 \end{bmatrix}$