If $A = [a_{ij}]$ be a $3 \times 3$ matrix, where $a_{ij} = i - 3j$, then which of the following is $false$? (A) $a_{11} a_{31}$ (D) $a_{31} = 0$

> [!TLDR] > We construct the $3 \times 3$ matrix $A$ using the given rule $a_{ij} = i - 3j$ and then evaluate each option. The statement $a_{13} > a_{31}$ is found to be false. When we are given a matrix $A = [a_{ij}]$, it means that $A$ is composed of elements where $a_{ij}$ refers to the element located at the $i$-th row and $j$-th column. The problem defines a $3 \times 3$ matrix, which means it has 3 rows and 3 columns. The indices $i$ and $j$ will therefore range from 1 to 3. The core idea here is to systematically determine each element of the matrix using the provided formula $a_{ij} = i - 3j$. Once the matrix elements are known, we can directly check the truthfulness of each given statement. 1. **Understand the Matrix Structure and Element Definition:** A $3 \times 3$ matrix $A$ has elements $a_{ij}$ where $i \in \{1, 2, 3\}$ represents the row number and $j \in \{1, 2, 3\}$ represents the column number. The rule for each element is given by: > [!FORMULA] > $a_{ij} = i - 3j$ 2. **Calculate Each Element of the Matrix:** We will substitute the values of $i$ and $j$ into the formula to find each element: * For the first row ($i=1$): * $a_{11} = 1 - 3(1) = 1 - 3 = -2$ * $a_{12} = 1 - 3(2) = 1 - 6 = -5$ * $a_{13} = 1 - 3(3) = 1 - 9 = -8$ * For the second row ($i=2$): * $a_{21} = 2 - 3(1) = 2 - 3 = -1$ * $a_{22} = 2 - 3(2) = 2 - 6 = -4$ * $a_{23} = 2 - 3(3) = 2 - 9 = -7$ * For the third row ($i=3$): * $a_{31} = 3 - 3(1) = 3 - 3 = 0$ * $a_{32} = 3 - 3(2) = 3 - 6 = -3$ * $a_{33} = 3 - 3(3) = 3 - 9 = -6$ 3. **Construct the Matrix A:** Now we can write down the complete matrix $A$: $$A = \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{pmatrix} = \begin{pmatrix} -2 & -5 & -8 \\ -1 & -4 & -7 \\ 0 & -3 & -6 \end{pmatrix}$$ 4. **Evaluate Each G

If A = [aᵢⱼ] be a 3 × 3 matrix, where aᵢⱼ = i - 3j, the — CBSE Class 12 Mathematics 2024 PYQ

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A = [a_{ij}]

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Concept understanding — Matrix Construction

Matrix Construction: Building a Grid of Numbers

A teacher recording attendance for $30$ students over $5$ days could keep separate lists — but that is messy. Instead, draw a grid: rows for students, columns for days, each cell a $1$ (present) or $0$ (absent). That grid is a matrix. Constructing a matrix means deciding its shape and what number sits in each cell.

Why a Grid?

Every cell of a matrix has a unique address $(i, j)$ — row $i$ , column $j$ — so the entry in row $2$ , column $3$ is written $a_{23}$ . A grid beats a plain list because so many problems have two natural dimensions: a system of equations (equation $\times$ variable), a digital image (row $\times$ column of pixels), or a network (source node $\times$ destination node). The grid lets operations act on both dimensions at once.

The Precise Form

A matrix $A$ of order $m \times n$ (" $m$ by $n$ ") has $m$ rows and $n$ columns:

A = \begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn} \end{pmatrix}, \qquad A = [a_{ij}]_{m \times n}.

Each $a_{ij}$ is an entry: the first index $i$ is the row, the second $j$ is the column.

How You Construct One

To build a matrix you specify:

Dimensions — how many rows $m$ and columns $n$ .
Entry rule — what number fills each cell: an explicit list, a formula in $i$ and $j$ , or data from a problem.
Placement — order matters; swapping rows or columns gives a different matrix.

Explicit: a $2 \times 3$ matrix with rows $(1, 0, -2)$ and $(3, 5, 7)$ is

A = \begin{pmatrix} 1 & 0 & -2 \\ 3 & 5 & 7 \end{pmatrix}.

Formula-based: for a $3 \times 3$ matrix with $a_{ij} = i^2 - j$ , we get $a_{11} = 0$ , $a_{12} = -1$ , $a_{21} = 3$ , giving

$A = \begin{pmatrix} 0 & -1 & -2 \\ 3 & 2 & 1 \\ 8 & 7 & 6 \end{pmatrix}.$ …

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We construct the $3 \times 3$ matrix $A$ using the given rule $a_{ij} = i - 3j$ and then evaluate each option. The statement $a_{13} > a_{31}$ is found to be false.

When we are given a matrix $A = [a_{ij}]$ , it means that $A$ is composed of elements where $a_{ij}$ refers to the element located at the $i$ -th row and $j$ -th column. The problem defines a $3 \times 3$ matrix, which means it has 3 rows and 3 columns. The indices $i$ and $j$ will therefore range from 1 to 3.

The core idea here is to systematically determine each element of the matrix using the provided formula $a_{ij} = i - 3j$ . Once the matrix elements are known, we can directly check the truthfulness of each given statement.

Understand the Matrix Structure and Element Definition:

A $3 \times 3$ matrix $A$ has elements $a_{ij}$ where $i \in \{1, 2, 3\}$ represents the row number and $j \in \{1, 2, 3\}$ represents the column number.

The rule for each element is given by:
$a_{ij} = i - 3j$
Calculate Each Element of the Matrix:

We will substitute the values of $i$ and $j$ into the formula to find each element:
- For the first row ( $i=1$ $i = 1$ ):
  - $a_{11} = 1 - 3(1) = 1 - 3 = -2$
  - $a_{12} = 1 - 3(2) = 1 - 6 = -5$
  - $a_{13} = 1 - 3(3) = 1 - 9 = -8$
- For the second row ( $i=2$ $i = 2$ ):
  - $a_{21} = 2 - 3(1) = 2 - 3 = -1$
  - $a_{22} = 2 - 3(2) = 2 - 6 = -4$
  - $a_{23} = 2 - 3(3) = 2 - 9 = -7$
- For the third row ( $i=3$ $i = 3$ ):
  - $a_{31} = 3 - 3(1) = 3 - 3 = 0$
  - $a_{32} = 3 - 3(2) = 3 - 6 = -3$
  - $a_{33} = 3 - 3(3) = 3 - 9 = -6$
Construct the Matrix A:

Now we can write down the complete matrix $A$ :

A = \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{pmatrix} = \begin{pmatrix} -2 & -5 & -8 \\ -1 & -4 & -7 \\ 0 & -3 & -6 \end{pmatrix}

Evaluate Each Given Option: …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 16 on this concept.

CBSE 2026Set ANNUAL1 mark
$2 \times 2$
›Reveal solutionSolution

Substitute $i,j=1,2$ into $a_{ij}=\dfrac{(i+j)^2}{2}$ to build each entry.

$a_{11}=\dfrac{(1+1)^2}{2}=\dfrac{4}{2}=2$

$a_{12}=\dfrac{(1+2)^2}{2}=\dfrac{9}{2}$

$a_{21}=\dfrac{(2+1)^2}{2}=\dfrac{9}{2}$

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. A matrix of order 2\times2 whose elements are given by a_ij = (i+j)²/3 is ................. .$
(a) [[4/3, 3], [3, 16/3]]
(b) [[4/3, 2/3], [2/3, 16/3]]
(c) [[2/3, 3], [3, 13/3]]
(d) [[4/3, 1/3], [2/3, 14/3]]
›Reveal solutionSolution

Substitute $i,j=1,2$ into $a_{ij}=(i+j)^2/3$ to build each entry of the matrix.

For a $2\times2$ matrix, the indices run $i,j\in\{1,2\}$ .

$a_{11}=(1+1)^2/3=4/3$

$a_{12}=(1+2)^2/3=9/3=3$

$a_{21}=(2+1)^2/3=9/3=3$

$a_{22}=(2+2)^2/3=16/3$

…

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CBSE 2025Set X11 markMCQ
$2 \times 2$
(a) $\begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix}$
(b) $\begin{bmatrix} 1 & 0 \\ 3 & 2 \end{bmatrix}$
(c) $\begin{bmatrix} 1 & 1 \\ 2 & 2 \end{bmatrix}$
(d) $\begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix}$
›Reveal solutionSolution

Building a matrix from a rule $a_{ij}=2i-j$ — correct option is (b).

Each entry is computed by substituting its row index $i$ and column index $j$ into $a_{ij}=2i-j$ . This gives $a_{11}=2(1)-1=1$ , $a_{12}=2(1)-2=0$ , $a_{21}=2(2)-1=3$ , and $a_{22}=2(2)-2=2$ . Assembling these e …
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CBSE 2025Set ANNUAL1 mark
$a_{23}$
›Reveal solutionSolution

Substitute the row/column indices directly into $a_{ij}=\frac{|i-j|}{2}$ .

The entry $a_{ij}$ is defined by $a_{ij}=\dfrac{|i-j|}{2}$ .

For $a_{23}$ we take $i=2,\ j=3$ :

$a_{23}=\frac{|2-3|}{2}=\frac{|-1|}{2}=\frac12.$

For $a_{32}$ we take $i=3,\ j=2$ : …

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CBSE 2024Set 65/2/11 markMCQ
$3 \times 3$
›Reveal solutionSolution
A scalar matrix has all diagonal entries equal and all off-diagonal entries zero. With sum of all 9 elements = 9, the diagonal entry is 3, so the product of all elements is $3 \times 3 \times 3 \times 0 \times \dots = 0$ .

Concept & Intuition

A scalar matrix is a special kind of diagonal matrix. In a diagonal matrix, only the entries on the main diagonal can be non-zero; everything else is zero. A scalar matrix goes one step further: all the diagonal entries are the same number. So a $3 \times 3$ scalar matrix looks like this:
$\begin{pmatrix} k & 0 & 0 \\ 0 & k & 0 \\ 0 & 0 & k \end{pmatrix}$
where $k$ is some constant (could be any real number, including zero). The key insight: because there are six zeros in the matrix, any product that includes any of those zeros will be zero — unless every single element is non-zero, which is impossible here. So the product of all nine elements is almost certainly zero, unless the diagonal entry itself is zero (which would also give zero). The only way the product could be non-zero is if there were no zeros at all — but a scalar matrix always has zeros off the diagonal. So the answer must be zero, regardless of $k$ . Let's verify with the given sum condition.

Step-by-step solution

Write the general form of a $3 \times 3$ scalar matrix. Let the common diagonal entry be $k$ . Then the matrix is:

$A = \begin{pmatrix} k & 0 & 0 \\ 0 & k & 0 \\ 0 & 0 & k \end{pmatrix}$

Find the sum of all nine elements. The sum is: $k + 0 + 0 + 0 + k + 0 + 0 + 0 + k = 3k$ . The problem states this sum equals 9, so:

$3k = 9 \quad \Rightarrow \quad k = 3$

Now list all nine elements explicitly. They are: $3, 0, 0, 0, 3, 0, 0, 0, 3$ . …
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CBSE 2024Set 65/2/11 markMCQ
$A = [a_{ij}]$
›Reveal solutionSolution
We construct the $3 \times 3$ matrix $A$ using the given rule $a_{ij} = i - 3j$ and then evaluate each option. The statement $a_{13} > a_{31}$ is found to be false.

When we are given a matrix $A = [a_{ij}]$ , it means that $A$ is composed of elements where $a_{ij}$ refers to the element located at the $i$ -th row and $j$ -th column. The problem defines a $3 \times 3$ matrix, which means it has 3 rows and 3 columns. The indices $i$ and $j$ will therefore range from 1 to 3.

The core idea here is to systematically determine each element of the matrix using the provided formula $a_{ij} = i - 3j$ . Once the matrix elements are known, we can directly check the truthfulness of each given statement.

Understand the Matrix Structure and Element Definition:

A $3 \times 3$ matrix $A$ has elements $a_{ij}$ where $i \in \{1, 2, 3\}$ represents the row number and $j \in \{1, 2, 3\}$ represents the column number.

The rule for each element is given by:
$a_{ij} = i - 3j$

Calculate Each Element of the Matrix:

We will substitute the values of $i$ and $j$ into the formula to find each element:

For the first row ( $i=1$ ):

$a_{11} = 1 - 3(1) = 1 - 3 = -2$

$a_{12} = 1 - 3(2) = 1 - 6 = -5$

$a_{13} = 1 - 3(3) = 1 - 9 = -8$

For the second row ( $i=2$ ):

$a_{21} = 2 - 3(1) = 2 - 3 = -1$

$a_{22} = 2 - 3(2) = 2 - 6 = -4$

$a_{23} = 2 - 3(3) = 2 - 9 = -7$

For the third row ( $i=3$ ):

$a_{31} = 3 - 3(1) = 3 - 3 = 0$

$a_{32} = 3 - 3(2) = 3 - 6 = -3$

$a_{33} = 3 - 3(3) = 3 - 9 = -6$

Construct the Matrix A:

Now we can write down the complete matrix $A$ :

$A = \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{pmatrix} = \begin{pmatrix} -2 & -5 & -8 \\ -1 & -4 & -7 \\ 0 & -3 & -6 \end{pmatrix}$

Evaluate Each Given Option: …
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CBSE 2024Set A11 markMCQ
$A = [a_{ij}]$
(a) $\begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix}$
(b) $\begin{bmatrix} 1 & \frac{1}{2} \\ 2 & 1 \end{bmatrix}$
(c) $\begin{bmatrix} 0 & 2 \\ \frac{1}{2} & 0 \end{bmatrix}$
(d) $\begin{bmatrix} 1 & 2 \\ \frac{1}{2} & 1 \end{bmatrix}$
›Reveal solutionSolution

Evaluate $a_{ij}=i/j$ at each position — the matrix is $\begin{bmatrix}1&1/2\\2&1\end{bmatrix}$ , so (b). …

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CBSE 2024Set ANNUAL1 mark
$2 \times 2$
›Reveal solutionSolution

Compute each entry from the given rule; the alternative equates corresponding entries of two equal matrices.

We construct $A=[a_{ij}]$ of order $2\times2$ with $a_{ij}=\dfrac{(i+2j)^2}{2}$ .

$a_{11}=\dfrac{(1+2\cdot1)^2}{2}=\dfrac{9}{2}, \qquad a_{12}=\dfrac{(1+2\cdot2)^2}{2}=\dfrac{25}{2},$

$a_{21}=\dfrac{(2+2\cdot1)^2}{2}=\dfrac{16}{2}=8, \qquad a_{22}=\dfrac{(2+2\cdot2)^2}{2}=\dfrac{36}{2}=18.$

Hence

$A=\begin{bmatrix} \dfrac{9}{2} & \dfrac{25}{2} \\[6pt] 8 & 18 \end{bmatrix}.$

…

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CBSE 2022Set TERM11 markMCQ
$Q. The element a₂₁ of a 3 \times 2 matrix A = (a_ij) whose a_ij = |i - 3j| / 2 is :$
(a) 5/2
(b) -1/2
(c) 1/2
(d) None of these
›Reveal solutionSolution

Substitute $i=2,\,j=1$ into $a_{ij}=\dfrac{|i-3j|}{2}$ .

…

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CBSE 2022Set ANNUAL1 mark
$2 \times 2$
›Reveal solutionSolution

Evaluate $a_{ij} = \tfrac12|-3i+j|$ for each of the four positions.

For a $2\times 2$ matrix $A = [a_{ij}]$ with $a_{ij} = \dfrac{1}{2}\lvert -3i + j\rvert$ :

$a_{11} = \tfrac12\lvert -3(1)+1\rvert = \tfrac12\lvert -2\rvert = 1,$

$a_{12} = \tfrac12\lvert -3(1)+2\rvert = \tfrac12\lvert -1\rvert = \tfrac12,$

$a_{21} = \tfrac12\lvert -3(2)+1\rvert = \tfrac12\lvert -5\rvert = \tfrac52,$

$a_{22} = \tfrac12\lvert -3(2)+2\rvert = \tfrac12\lvert -4\rvert = 2.$

…

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CBSE 2021Set NC1 mark
$2 \times 2$
›Reveal solutionSolution

Compute each entry $a_{ij}$ directly from the given formula for $i,j=1,2$ .

We need $A=[a_{ij}]_{2\times2}$ with $a_{ij}=\dfrac{(i+j)^2}{2}$ .

$a_{11}=\frac{(1+1)^2}{2}=\frac{4}{2}=2$

$a_{12}=\frac{(1+2)^2}{2}=\frac{9}{2}$

$a_{21}=\frac{(2+1)^2}{2}=\frac{9}{2}$

$a_{22}=\frac{(2+2)^2}{2}=\frac{16}{2}=8$

So

$A=\begin{bmatrix} 2 & \dfrac{9}{2} \\[4pt] \dfrac{9}{2} & 8 \end{bmatrix}$

…

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CBSE 2020Set 65/3/11 mark
$2 \times 2$
›Reveal solutionSolution
To construct a $2 \times 2$ matrix $A = [a_{ij}]$ with elements $a_{ij} = |(i)^2 - j|$ , we calculate each element by substituting its row ( $i$ ) and column ( $j$ ) indices into the given formula. The resulting matrix is $\begin{pmatrix} 0 & 1 \\ 3 & 2 \end{pmatrix}$ .

When we talk about constructing a matrix, especially when its elements are defined by a rule, we are essentially given a blueprint for how each entry in the matrix should be calculated. A matrix is a rectangular array of numbers, and each number's position is uniquely identified by its row and column indices.

For a matrix $A = [a_{ij}]$ , the subscript $i$ always refers to the row number, and $j$ refers to the column number. So, $a_{12}$ means the element in the first row and second column, and $a_{21}$ means the element in the second row and first column. The problem specifies a $2 \times 2$ matrix, which means it has 2 rows and 2 columns. This tells us exactly how many elements we need to calculate and where each one belongs.

The rule $a_{ij} = |(i)^2 - j|$ provides a formula to find the value of any element $a_{ij}$ once we know its row ( $i$ ) and column ( $j$ ). The absolute value function, denoted by $| \cdot |$ , means we take the positive value of the expression inside it. For example, $|-3| = 3$ and $|5| = 5$ .

Let's construct the matrix step by step.

Understand the structure of a $2 \times 2$ matrix: A general $2 \times 2$ matrix $A$ looks like this:

$A = \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix}$
Here, $i$ can take values $1, 2$ (for rows) and $j$ can take values $1, 2$ (for columns). We need to calculate each of these four elements using the given rule $a_{ij} = |(i)^2 - j|$.

2. Calculate the element $a_{11}$ :

For $a_{11}$ , we have $i=1$ and $j=1$ . Substitute these values into the formula:
$a_{11} = |(1)^2 - 1| = |1 - 1| = |0| = 0$

Calculate the element $a_{12}$ : …
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