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  • Construct a_2x2 matrix where i) a_ij =i -2j /2

Construct a_{2\times2 } matrix where
(i) a_{ij} =\frac{\left ( \hat i - 2 \hat j \right )}{2}
(ii) a_{ij} = |- 2i + 3j|

Answers (1)

We know that,

A matrix, us a rectangular formation in which symbols, numbers, alphabets and expressions are arranged in rows and columns. 

Also,

We know that, the notation A = [a\textsubscript{ij}]\textsubscript{m$ \times $ m} indicates that A is a matrix having the order m \times n, also 1 \leq i \leq m, 1 \leq j \leq n; i, j \in \mathbb{N}.

(i).We need to construct a matrix, a\textsubscript{2$ \times $ 2}, where

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

For a\textsubscript{2$ \times $ 2},

1 \leq i \leq m

\Rightarrow 1 \leq i \leq 2 [ \because m = 2]

And,

\\ 1 \leq j \leq n\\ \\ \Rightarrow 1 \leq j \leq 2 \ [\because n = 2]

Put i = 1 and j = 1.

\\ \mathrm{a}_{11}=\frac{(1-2(1))^{2}}{2} \\ \Rightarrow \mathrm{a}_{11}=\frac{(1-2)^{2}}{2} \\ \Rightarrow \mathrm{a}_{11}=\frac{(-1)^{2}}{2} \\ \Rightarrow \mathrm{a}_{11}=\frac{1}{2} \\ \text { Put } \mathrm{i}=1 \text { and } \mathrm{j}=2 \\ \mathrm{a}_{12}=\frac{(1-2(2))^{2}}{2} \\ \Rightarrow \mathrm{a}_{12}=\frac{(1-4)^{2}}{2}

\\ \Rightarrow a_{12}=\frac{(-3)^{2}}{2} \\ \Rightarrow a_{12}=\frac{9}{2} \\ \text { Put } i=2 \text { and } j=1 \\ a_{21}=\frac{(2-2(1))^{2}}{2} \\ \Rightarrow a_{21}=\frac{(2-2)^{2}}{2} \\ \Rightarrow a_{21}=\frac{0}{2} \\ \Rightarrow a_{21}=0 \\ \text { Put } i=2 \text { and } j=2 \\ a_{22}=\frac{(2-2(2))^{2}}{2}

\\ \Rightarrow \mathrm{a}_{22}=\frac{(2-4)^{2}}{2} \\ \Rightarrow \mathrm{a}_{22}=\frac{(-2)^{2}}{2} \\ \Rightarrow \mathrm{a}_{22}=\frac{4}{2} \\ \Rightarrow \mathrm{a}_{22}=2

Let the matrix formed be named A.

\begin{aligned} &A=\left[\begin{array}{ll} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array}\right]\\ &\text { By substituting the values of } a_{11}, a_{12}, a_{21} \text { and } a_{22}, \end{aligned} the matrix formed is 

A=\left[\begin{array}{ll} \frac{1}{2} & \frac{9}{2} \\ 0 & 2 \end{array}\right]

(ii). We need to construct a matrix, a\textsubscript{2$ \times $ 2}, where

a \textsubscript {ij} = \vert -2i + 3j \vert

For a\textsubscript{2$ \times $ 2},

1 \leq i \leq m

\Rightarrow 1 \leq i \leq 2 [ \because m = 2]

And,

\\ 1 \leq j \leq n \\ \Rightarrow 1 \leq j \leq 2 \ [\because n = 2]

Put i = 1 and j = 1.

\\ a_{11} = \vert -2(1) + 3(1) \vert \\ \Rightarrow a_{11} = \vert -2 + 3 \vert \\ \Rightarrow a_{11} = \vert 1 \vert \\ \Rightarrow a_{11} = 1

Put i = 1 and j = 2.

\\ a_{12} = \vert -2(1) + 3(2) \vert \\ \Rightarrow a_{12} = \vert -2 + 6 \vert \\ \Rightarrow a_{12} = \vert 4 \vert \\ \Rightarrow a_{12} = 4

Put i = 2 and j = 1.

\\ a_{21} = \vert -2(2) + 3(1) \vert \\ \Rightarrow a_{21} = \vert -4 + 3 \vert \\ \Rightarrow a_{21} = \vert -1 \vert \\ \Rightarrow a_{21} = 1

Put i = 2 and j = 2. .

\\ a_{22} = \vert -2(2) + 3(2) \vert \\ \Rightarrow a_{22} = \vert -4 + 6 \vert \\ \Rightarrow a_{22} = \vert 2 \vert \\ \Rightarrow a_{22} = 2

Let the matrix formed be A.

\begin{aligned} &A=\left[\begin{array}{ll} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array}\right]\\ &\text { By substituting the values of } a_{11}, a_{12}, a_{21} \text { and } a_{22}, \text { the matrix formed is }\\ &A=\left[\begin{array}{ll} 1 & 4 \\ 1 & 2 \end{array}\right] \end{aligned}

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