If aᵢⱼ = (1)/(2)(3i - 2j) and A = [aᵢⱼ]2× 2 is (1) – Class 11 Mathematics

Exercise 7.5 · Q1

a_{ij} = \dfrac{1}{2}(3i - 2j)

(1)

\begin{bmatrix} \dfrac{1}{2} & 2 \\ -\dfrac{1}{2} & 1 \end{bmatrix}

(2)

\begin{bmatrix} \dfrac{1}{2} & -\dfrac{1}{2} \\ 2 & 1 \end{bmatrix}

(3)

\begin{bmatrix} 2 & 2 \\ \dfrac{1}{2} & -\dfrac{1}{2} \end{bmatrix}

(4)

\begin{bmatrix} -\dfrac{1}{2} & \dfrac{1}{2} \\ 1 & 2 \end{bmatrix}

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✓ Free question

Concept understanding — Types of Matrices

A matrix is a rectangular array of numbers (or real-valued functions) arranged in rows and columns, enclosed in square brackets. If a matrix $A$ has $m$ rows and $n$ columns, we say $A$ is of order $m \times n$ , and write $A = [a_{ij}]_{m\times n}$ , $1 \le i \le m,\ 1 \le j \le n$ , where $a_{ij}$ is the entry common to the $i$ th row and $j$ th column.

The book classifies matrices by shape and pattern of entries:

Row matrix: only one row, order $1 \times n$ .
Column matrix: only one column, order $m \times 1$ .
Zero (null/void) matrix $O$ : every entry is $0$ .
Square matrix: number of rows = number of columns, order $n \times n$ . The entries $a_{11}, a_{22}, \ldots, a_{nn}$ form the principal (main/leading) diagonal.
Diagonal matrix: a square matrix in which every off-diagonal entry is $0$ (i.e. $a_{ij}=0$ whenever $i\ne j$ ); the diagonal entries themselves may be anything, including $0$ .
Scalar matrix: a diagonal matrix whose diagonal entries are all equal to the same constant $c$ .
Unit (identity) matrix $I_n$ : a diagonal matrix whose diagonal entries are all $1$ . Every unit matrix is a scalar matrix (with $c=1$ ), and every square zero matrix is a (trivial) scalar/diagonal matrix.
Upper triangular matrix: a square matrix with every entry below the main diagonal equal to $0$ , i.e. $a_{ij}=0$ for all $i>j$ .
Lower triangular matrix: a square matrix with every entry above the main diagonal equal to $0$ , i.e. $a_{ij}=0$ for all $i<j$ .
Triangular matrix: a matrix that is either upper or lower triangular. A matrix that is simultaneously upper and lower triangular is necessarily a diagonal matrix.

Note

Matrices are commonly represented by capital letters $A, B, C, \ldots$ . In this chapter every entry is a real number or a real-valued function of real variables. Constructing a matrix from a rule $a_{ij} = f(i,j)$ simply means substituting every valid pair $(i,j)$ into $f$ and arranging the results in the $m\times n$ grid.

Substitute each pair $(i,j)$ into the rule $a_{ij} = \tfrac{1}{2}(3i - 2j)$ .

This question tests constructing a matrix from a general element formula.

Step 1. First row ( $i=1$ ): $a_{11} = \tfrac{1}{2}(3\cdot1 - 2\cdot1) = \tfrac{1}{2}(1) = \tfrac{1}{2}$ and $a_{12} = \tfrac{1}{2}(3\cdot1 - 2\cdot2) = \tfrac{1}{2}(-1) = -\tfrac{1}{2}$ .

Step 2. Second row ( $i=2$ ): $a_{21} = \tfrac{1}{2}(3\cdot2 - 2\cdot1) = \tfrac{1}{2}(4) = 2$ and $a_{22} = \tfrac{1}{2}(3\cdot2 - 2\cdot2) = \tfrac{1}{2}(2) = 1$ .

Step 3. Assemble: $A = \begin{bmatrix} 1/2 & -1/2 \\ 2 & 1 \end{bmatrix}$ .

✓Final answer

\begin{bmatrix} 1/2 & -1/2 \\ 2 & 1 \end{bmatrix}

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

CBSE 2023Set ANNUAL1 markMCQ
$\begin{bmatrix}1 & 0 & 0\\0 & 0 & 0\\0 & 0 & 5\end{bmatrix}$
(a) an upper triangular matrix
(b) a scalar matrix
(c) a lower triangular matrix
(d) a diagonal matrix
›Reveal solutionSolution
The matrix has all off-diagonal entries zero (so it IS diagonal, upper-triangular, and lower-triangular), but its diagonal entries $1,0,5$ are unequal, so it is NOT a scalar matrix.

Given matrix: $\begin{bmatrix}1&0&0\\0&0&0\\0&0&5\end{bmatrix}$ .

Upper triangular: all entries below the main diagonal are 0. True here.

Lower triangular: all entries above the main diagonal are 0. True here.

Diagonal matrix: all off-diagonal entries are 0 (both triangular). True here.

Scalar matrix: a diagonal matrix where every diagonal entry is the SAME constant $k$ . Here the diagonal is $1,0,5$ — not equal, so this is false.

✓Final answer
$"A scalar matrix" is the statement that is NOT true.$
CBSE 2022Set M1 mark
$Q. Define a diagonal matrix.$
›Reveal solutionSolution

A diagonal matrix is a square matrix with all off-diagonal entries zero.

A square matrix $A=[a_{ij}]$ is called a diagonal matrix if all its non-diagonal elements are zero, i.e. $a_{ij}=0$ whenever $i\neq j$ .

✓Final answer
$0$
CBSE 2022Set ANNUAL1 mark
$Q. Define an identity matrix.$
›Reveal solutionSolution

State the definition of an identity matrix.

A square matrix $A=[a_{ij}]_{n\times n}$ is called an identity matrix if

$a_{ij}=\begin{cases}1, & i=j\\ 0, & i e j\end{cases}$

i.e. all diagonal entries are $1$ and all non-diagonal entries are $0$ . It is denoted $I_n$ (or simply $I$ ). For example,

$I_3=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$

✓Final answer
$I_n$
CBSE 2022Set ANNUAL1 markMCQ
$A = \begin{bmatrix} 2 & 3 \\ a & 6 \end{bmatrix}$
(a) 6
(b) $-5$
(c) 3
(d) 4
›Reveal solutionSolution

A matrix is singular exactly when $\det A = 0$ . For $A = \begin{bmatrix} 2 & 3 \\ a & 6 \end{bmatrix}$ this forces $12 - 3a = 0$ , i.e. $a = 4$ .

For a $2\times 2$ matrix $\begin{bmatrix} p & q \\ r & s \end{bmatrix}$ the determinant is $ps - qr$ .

Here

$\det A = (2)(6) - (3)(a) = 12 - 3a.$

The matrix is singular, which means its determinant vanishes:

$12 - 3a = 0.$

Solving, $3a = 12$ , so $a = 4$ .

✓Final answer
$a = 4$
CBSE 2019Set ANNUAL1 markMCQ
$\begin{bmatrix}1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 5\end{bmatrix}$
(a) an upper triangular matrix
(b) a lower triangular matrix
(c) a scalar matrix
(d) a diagonal matrix
›Reveal solutionSolution

The matrix has zero everywhere off the main diagonal, making it diagonal (and so also upper- and lower-triangular), but its diagonal entries $1,0,5$ are unequal, so it is NOT a scalar matrix.

The matrix is $\begin{bmatrix}1&0&0\\0&0&0\\0&0&5\end{bmatrix}$ .

A matrix is a diagonal matrix if every off-diagonal entry is 0 — true here, since all entries except the $(1,1)$ , $(2,2)$ , $(3,3)$ positions are already 0. This makes (d) true.

A diagonal matrix is automatically both upper triangular (nothing below the diagonal) and lower triangular (nothing above the diagonal), making (a) and (b) both true.

A scalar matrix additionally requires all diagonal entries to be equal to the same constant $k$ . Here the diagonal entries are $1, 0, 5$ — not all equal — so it is NOT a scalar matrix, making (c) the false (i.e. "not true") statement.

✓Final answer
$The correct option is (c) a scalar matrix.$
CBSE 2018Set ANNUAL1 mark
$Q. Define identity matrix.$
›Reveal solutionSolution

definition recall

The identity matrix $I_n$ of order $n$ is the square matrix in which every diagonal element is $1$ and every off-diagonal element is $0$ , i.e. $I=[a_{ij}]$ where $a_{ij}=1$ if $i=j$ and $a_{ij}=0$ if $i\ne j$ . It satisfies $AI=IA=A$ for any compatible matrix $A$ .

✓Final answer
$1$

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Q.If aij=12(3i−2j)a_{ij} = \dfrac{1}{2}(3i - 2j)aij​=21​(3i−2j) and A=[aij]2×2A = [a_{ij}]_{2\times 2}A=[aij​]2×2​ is

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Q.If $a_{ij} = \dfrac{1}{2}(3i - 2j)$ and $A = [a_{ij}]_{2\times 2}$ is

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