Question

If$A={\left[a_{\mathrm{ij}}\right]}_{2\times 2}$, where $a_{\mathrm{ij}}=i+j$, then $A$ is equal to

• A. $\mathrm{\left[}\begin{array}{cc}\mathrm{1}& \mathrm{1}\\ \mathrm{2}& \mathrm{2}\end{array}\mathrm{\right]}$
• B. $\mathrm{\left[}\begin{array}{cc}\mathrm{1}& \mathrm{2}\\ \mathrm{1}& \mathrm{2}\end{array}\mathrm{\right]}$
• C. $\mathrm{\left[}\begin{array}{cc}\mathrm{1}& \mathrm{2}\\ \mathrm{3}& \mathrm{4}\end{array}\mathrm{\right]}$
• D. $\mathrm{\left[}\begin{array}{cc}\mathrm{2}& \mathrm{3}\\ \mathrm{3}& \mathrm{4}\end{array}\mathrm{\right]}$

Answer 1

The correct option is D

$\mathrm{\left[}\begin{array}{cc}\mathrm{2}& \mathrm{3}\\ \mathrm{3}& \mathrm{4}\end{array}\mathrm{\right]}$

Explanation for correct option:

$$\begin{gathered} A={\left[{\mathrm{a}}_{\mathrm{ij}}\right]}_{2\times 2} \\ =\left[\begin{array}{cc}{\mathrm{a}}_{11}& {\mathrm{a}}_{12}\\ {\mathrm{a}}_{21}& {\mathrm{a}}_{22}\end{array}\right] \end{gathered}$$

$a_{\mathrm{ij}}=i+j$

so,

$$\begin{gathered} a_{11}=1+1 \\ =2 \\ {a}_{12}=1+2 \\ =3 \end{gathered}$$

$$\begin{gathered} a_{21}=2+1 \\ =3 \\ {a}_{22}=2+2 \\ =4 \end{gathered}$$

$\mathit{A}\mathbf{=}\left[\begin{array}{cc}2& 3\\ 3& 4\end{array}\right]$

Hence, the correct option is (D).