Question

Construct a 2 × 2 matrix whose elements a_ij are given by:
(i) $\frac{(i+j{)}^2}{2}$

(ii) $a_{ij}=\frac{(i-j{)}^2}{2}$

(iii) $a_{ij}=\frac{(i-2j{)}^2}{2}$

(iv) $a_{ij}=\frac{(2i+j{)}^2}{2}$

(v) $a_{ij}=\frac{\left|2i-3i\right|}{2}$

(vi) $a_{ij}=\frac{\left|-3i+j\right|}{2}$

(vii) $a_{ij}=e^{2ix}\sin \left(xj\right)$

Answer 1

$\left(i\right)$

$\frac{{\left(i+j\right)}^2}{2}$

Here,

$$\begin{gathered} a_{11}=\frac{{\left(1+1\right)}^2}{2}=\frac{{\left(2\right)}^2}{2}=\frac{4}{2}=2,a_{12}=\frac{{\left(1+2\right)}^2}{2}=\frac{{\left(3\right)}^2}{2}=\frac{9}{2} \\ {a}_{21}=\frac{{\left(2+1\right)}^2}{2}=\frac{{\left(3\right)}^2}{2}=\frac{9}{2},a_{22}=\frac{{\left(2+2\right)}^2}{2}=\frac{{\left(4\right)}^2}{2}=\frac{16}{2}=8 \end{gathered}$$

So, the required matrix is $\left[\begin{array}{c}2\frac{9}{2}\\ \frac{9}{2}8\end{array}\right]$.

$\left(ii\right)$

$a_{ij}=\frac{{\left(i-j\right)}^2}{2}$

Here,

$$\begin{gathered} a_{11}=\frac{{\left(1-1\right)}^2}{2}=\frac{{\left(0\right)}^2}{2}=\frac{0}{2}=0,a_{12}=\frac{{\left(1-2\right)}^2}{2}=\frac{{\left(-1\right)}^2}{2}=\frac{1}{2} \\ {a}_{21}=\frac{{\left(2-1\right)}^2}{2}=\frac{{\left(1\right)}^2}{2}=\frac{1}{2},a_{22}=\frac{{\left(2-2\right)}^2}{2}=\frac{{\left(0\right)}^2}{2}=\frac{0}{2}=0 \end{gathered}$$

So, the required matrix is $\left[\begin{array}{cc}0& \frac{1}{2}\\ \frac{1}{2}& 0\end{array}\right]$.

$\left(iii\right)$
$a_{ij}=\frac{{\left(i-2j\right)}^2}{2}$

Here,

$$\begin{gathered} a_{11}=\frac{{\left[1-2\left(1\right)\right]}^2}{2}=\frac{{\left(1-2\right)}^2}{2}=\frac{{\left(-1\right)}^2}{2}=\frac{1}{2},a_{12}=\frac{{\left[1-2\left(2\right)\right]}^2}{2}=\frac{{\left(1-4\right)}^2}{2}=\frac{{\left(-3\right)}^2}{2}=\frac{9}{2} \\ {a}_{21}=\frac{{\left[2-2\left(1\right)\right]}^2}{2}=\frac{{\left(2-2\right)}^2}{2}=\frac{0}{2}=0,a_{22}=\frac{{\left[2-2\left(2\right)\right]}^2}{2}=\frac{{\left(2-4\right)}^2}{2}=\frac{{\left(-2\right)}^2}{2}=\frac{4}{2}=2 \end{gathered}$$

So, the required matrix is $\left[\begin{array}{cc}\frac{1}{2}& \frac{9}{2}\\ 0& 2\end{array}\right]$.

$\left(iv\right)$
$a_{ij}=\frac{{\left(2i+j\right)}^2}{2}$

Here,

$$\begin{gathered} a_{11}=\frac{{\left[2\left(1\right)+1\right]}^2}{2}=\frac{{\left(2+1\right)}^2}{2}=\frac{{\left(3\right)}^2}{2}=\frac{9}{2},a_{12}=\frac{{\left[2\left(1\right)+2\right]}^2}{2}=\frac{{\left(4\right)}^2}{2}=\frac{16}{2}=8 \\ {a}_{21}=\frac{{\left[2\left(2\right)+1\right]}^2}{2}=\frac{{\left(4+1\right)}^2}{2}=\frac{{\left(5\right)}^2}{2}=\frac{25}{2},a_{22}=\frac{{\left[2\left(2\right)+2\right]}^2}{2}=\frac{{\left(4+2\right)}^2}{2}=\frac{{\left(6\right)}^2}{2}=\frac{36}{2}=18 \end{gathered}$$

So, the required matrix is $\left[\begin{array}{cc}\frac{9}{2}& 8\\ \frac{25}{2}& 18\end{array}\right]$.

$\left(v\right)$ $a_{ij}=\frac{\left|2i-3j\right|}{2}$

Here,

$$\begin{gathered} a_{11}=\frac{\left|2\left(1\right)-3\left(1\right)\right|}{2}=\frac{\left|2-3\right|}{2}=\frac{\left|-1\right|}{2}=\frac{1}{2},a_{12}=\frac{\left|2\left(1\right)-3\left(2\right)\right|}{2}=\frac{\left|2-6\right|}{2}=\frac{\left|-4\right|}{2}=2 \\ {a}_{21}=\frac{\left|2\left(2\right)-3\left(1\right)\right|}{2}=\frac{\left|4-3\right|}{2}=\frac{1}{2},a_{22}=\frac{\left|2\left(2\right)-3\left(2\right)\right|}{2}=\frac{\left|4-6\right|}{2}=\frac{\left|-2\right|}{2}=1 \end{gathered}$$

So, the required matrix is $\left[\begin{array}{cc}\frac{1}{2}& 2\\ \frac{1}{2}& 1\end{array}\right]$.

$\left(vi\right)$
$a_{ij}=\frac{\left|-3i+j\right|}{2}$

Here,

$$\begin{gathered} a_{11}=\frac{\left|-3\left(1\right)+1\right|}{2}=\frac{\left|-3+1\right|}{2}=\frac{\left|-2\right|}{2}=1,a_{12}=\frac{\left|-3\left(1\right)+2\right|}{2}=\frac{\left|-3+2\right|}{2}=\frac{\left|-1\right|}{2}=\frac{1}{2} \\ {a}_{21}=\frac{\left|-3\left(2\right)+1\right|}{2}=\frac{\left|-6+1\right|}{2}=\frac{\left|-5\right|}{2}=\frac{5}{2},a_{22}=\frac{\left|-3\left(2\right)+2\right|}{2}=\frac{\left|-6+2\right|}{2}=\frac{\left|-4\right|}{2}=2 \end{gathered}$$

So, the required matrix is $\left[\begin{array}{cc}1& \frac{1}{2}\\ \frac{5}{2}& 2\end{array}\right]$.

(vii) $a_{ij}=e^{2ix}\sin \left(xj\right)$

Here,

$$\begin{gathered} a_{11}=e^{2\times 1\times x}\sin \left(x\times 1\right)=e^{2x}\sin \left(x\right),a_{12}=e^{2\times 1\times x}\sin \left(x\times 2\right)=e^{2x}\sin \left(2x\right) \\ {a}_{21}=e^{2\times 2\times x}\sin \left(x\times 1\right)=e^{4x}\sin \left(x\right),a_{22}=e^{2\times 2\times x}\sin \left(x\times 2\right)=e^{4x}\sin \left(2x\right) \end{gathered}$$

So, the required matrix is $\left[\begin{array}{cc}{e}^{2x}\sin \left(x\right)& e^{2x}\sin \left(2x\right)\\ e^{4x}\sin \left(x\right)& e^{4x}\sin \left(2x\right)\end{array}\right]$.