Question

Order of $\left[\begin{array}{lll}2& 7& 4\\ 3& 1& 0\end{array}\right]\left[\begin{array}{l}5\\ 4\\ 3\end{array}\right]$ is:

• (A) $3\times 1$
• (B) $2\times 1$
• (C) $2\times 3$
• (D) $3\times 3$

Answer 1

The Correct Option is B

Explanation:
Given,First matrix $=\left[\begin{array}{ccc}2& 7& 4\\ 3& 1& 0\end{array}\right]$Number of rows $\left(m_1\right)=2$Number of columns $\left(n_1\right)=3$So, order of first matrix $=2\times 3$And second matrix $=\left[\begin{array}{l}5\\ 4\\ 3\end{array}\right]$Number of rows $\left(m_2\right)=3$Number of columns $\left(n_2\right)=1$So, order of first matrix $=3\times 1$As we know,Multiplication of matrices is possible if Number of column $(n_1)$ in the first matrix $=$ Number of rows $(m_2)$ in the second matrixi.e., $n_1=m_2$$\therefore n_1=m_2=3$So, $\left[\begin{array}{lll}2& 7& 4\\ 3& 1& 0\end{array}\right]\left[\begin{array}{l}5\\ 4\\ 3\end{array}\right]$ is possible.As we know,Multiplication of matrices is possible if the order of the new matrix is a row of the first matrix and column of the second matrix.i.e., $m_1\times n_2$Now, Order of new matrix formed by multiplication $=m_1\times n_2=2\times 1$Hence, the correct option is (B).