Q.If $\begin{vmatrix} 3 & -4 \\ m & 5 \end{vmatrix} = 3$, then value of $m$ is:

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1. CBSE 2026Set 65/2/11 markMCQ
   Q.If $\begin{vmatrix} -1 & -2 & 5 \\ -2 & a & -1 \\ 0 & 4 & 2a \end{vmatrix} = -86$, then the sum of all possible values of $a$ is (A) 4 (B) 5 (C) -4 (D) 9

   ›Reveal solutionSolution

   Expand the determinant along the first column, set it equal to $-86$, and solve the resulting quadratic. The sum of roots is -4.

   When a determinant equals a specific value, we compute the determinant algebraically (treating any unknowns as variables), then solve the resulting equation. The determinant of a $3 \times 3$ matrix can be found by cofactor expansion along any row or column; choosing the column or row with the most zeros minimizes arithmetic.

   Here the first column has a zero in position $(3,1)$, so expanding along the first column is efficient.

   Solution

   1. Expand along the first column

      The determinant is:

   $\begin{vmatrix} -1 & -2 & 5 \\ -2 & a & -1 \\ 0 & 4 & 2a \end{vmatrix} = (-1) \cdot \begin{vmatrix} a & -1 \\ 4 & 2a \end{vmatrix} - (-2) \cdot \begin{vmatrix} -2 & 5 \\ 4 & 2a \end{vmatrix} + 0 \cdot \begin{vmatrix} -2 & 5 \\ a & -1 \end{vmatrix}$

   The signs alternate: $+, -, +$ down the column, and we multiply each by the element in that position.

   2. Compute the $2 \times 2$ determinants

      For the first minor:

   $\begin{vmatrix} a & -1 \\ 4 & 2a \end{vmatrix} = a(2a) - (-1)(4) = 2a^2 + 4$

   For the second minor:

   $\begin{vmatrix} -2 & 5 \\ 4 & 2a \end{vmatrix} = (-2)(2a) - (5)(4) = -4a - 20$

   3. Substitute back

      $\text{Det} = (-1)(2a^2 + 4) + 2(-4a - 20)$ …

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2. CBSE 2026Set A1 markMCQ
   Q.$\begin{vmatrix} x & 4 \\ 2 & 2x \end{vmatrix} = 0 \Rightarrow x =$

   (a) $\pm 2$

   (b) $\pm 1$

   (c) $\pm 3$

   (d) $0$

   ›Reveal solutionSolution

   Expanding the determinant: $2x^2-8=0$, so $x=\pm 2$.

   Expand:

   $\begin{vmatrix} x & 4 \\ 2 & 2x \end{vmatrix} = x\cdot 2x - 4\cdot 2 = 2x^2 - 8.$

   Set equal to $0$: …

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3. CBSE 2026Set ANNUAL1 markMCQ
   Q.If $\begin{vmatrix} x & 3 \\ 8 & x \end{vmatrix} = \begin{vmatrix} 6 & 2 \\ 18 & 2 \end{vmatrix}$ then $x =$

   (a) $\sqrt{24}$

   (b) $-\sqrt{24}$

   (c) $\pm\sqrt{24}$

   (d) None of these

   ›Reveal solutionSolution

   Expand both $2\times2$ determinants and equate; the resulting value of $x$ does not match the listed options.

   LHS: $\begin{vmatrix} x & 3 \\ 8 & x \end{vmatrix} = x^2 - 24$

   RHS: $\begin{vmatrix} 6 & 2 \\ 18 & 2 \end{vmatrix} = 6(2) - 2(18) = 12-36 = -24$

   Setting LHS = RHS:

   $x^2 - 24 = -24$

   $x^2 = 0$

   …

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4. CBSE 2026Set ANNUAL1 markMCQ
   Q.$\begin{vmatrix} 2 & 4 \\ 5 & 1 \end{vmatrix} = \begin{vmatrix} 2x & 4 \\ 6 & x \end{vmatrix}$, the possible value of $x$ is/are:

   (a) 3

   (b) $\sqrt{3}$

   (c) $-\sqrt{3}$

   (d) $\sqrt{3}, -\sqrt{3}$

   ›Reveal solutionSolution

   Evaluate both determinants and equate them to solve for $x$.

   LHS: $\begin{vmatrix}2 & 4\\ 5 & 1\end{vmatrix}=2(1)-4(5)=2-20=-18$

   RHS: $\begin{vmatrix}2x & 4\\ 6 & x\end{vmatrix}=2x(x)-4(6)=2x^2-24$

   …

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5. CBSE 2026Set ANNUAL1 markMCQ
   Q.If |x 0; 1 x| = |16 0; 8 4| (2×2 determinants) then value of x is:

   (a) 3

   (b) 2

   (c) 4

   (d) 8

   ›Reveal solutionSolution

   Expand both $2\times2$ determinants and equate them to get $x^2 = 64$.

   For a $2\times2$ determinant $\begin{vmatrix}a & b\\ c& d\end{vmatrix} = ad-bc$.

   Left side: $\begin{vmatrix}x & 0\\ 1 & x\end{vmatrix} = x\cdot x - 0\cdot 1 = x^2$

   …

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6. CBSE 2026Set ANNUAL1 markMCQ
   Q.If the determinant \begin{vmatrix}2x & 4\ 2 & 1\end{vmatrix} = 0, then the value of x will be:

   (a) 2

   (b) 4

   (c) 6

   (d) 8

   ›Reveal solutionSolution

   Expand the $2\times2$ determinant and solve the resulting linear equation for $x$.

   Working:

   $\begin{vmatrix}2x & 4\\ 2 & 1\end{vmatrix} = (2x)(1) - (4)(2) = 2x - 8$

   …

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7. CBSE 2026Set ANNUAL1 markMCQ
   Q.If $\begin{vmatrix} 3 & x \\ x & 1 \end{vmatrix} = \begin{vmatrix} 3 & 2 \\ 4 & 1 \end{vmatrix}$, then the value of $x$ is

   (a) $\pm 2\sqrt{2}$

   (b) $\pm\sqrt{2}$

   (c) 2

   (d) -2

   ›Reveal solutionSolution

   Expand both $2\times2$ determinants and equate, then solve the resulting quadratic in $x$.

   Left-hand side:

   $\begin{vmatrix} 3 & x \\ x & 1 \end{vmatrix} = 3(1) - x(x) = 3 - x^2$

   Right-hand side: …

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8. CBSE 2025Set E1 markMCQ
   Q.$\begin{vmatrix} x & 15 \\ 4 & 4 \end{vmatrix} = 0 \ \Rightarrow x =$

   (a) $15$

   (b) $-15$

   (c) $12$

   (d) $60$

   ›Reveal solutionSolution

   Expand the determinant, set it to zero and solve for $x$; $x=15$.

   $\begin{vmatrix} x & 15 \\ 4 & 4 \end{vmatrix}=(x)(4)-(15)(4)=4x-60.$

   …

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9. CBSE 2025Set A1 markMCQ
   Q.If $\begin{vmatrix}1 & x \\ x & 1\end{vmatrix} = \begin{vmatrix}0 & 2 \\ 1 & 2\end{vmatrix}$, then the value of $x$ is:

   (a) $0$

   (b) $\pm 1$

   (c) $\pm\sqrt{3}$

   (d) $\pm\sqrt{2}$

   ›Reveal solutionSolution

   Expand both $2\times2$ determinants and equate.

   Left side: $\begin{vmatrix}1 & x \\ x & 1\end{vmatrix} = 1(1) - x(x) = 1 - x^2$

   Right side: $\begin{vmatrix}0 & 2 \\ 1 & 2\end{vmatrix} = 0(2) - 2(1) = -2$

   …

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10. CBSE 2025Set ANNUAL1 mark
    Q.If $\begin{vmatrix}2 & 1\\ 1 & 2\end{vmatrix}-k\begin{vmatrix}1 & 0\\ 0 & 1\end{vmatrix}=0$, then $k=$ _____.

    ›Reveal solutionSolution

    Evaluate both determinants and solve the resulting linear equation for $k$.

    …

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11. CBSE 2025Set ANNUAL1 markMCQ
    Q.The value of $x$ for which the matrix $A = \begin{bmatrix} x & 2 \\ 2 & 4 \end{bmatrix}$ is a singular matrix, is

    (a) $1$

    (b) $0$

    (c) $-1$

    (d) $2$

    ›Reveal solutionSolution

    A singular matrix has determinant zero; set |A| = 0 and solve for x.

    $A = \begin{bmatrix} x & 2 \\ 2 & 4 \end{bmatrix}$

    $|A| = x(4) - 2(2) = 4x - 4$

    …

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12. CBSE 2025Set ANNUAL1 markMCQ
    Q.If $\begin{vmatrix} 2 & 3 \\ 4 & 5 \end{vmatrix} = \begin{vmatrix} x & 3 \\ 2x & 5 \end{vmatrix}$ then $x =$

    (a) $2$

    (b) $4$

    (c) $0$

    (d) $1$

    ›Reveal solutionSolution

    Evaluate both 2×2 determinants and equate them.

    $\begin{vmatrix} 2 & 3 \\ 4 & 5 \end{vmatrix} = 2(5)-3(4) = 10-12 = -2$

    …

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