Question

Construct $2\times 2$ matrix A whose elements are $a_{ij}=\frac{1}{2}(i-3j)$. Hence $3A=$

• A. $\left[\begin{array}{cc}0& 1\\ 2& 3\end{array}\right]$
• B. $\left[\begin{array}{cc}1& 2\\ 2& 4\end{array}\right]$
• C. $\left[\begin{array}{cc}-3& \frac{-15}{2}\\ \frac{-3}{2}& -6\end{array}\right]$
• D. $\left[\begin{array}{cc}3& \frac{-15}{2}\\ \frac{3}{2}& 6\end{array}\right]$

Answer 1

The correct option is C $\left[\begin{array}{cc}-3& \frac{-15}{2}\\ \frac{-3}{2}& -6\end{array}\right]$

Given elements are of form

$a_{ij}=\frac{1}{2}(i-3j)$ in matrix A

$i\to$row number

$j\to$column number

Number in first row and first column

$a_{11}=\frac{1}{2}(1-3\left(1\right))=\frac{1}{2}(-2)=-1$

Number in first row and second column

$a_{12}=\frac{1}{2}(2-3)=\frac{-1}{2}$

Number in second row and first column

$a_{21}=\frac{1}{2}(1-6)=\frac{-5}{2}$

Number in second row and second column

$a_{22}=\frac{1}{2}(2-6)=-2$

$A=\left[\begin{array}{cc}-1& \frac{-5}{2}\\ \frac{-1}{2}& -2\end{array}\right]$ Then

$3A=\left[\begin{array}{cc}-3& \frac{-15}{2}\\ \frac{-3}{2}& -6\end{array}\right]$

$\therefore$ Option C will be correct.