Construct a $2 \times 2$ matrix where (i) $a_{ij} = \dfrac{(i-2j)^2}{2}$ (ii) $a_{ij} = |-2i+3j|$

Construct a 2 × 2 matrix where (i) aᵢⱼ = ((i-2j)²)/(2) (ii) aᵢⱼ = |-2i+3j|

NCERT Exemplar · Q3

2 \times 2

(i)

a_{ij} = \dfrac{(i-2j)^2}{2}

(ii)

a_{ij} = |-2i+3j|

Andhra Pradesh BieapShort· 2mImportance★★★★★

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✓ Free question

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}.

Watch out

Do not confuse $a_{ij}$ with $a_{ji}$ . The first index is always the row, the second the column — so $a_{23}$ is row $2$ , column $3$ .

Important

A matrix is not just a set of numbers — it is an ordered arrangement. The same numbers placed differently give a different matrix. When a problem says "construct $A = [a_{ij}]$ where $a_{ij} = \dots$ ", fix the dimensions first, then fill the cells one by one using the rule.

Constructing a matrix from a given formula for its entries, such as aᵢⱼ = i² − j, is a standard NCERT exercise type in the CBSE Class 12 Matrices chapter, and "construct a 3x3 matrix whose elements are given by formula" is a frequently searched question format. This skill is regularly tested in board exams as a straightforward, formula-substitution-based question.

Evaluate each formula at $(i,j)=(1,1),(1,2),(2,1),(2,2)$ . Part (i): $\begin{pmatrix} \tfrac{1}{2} & \tfrac{9}{2} \\0 & 2 \end{pmatrix}$ . Part (ii): $\begin{pmatrix} 1 & 4 \\1 & 2 \end{pmatrix}$ .

A $2\times 2$ matrix has entries $a_{ij}$ where $i$ is the row ( $1,2$ ) and $j$ is the column ( $1,2$ ). We simply substitute each $(i,j)$ into the given rule.

Part (i): $a_{ij}=\dfrac{(i-2j)^2}{2}$

a_{11}=\frac{(1-2)^2}{2}=\frac{1}{2},\qquad a_{12}=\frac{(1-4)^2}{2}=\frac{9}{2},

a_{21}=\frac{(2-2)^2}{2}=0,\qquad a_{22}=\frac{(2-4)^2}{2}=\frac{4}{2}=2.

A=\begin{pmatrix} \tfrac{1}{2} & \tfrac{9}{2} \\[2pt] 0 & 2 \end{pmatrix}.

Part (ii): $a_{ij}=|-2i+3j|$

a_{11}=|-2+3|=1,\qquad a_{12}=|-2+6|=4,

a_{21}=|-4+3|=|-1|=1,\qquad a_{22}=|-4+6|=2.

A=\begin{pmatrix} 1 & 4 \\1 & 2 \end{pmatrix}.

✓Final answer

\begin{pmatrix} \tfrac{1}{2} & \tfrac{9}{2} \\[2pt] 0 & 2 \end{pmatrix}

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

AP EAPCET 2023Set eng-2023-05-16-AN1 markMCQ
$a_{ij}$
›Reveal solutionSolution
Plugging $i,j\in\{1,2,3\}$ into $a_{ij}=\frac13|i-5j|$ term by term reproduces the matrix in option (B) exactly.

Concept and Intuition

This is a direct construction problem: build the matrix entry-by-entry from the given rule and match against the options, watching the fractions carefully (many wrong options differ only by a dropped $\frac13$ or a sign/arithmetic slip in a single entry).

Step-by-Step Solution

$i=1$ : $a_{11}=\frac13|1-5|=\frac43$ , $a_{12}=\frac13|1-10|=\frac93=3$ , $a_{13}=\frac13|1-15|=\frac{14}3$ .

$i=2$ : $a_{21}=\frac13|2-5|=1$ , $a_{22}=\frac13|2-10|=\frac83$ , $a_{23}=\frac13|2-15|=\frac{13}3$ .

$i=3$ : $a_{31}=\frac13|3-5|=\frac23$ , $a_{32}=\frac13|3-10|=\frac73$ , $a_{33}=\frac13|3-15|=\frac{12}3=4$ .

Assembling: $\begin{bmatrix}\frac43&3&\frac{14}3\\1&\frac83&\frac{13}3\\\frac23&\frac73&4\end{bmatrix}$ — this is exactly option (B).

Common Mistakes

Dropping the $\frac13$ factor on some entries (as options A, D do).

Arithmetic slip in $|1-15|=14$ vs. miscounting as $10$ (option C's error).

✓Final answer
$\begin{bmatrix}\frac43&3&\frac{14}3\\1&\frac83&\frac{13}3\\\frac23&\frac73&4\end{bmatrix}$

ANSWER: B
AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
$A = \begin{bmatrix} x & y & 0 \\ -3 & 1 & 2 \\ 1 & -2 & z \end{bmatrix}$
›Reveal solutionSolution
This tests cofactor expansion to recover unknown matrix entries, then direct matrix multiplication. The answer is (C).

Concept and Intuition

A cofactor $C_{ij}=(-1)^{i+j}M_{ij}$ where $M_{ij}$ is the minor obtained by deleting row $i$ and column $j$ . Being told three cofactor values turns the problem into three simple linear equations for $x,y,z$ — once those are known, $AB$ is pure computation.

Step-by-Step Solution

Matrix: $A=\begin{bmatrix}x&y&0\\-3&1&2\\1&-2&z\end{bmatrix}$ .

Cofactor of $z$ (entry $(3,3)$ ): delete row 3, column 3 $\Rightarrow$ minor $=\begin{vmatrix}x&y\\-3&1\end{vmatrix}=x+3y$ . Sign $(-1)^{3+3}=+1$ . Given $=9$ :

$x+3y=9.$

Cofactor of "1" in row 3 — the entry "1" sits at position $(3,1)$ . Delete row 3, column 1 $\Rightarrow$ minor $=\begin{vmatrix}y&0\\1&2\end{vmatrix}=2y$ . Sign $(-1)^{3+1}=+1$ . Given $=4$ :

$2y=4\Rightarrow y=2.$
Substituting into step 2: $x+6=9\Rightarrow x=3$ .

4. Cofactor of $x$ (entry $(1,1)$ ): delete row 1, column 1 $\Rightarrow$ minor $=\begin{vmatrix}1&2\\-2&z\end{vmatrix}=z+4$ . Sign $(-1)^{1+1}=+1$ . Given $=3$ :
$z+4=3\Rightarrow z=-1.$

So $A=\begin{bmatrix}3&2&0\\-3&1&2\\1&-2&-1\end{bmatrix}$ , $B=\begin{bmatrix}1&-2&-2\\2&0&1\\2&1&0\end{bmatrix}$ .

Multiply row by row:

Row 1: $(3\cdot1+2\cdot2+0\cdot2,\ 3\cdot(-2)+2\cdot0+0\cdot1,\ 3\cdot(-2)+2\cdot1+0\cdot0) = (7,-6,-4)$ .

Row 2: $(-3\cdot1+1\cdot2+2\cdot2,\ -3\cdot(-2)+1\cdot0+2\cdot1,\ -3\cdot(-2)+1\cdot1+2\cdot0) = (3,8,7)$ .

Row 3: $(1\cdot1+(-2)\cdot2+(-1)\cdot2,\ 1\cdot(-2)+(-2)\cdot0+(-1)\cdot1,\ 1\cdot(-2)+(-2)\cdot1+(-1)\cdot0) = (-5,-3,-4)$ .

So $AB=\begin{bmatrix}7&-6&-4\\3&8&7\\-5&-3&-4\end{bmatrix}$ .

Common Mistakes

Mixing up which entry "1 in the 3rd row" refers to — the 3rd row is $(1,-2,z)$ , so the plain "1" is the $(3,1)$ entry, not $(3,2)$ .

Forgetting the sign $(-1)^{i+j}$ when the minor position has odd $i+j$ (not an issue here since all three positions used have even $i+j$ , but always check).

Arithmetic slips while multiplying — recompute each dot product independently.

✓Final answer
$\begin{bmatrix} 7 & -6 & -4 \\ 3 & 8 & 7 \\ -5 & -3 & -4 \end{bmatrix}$

ANSWER: C
AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
$-1$
›Reveal solutionSolution
A row-swap is a determinant-negating, entry-preserving involution — so it pairs up every det- $1$ matrix with a unique det- $(-1)$ matrix, forcing $|B|=|C|$ .

Concept and Intuition

Rather than trying to count $B$ and $C$ directly (hard — most $0/1$ matrices have determinant $0$ , and enumerating the nonzero-determinant ones by brute force is tedious), look for a structure-preserving bijection between them. Swapping two rows of any matrix is a classical determinant-sign-flipping operation, and critically it never changes what the entries are (only their row positions) — so a $0/1$ matrix stays a $0/1$ matrix after the swap.

Step-by-Step Solution

Let $M \in B$ , so $M$ is a $3\times3$ matrix with entries in $\{0,1\}$ and $\det(M)=1$ .

Define $\sigma(M)$ = the matrix obtained by swapping (say) rows 1 and 2 of $M$ .

Swapping two rows of any matrix multiplies its determinant by $-1$ : $\det(\sigma(M)) = -\det(M) = -1$ . So $\sigma(M)\in C$ .

Since $\sigma(M)$ only rearranges $M$ 's entries (no entry is changed, only relocated), $\sigma(M)$ is still a valid $0/1$ matrix.

$\sigma$ is an involution: swapping rows 1,2 twice returns the original matrix, $\sigma(\sigma(M))=M$ . So $\sigma:B\to C$ is a bijection (it has an inverse, namely itself).

A bijection between two finite sets means they have exactly the same cardinality: $|B|=|C|$ .

This directly rules out (D) (which claims $|B|=2|C|$ ) and (A) (since most $0/1$ matrices have $\det=0$ , so $A \ne B\cup C$ ), and (B) is false since $\sigma$ explicitly produces elements of $C$ (e.g. swap two rows of the identity matrix, $\det=1$ , to get a matrix with $\det=-1$ ), so $C$ is non-empty.

Common Mistakes

Assuming that because "det $=1$ " is somehow more "special" than "det $=-1$ ," there ought to be fewer or more such matrices — the row-swap argument shows the two counts are forced to be exactly equal, with no need to compute either count.

Trying to brute-force enumerate all $2^9=512$ matrices instead of spotting the symmetry argument.

✓Final answer
$The correct option is (C) — B and C contain the same number of elements.$

ANSWER: C

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Q.Construct a $2 \times 2$ matrix where

Concept understanding — Matrix Construction

Matrix Construction: Building a Grid of Numbers

Why a Grid?

The Precise Form

How You Construct One

Part (i): $a_{ij}=\dfrac{(i-2j)^2}{2}$

Part (ii): $a_{ij}=|-2i+3j|$

Method: Constructing a matrix from a formula $a_{ij}=f(i,j)$

Steps

Common Mistakes

Concept and Intuition

Step-by-Step Solution

Common Mistakes

Concept and Intuition

Step-by-Step Solution

Common Mistakes

Concept and Intuition

Step-by-Step Solution

Common Mistakes

Q.Construct a 2×22 \times 22×2 matrix where

Part (i): aij=(i−2j)22a_{ij}=\dfrac{(i-2j)^2}{2}aij​=2(i−2j)2​

Part (ii): aij=∣−2i+3j∣a_{ij}=|-2i+3j|aij​=∣−2i+3j∣

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Q.Construct a $2 \times 2$ matrix where

Part (i): $a_{ij}=\dfrac{(i-2j)^2}{2}$

Part (ii): $a_{ij}=|-2i+3j|$

Unlock everything free for 14 days