Construct a 2 × 2 matrix whose elements aij are given by:
Question:

Construct a $2 \times 2$ matrix whose elements $a_{i j}$ are given by:

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

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

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

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

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

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

(vii) $a_{i j}=e^{2 i x} \sin (x j)$

Solution:

$(i)$

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

Here,

$a_{11}=\frac{(1+1)^{2}}{2}=\frac{(2)^{2}}{2}=\frac{4}{2}=2, a_{12}=\frac{(1+2)^{2}}{2}=\frac{(3)^{2}}{2}=\frac{9}{2}$

$a_{21}=\frac{(2+1)^{2}}{2}=\frac{(3)^{2}}{2}=\frac{9}{2}, a_{22}=\frac{(2+2)^{2}}{2}=\frac{(4)^{2}}{2}=\frac{16}{2}=8$

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

(ii)

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

Here,

$a_{11}=\frac{(1-1)^{2}}{2}=\frac{(0)^{2}}{2}=\frac{0}{2}=0, a_{12}=\frac{(1-2)^{2}}{2}=\frac{(-1)^{2}}{2}=\frac{1}{2}$

$a_{21}=\frac{(2-1)^{2}}{2}=\frac{(1)^{2}}{2}=\frac{1}{2}, a_{22}=\frac{(2-2)^{2}}{2}=\frac{(0)^{2}}{2}=\frac{0}{2}=0$

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

$(i i i)$

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

Here,

$a_{11}=\frac{\lfloor 1-2(1)\rfloor^{2}}{2}=\frac{(1-2)^{2}}{2}=\frac{(-1)^{2}}{2}=\frac{1}{2}, a_{12}=\frac{[1-2(2)]^{2}}{2}=\frac{(1-4)^{2}}{2}=\frac{(-3)^{2}}{2}=\frac{9}{2}$

$a_{21}=\frac{[2-2(1)]^{2}}{2}=\frac{(2-2)^{2}}{2}=\frac{0}{2}=0, a_{22}=\frac{[2-2(2)]^{2}}{2}=\frac{(2-4)^{2}}{2}=\frac{(-2)^{2}}{2}=\frac{4}{2}=2$

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

$(i v)$

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

Here,

$a_{11}=\frac{[2(1)+1]^{2}}{2}=\frac{(2+1)^{2}}{2}=\frac{(3)^{2}}{2}=\frac{9}{2}, a_{12}=\frac{[2(1)+2]^{2}}{2}=\frac{(4)^{2}}{2}=\frac{16}{2}=8$

$a_{21}=\frac{[2(2)+1]^{2}}{2}=\frac{(4+1)^{2}}{2}=\frac{(5)^{2}}{2}=\frac{25}{2}, a_{22}=\frac{[2(2)+2]^{2}}{2}=\frac{(4+2)^{2}}{2}=\frac{(6)^{2}}{2}=\frac{36}{2}=18$

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

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

Here,

$a_{11}=\frac{|2(1)-3(1)|}{2}=\frac{|2-3|}{2}=\frac{|-1|}{2}=\frac{1}{2}, a_{12}=\frac{|2(1)-3(2)|}{2}=\frac{|2-6|}{2}=\frac{|-4|}{2}=2$

$a_{21}=\frac{|2(2)-3(1)|}{2}=\frac{|4-3|}{2}=\frac{1}{2}, a_{22}=\frac{|2(2)-3(2)|}{2}=\frac{|4-6|}{2}=\frac{|-2|}{2}=1$

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

(vi)

$a_{i j}=\frac{|-3 i+j|}{2}$

Here,

$a_{11}=\frac{|-3(1)+1|}{2}=\frac{|-3+1|}{2}=\frac{|-2|}{2}=1, a_{12}=\frac{|-3(1)+2|}{2}=\frac{|-3+2|}{2}=\frac{|-1|}{2}=\frac{1}{2}$

$a_{21}=\frac{|-3(2)+1|}{2}=\frac{|-6+1|}{2}=\frac{|-5|}{2}=\frac{5}{2}, a_{22}=\frac{|-3(2)+2|}{2}=\frac{|-6+2|}{2}=\frac{|-4|}{2}=2$

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

(vii) $a_{i j}=e^{2 i x} \sin (x j)$

Here,

$a_{11}=e^{2 \times 1 \times x} \sin (x \times 1)=e^{2 x} \sin (x), a_{12}=e^{2 \times 1 \times x} \sin (x \times 2)=e^{2 x} \sin (2 x)$

$a_{21}=e^{2 \times 2 \times x} \sin (x \times 1)=e^{4 x} \sin (x), a_{22}=e^{2 \times 2 \times x} \sin (x \times 2)=e^{4 x} \sin (2 x)$

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

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