Q.If the sum of all the elements of a $3 \times 3$ scalar matrix is 9, then the product of all its elements is: (A) $0$ (B) $9$ (C) $27$ (D) $729$

Showing the 12 most recent of 16 on this concept.

1. CBSE 2026Set ANNUAL1 mark
   Q.Construct a $2 \times 2$ matrix $A$, $A = [a_{ij}]$, where $a_{ij} = \frac{(i+j)^2}{2}$.

   ›Reveal solutionSolution

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

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

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

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

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

   ✓Final answer

   $A=\begin{bmatrix}2 & \dfrac{9}{2}\\ \dfrac{9}{2} & 8\end{bmatrix}$

2. CBSE 2026Set ANNUAL1 markMCQ
   Q.A matrix of order 2×2 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 $i,j\in\{1,2\}$.

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

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

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

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

   So $A=\begin{bmatrix}4/3 & 3\\ 3 & 16/3\end{bmatrix}$.

   ✓Final answer

   Option (a): $A=\begin{bmatrix}4/3 & 3\\ 3 & 16/3\end{bmatrix}$.

3. CBSE 2025Set X11 markMCQ
   Q.For a $2 \times 2$ matrix $A = [a_{ij}]$ whose elements are given by $a_{ij} = 2i - j$ then $A$ is equal to

   (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 entries yields $A=\begin{bmatrix}1&0\\3&2\end{bmatrix}$.

   ✓Final answer

   (b) $\begin{bmatrix} 1 & 0 \\ 3 & 2 \end{bmatrix}$

4. CBSE 2025Set ANNUAL1 mark
   Q.Write the elements $a_{23}$ and $a_{32}$ of a $3 \times 3$ matrix $A = [a_{ij}]$, whose elements $a_{ij}$ are given by $a_{ij} = \dfrac{|i-j|}{2}$.

   ›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 $i=3,\ j=2$:

   $a_{32}=\frac{|3-2|}{2}=\frac{|1|}{2}=\frac12.$

   (Because $|i-j|=|j-i|$, these two symmetric entries are necessarily equal.)

   ✓Final answer

   $a_{23}=\dfrac12$ and $a_{32}=\dfrac12$

5. CBSE 2024Set 65/2/11 markMCQ
   Q.If the sum of all the elements of a $3 \times 3$ scalar matrix is 9, then the product of all its elements is: (A) $0$ (B) $9$ (C) $27$ (D) $729$

   ›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

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

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

   2. 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:

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

   3. Now list all nine elements explicitly.

      They are: $3, 0, 0, 0, 3, 0, 0, 0, 3$.

   4. Compute the product of all nine elements.

      The product is $3 \times 0 \times 0 \times 0 \times 3 \times 0 \times 0 \times 0 \times 3$.

      Since multiplication by zero gives zero, the entire product is $0$.

   Watch out

   A common mistake is to think only of the diagonal entries and multiply $3 \times 3 \times 3 = 27$, forgetting the six zeros that are also part of the matrix. The product of all elements includes every entry, not just the diagonal.

   Tip

   For any scalar matrix of size $n \times n$ where $n \ge 2$, the product of all its elements is always $0$, because there is at least one zero off the diagonal. The sum condition here only confirms the diagonal value, but doesn't change the zero product.

   ✓Final answer

   The product of all its elements is $\boxed{0}$, which corresponds to option (A).

6. CBSE 2024Set 65/2/11 markMCQ
   Q.If $A = [a_{ij}]$ be a $3 \times 3$ matrix, where $a_{ij} = i - 3j$, then which of the following is $false$? (A) $a_{11} < 0$ (B) $a_{12} + a_{21} = -6$ (C) $a_{13} > a_{31}$ (D) $a_{31} = 0$

   ›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.

   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:

      $a_{ij} = i - 3j$

   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 Given Option:

      We will check each statement against the calculated values. The question asks for the false statement.

      • (A) $a_{11} < 0$

        From our calculations, $a_{11} = -2$.

        Is $-2 < 0$? Yes, this statement is true.

      • (B) $a_{12} + a_{21} = -6$

        From our calculations, $a_{12} = -5$ and $a_{21} = -1$.

        Is $-5 + (-1) = -6$? Yes, $-6 = -6$, so this statement is true.

      • (C) $a_{13} > a_{31}$

        From our calculations, $a_{13} = -8$ and $a_{31} = 0$.

        Is $-8 > 0$? No, $-8$ is less than $0$. This statement is false.

      • (D) $a_{31} = 0$

        From our calculations, $a_{31} = 0$.

        Is $0 = 0$? Yes, this statement is true.

   5. Identify the False Statement:

      Based on the evaluation, option (C) is the false statement.

   ✓Final answer

   The false statement is (C) $a_{13} > a_{31}$.

7. CBSE 2024Set A11 markMCQ
   Q.If $A = [a_{ij}]$ is a $2 \times 2$ matrix whose elements are given by $a_{ij} = \dfrac{i}{j}$ then $A$ is

   (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).

   Using $a_{ij}=\dfrac{i}{j}$: $a_{11}=\tfrac{1}{1}=1$, $a_{12}=\tfrac{1}{2}$, $a_{21}=\tfrac{2}{1}=2$, $a_{22}=\tfrac{2}{2}=1$. Hence $A=\begin{bmatrix}1&\tfrac12\\2&1\end{bmatrix}$.

   ✓Final answer

   (b) $\begin{bmatrix} 1 & \frac{1}{2} \\ 2 & 1 \end{bmatrix}$

8. CBSE 2024Set ANNUAL1 mark
   Q.Construct a $2 \times 2$ matrix $A = [a_{ij}]$, where $a_{ij} = \dfrac{(i+2j)^2}{2}$. OR Find the values of $x$ and $y$, if $\begin{bmatrix} 3x+y & -y \\ 2x-y & 3 \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 4 & 3 \end{bmatrix}$

   ›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

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

   ✓Final answer

   $A = \begin{bmatrix} \dfrac{9}{2} & \dfrac{25}{2} \\[4pt] 8 & 18 \end{bmatrix}$

   Alternative (Or):

   Two equal matrices have equal corresponding entries; solve the resulting equations.

   Given $\begin{bmatrix} 3x+y & -y \\ 2x-y & 3 \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 4 & 3 \end{bmatrix}$.

   Equating corresponding entries:

   $-y = 2 \implies y = -2.$

   $3x + y = 1 \implies 3x - 2 = 1 \implies 3x = 3 \implies x = 1.$

   Check with the entry $2x-y$: $2(1)-(-2) = 2+2 = 4$ ✓ (matches).

   ✓Final answer

   $\boldsymbol{x = 1,\ y = -2}$

9. CBSE 2022Set TERM11 markMCQ
   Q.The element a₂₁ of a 3 × 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 $a_{ij}=\dfrac{|i-3j|}{2}$.

   $a_{21}=\dfrac{|2-3(1)|}{2}=\dfrac{|2-3|}{2}=\dfrac{|-1|}{2}=\dfrac12$.

   ✓Final answer

   $a_{21}=\dfrac12$ — option (c).

10. CBSE 2022Set ANNUAL1 mark
    Q.Construct a $2 \times 2$ matrix whose elements are given by $a_{ij} = \dfrac{1}{2}|-3i+j|$ OR Find $AB$, if $A = \begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix}$

    ›Reveal solutionSolution

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

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

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

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

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

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

    ✓Final answer

    $A = \begin{bmatrix} 1 & \tfrac12 \\[2pt] \tfrac52 & 2 \end{bmatrix}.$

    Alternative (Or):

    Multiply the two $2\times 2$ matrices row-by-column.

    With $A = \begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix}$,

    $AB = \begin{bmatrix} 2(1)+4(-2) & 2(3)+4(5) \\ 3(1)+2(-2) & 3(3)+2(5) \end{bmatrix} = \begin{bmatrix} 2-8 & 6+20 \\ 3-4 & 9+10 \end{bmatrix} = \begin{bmatrix} -6 & 26 \\ -1 & 19 \end{bmatrix}.$

    ✓Final answer

    $AB = \begin{bmatrix} -6 & 26 \\ -1 & 19 \end{bmatrix}.$

11. CBSE 2021Set NC1 mark
    Q.Construct a $2 \times 2$ matrix $A = [a_{ij}]$, whose elements are given by $a_{ij} = \dfrac{(i+j)^2}{2}$ OR Find the value of $AB$ when $A = [1\ 2\ 3\ 4]$ and $B = \begin{bmatrix} 1 \\ 2 \\ 3 \\ 4 \end{bmatrix}$

    ›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

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

    ✓Final answer

    $A=\begin{bmatrix}2 & 9/2\\ 9/2 & 8\end{bmatrix}$

    Alternative (Or): Find $AB$ when $A=[1\ 2\ 3\ 4]$ and $B=\begin{bmatrix}1\\2\\3\\4\end{bmatrix}$.

    $A$ is a $1\times4$ row matrix and $B$ is a $4\times1$ column matrix, so $AB$ is a well-defined $1\times1$ matrix -- the dot product of the two lists of entries.

    $AB = [1\ 2\ 3\ 4]\begin{bmatrix}1\\2\\3\\4\end{bmatrix} = [\,1(1)+2(2)+3(3)+4(4)\,] = [\,1+4+9+16\,] = [30]$

    ✓Final answer

    $AB=[30]$ (a $1\times1$ matrix)

12. CBSE 2020Set 65/3/11 mark
    Q.Construct a $2 \times 2$ matrix $A = [a_{ij}]$ whose elements are given by $a_{ij} = |(i)^2 - j|$.

    ›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.

    1. 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:

    $a_{11} = |(1)^2 - 1| = |1 - 1| = |0| = 0$

    3. Calculate the element $a_{12}$: For $a_{12}$, we have $i=1$ and $j=2$. Substitute these values into the formula:

    $a_{12} = |(1)^2 - 2| = |1 - 2| = |-1| = 1$

    4. Calculate the element $a_{21}$: For $a_{21}$, we have $i=2$ and $j=1$. Substitute these values into the formula:

    $a_{21} = |(2)^2 - 1| = |4 - 1| = |3| = 3$

    5. Calculate the element $a_{22}$: For $a_{22}$, we have $i=2$ and $j=2$. Substitute these values into the formula:

    $a_{22} = |(2)^2 - 2| = |4 - 2| = |2| = 2$

    6. Assemble the matrix: Now that we have calculated all four elements, we place them into their respective positions in the $2 \times 2$ matrix:

    $A = \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 3 & 2 \end{pmatrix}$

    ✓Final answer

    The $2 \times 2$ matrix $A$ whose elements are given by $a_{ij} = |(i)^2 - j|$ is $\boxed{\begin{pmatrix} 0 & 1 \\ 3 & 2 \end{pmatrix}}$.