# Construct a $2\times 2$ matrix $A={{\left[ {{a}_{ij}} \right]}_{2\times 2}}$ , whose elements are ${{a}_{ij}}$ and ${{a}_{ij}}=\dfrac{{{\left( i+j \right)}^{2}}}{2}$ .

Answer

Verified

360.6k+ views

Hint: At first construct the matrix by putting ${{a}_{ij}}$, with the actual values of $i,j$ . Then calculate the numerical values by using ${{a}_{ij}}=\dfrac{{{\left( i+j \right)}^{2}}}{2}$.

When some numbers are arranged in rows and columns and are surrounded on both sides by square brackets, we call it a matrix.

Here we have to construct a $2\times 2$ matrix named $A$ .

Now, this $2\times 2$ is known as the order of the matrix. Order of matrices helps us to know how many rows and how many columns there are in a matrix.

Generally we denote the order of a matrix by $m\times n$, where $m$ denotes the number of rows and $n$ denotes the number of columns. Here the order of the matrix is $2\times 2$ . That means this matrix has 2 rows and 2 columns.

Now, the elements of the matrix are generally denoted by ${{a}_{ij}}$. Where $ij$ help us to understand the actual position of the element.

In a matrix $A$, an element in row $i$ and column $j$ is represented by ${{a}_{ij}}$.

${{a}_{11}}$ is the element in 1st row and 1st column.

${{a}_{12}}$ is the element in 1st row and 2nd column.

${{a}_{21}}$ is the element in the 2nd row and 1st column.

${{a}_{22}}$ is the element in the 2nd row and 2nd column.

Therefore, $A=\left[ \begin{matrix}

{{a}_{11}} & {{a}_{12}} \\

{{a}_{21}} & {{a}_{22}} \\

\end{matrix} \right]$

Now let us substitute the values.

${{a}_{11}}=\dfrac{{{\left( 1+1 \right)}^{2}}}{2}=\dfrac{{{2}^{2}}}{2}=2$, here i is 1 and j is 1.

${{a}_{12}}=\dfrac{{{\left( 1+2 \right)}^{2}}}{2}=\dfrac{{{3}^{2}}}{2}=\dfrac{9}{2}$, here i is 1 and j is 2.

${{a}_{21}}=\dfrac{{{\left( 2+1 \right)}^{2}}}{2}=\dfrac{{{3}^{2}}}{2}=\dfrac{9}{2}$, here i is 2 and j is 1.

${{a}_{22}}=\dfrac{{{\left( 2+2 \right)}^{2}}}{2}=\dfrac{{{4}^{2}}}{2}=\dfrac{16}{2}=8$, here i is 2 and j is 2.

Therefore,

$A=\left[ \begin{matrix}

{{a}_{11}} & {{a}_{12}} \\

{{a}_{21}} & {{a}_{22}} \\

\end{matrix} \right]=\left[ \begin{matrix}

2 & \dfrac{9}{2} \\

\dfrac{9}{2} & 8 \\

\end{matrix} \right]$

Hence, $A=\left[ \begin{matrix}

2 & \dfrac{9}{2} \\

\dfrac{9}{2} & 8 \\

\end{matrix} \right]$

Note: We generally make mistakes to understand the position of an element in a matrix. Always check the position of the row first then the column.

__Complete step-by-step answer:__When some numbers are arranged in rows and columns and are surrounded on both sides by square brackets, we call it a matrix.

Here we have to construct a $2\times 2$ matrix named $A$ .

Now, this $2\times 2$ is known as the order of the matrix. Order of matrices helps us to know how many rows and how many columns there are in a matrix.

Generally we denote the order of a matrix by $m\times n$, where $m$ denotes the number of rows and $n$ denotes the number of columns. Here the order of the matrix is $2\times 2$ . That means this matrix has 2 rows and 2 columns.

Now, the elements of the matrix are generally denoted by ${{a}_{ij}}$. Where $ij$ help us to understand the actual position of the element.

In a matrix $A$, an element in row $i$ and column $j$ is represented by ${{a}_{ij}}$.

${{a}_{11}}$ is the element in 1st row and 1st column.

${{a}_{12}}$ is the element in 1st row and 2nd column.

${{a}_{21}}$ is the element in the 2nd row and 1st column.

${{a}_{22}}$ is the element in the 2nd row and 2nd column.

Therefore, $A=\left[ \begin{matrix}

{{a}_{11}} & {{a}_{12}} \\

{{a}_{21}} & {{a}_{22}} \\

\end{matrix} \right]$

Now let us substitute the values.

${{a}_{11}}=\dfrac{{{\left( 1+1 \right)}^{2}}}{2}=\dfrac{{{2}^{2}}}{2}=2$, here i is 1 and j is 1.

${{a}_{12}}=\dfrac{{{\left( 1+2 \right)}^{2}}}{2}=\dfrac{{{3}^{2}}}{2}=\dfrac{9}{2}$, here i is 1 and j is 2.

${{a}_{21}}=\dfrac{{{\left( 2+1 \right)}^{2}}}{2}=\dfrac{{{3}^{2}}}{2}=\dfrac{9}{2}$, here i is 2 and j is 1.

${{a}_{22}}=\dfrac{{{\left( 2+2 \right)}^{2}}}{2}=\dfrac{{{4}^{2}}}{2}=\dfrac{16}{2}=8$, here i is 2 and j is 2.

Therefore,

$A=\left[ \begin{matrix}

{{a}_{11}} & {{a}_{12}} \\

{{a}_{21}} & {{a}_{22}} \\

\end{matrix} \right]=\left[ \begin{matrix}

2 & \dfrac{9}{2} \\

\dfrac{9}{2} & 8 \\

\end{matrix} \right]$

Hence, $A=\left[ \begin{matrix}

2 & \dfrac{9}{2} \\

\dfrac{9}{2} & 8 \\

\end{matrix} \right]$

Note: We generally make mistakes to understand the position of an element in a matrix. Always check the position of the row first then the column.

Last updated date: 24th Sep 2023

â€¢

Total views: 360.6k

â€¢

Views today: 10.60k

Recently Updated Pages

What is the Full Form of DNA and RNA

What are the Difference Between Acute and Chronic Disease

Difference Between Communicable and Non-Communicable

What is Nutrition Explain Diff Type of Nutrition ?

What is the Function of Digestive Enzymes

What is the Full Form of 1.DPT 2.DDT 3.BCG

Trending doubts

How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Difference Between Plant Cell and Animal Cell

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

The equation xxx + 2 is satisfied when x is equal to class 10 maths CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Drive an expression for the electric field due to an class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

What is the past tense of read class 10 english CBSE