Q.Evaluate the determinant $\Delta = \begin{vmatrix} 1 & 2 & 4 \\ -1 & 3 & 0 \\ 4 & 1 & 0 \end{vmatrix}$.

Showing the 12 most recent of 59 on this concept.

1. CBSE 2026Set 65/1/11 markMCQ
   Q.If $\Delta_1 = \begin{vmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{vmatrix}$ and $\Delta_2 = \begin{vmatrix} 0 & 2 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 6 \end{vmatrix}$, then (A) $\Delta_1 = 2\Delta_2$ (B) $\Delta_2 = -2\Delta_1$ (C) $\Delta_1 = \Delta_2$ (D) $\Delta_2 = -\Delta_1$

   ›Reveal solutionSolution

   The first determinant is diagonal; the second requires one row interchange to reach diagonal form. Each interchange flips the sign; evaluating both determinants gives $\Delta_2 = -2\Delta_1$. The answer is (B).

   Why determinants change under row operations

   A determinant measures the signed volume of the parallelepiped spanned by the row vectors. When you swap two rows, you reflect the figure across a hyperplane—the volume stays the same in magnitude but the orientation reverses, flipping the sign.

   The diagonal determinant is the easiest to compute: the product of the diagonal entries. The second determinant looks scrambled, but a single row swap will bring it into a form we recognize.

   ---

   Step-by-step evaluation

   1. Compute $\Delta_1$ directly.

   The matrix is diagonal:

   $\Delta_1 = \begin{vmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{vmatrix} = 1 \cdot 2 \cdot 3 = 6.$

   2. Recognize the structure of $\Delta_2$.

   $\Delta_2 = \begin{vmatrix} 0 & 2 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 6 \end{vmatrix}.$

   The first two rows are out of order compared to a diagonal form. Swap rows 1 and 2 to bring the $1$ into the top-left position.

   3. Apply the row-interchange property.

   Swapping rows 1 and 2:

   $\Delta_2 = -\begin{vmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 6 \end{vmatrix}.$

   The negative sign comes from the single interchange.

   4. Evaluate the new diagonal determinant.

   $\begin{vmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 6 \end{vmatrix} = 1 \cdot 2 \cdot 6 = 12.$

   So $\Delta_2 = -12$.

   5. Relate $\Delta_2$ to $\Delta_1$.

   We have $\Delta_1 = 6$ and $\Delta_2 = -12$. Notice that

   $\Delta_2 = -12 = -2 \cdot 6 = -2\Delta_1.$

   This matches option (B).

   Watch out

   A common mistake is to forget the sign change from the row swap. Without it, you'd incorrectly conclude $\Delta_2 = 12$ and miss the negative relationship.

   Tip

   For small determinants, you can also expand along the first row or column. For $\Delta_2$, expanding along row 1 gives $0 \cdot C_{11} + 2 \cdot C_{12} + 0 \cdot C_{13}$, where $C_{12} = -\begin{vmatrix}1 & 0 \\ 0 & 6\end{vmatrix} = -6$, so $\Delta_2 = 2 \cdot (-6) = -12$.

   ✓Final answer

   The correct option is (B): $\Delta_2 = -2\Delta_1$.

2. CBSE 2024Set 65/1/11 markMCQ
   Q.$\begin{vmatrix} x+1 & x-1 \\ x^2+x+1 & x^2-x+1 \end{vmatrix}$ is equal to : (A) $2x^3$ (B) $2$ (C) $0$ (D) $2x^3 - 2$

   ›Reveal solutionSolution

   Expand the $2\times2$ determinant as $ad-bc$; the cube-sum and cube-difference collapse to a constant. The value is $2$, option (B).

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

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

   Use the standard factorisations $a^3+b^3=(a+b)(a^2-ab+b^2)$ and $a^3-b^3=(a-b)(a^2+ab+b^2)$ with $a=x,\ b=1$:

   $(x+1)(x^2-x+1)=x^3+1,\qquad (x-1)(x^2+x+1)=x^3-1$

   Therefore

   $\Delta=(x^3+1)-(x^3-1)=2.$

   The result is the constant $2$, independent of $x$.

   ✓Final answer

   $\boxed{2}$, option (B).

3. CBSE 2026Set A1 markMCQ
   Q.$\begin{vmatrix} 23 & 12 & 11 \\ 36 & 10 & 26 \\ 63 & 26 & 37 \end{vmatrix} =$

   (a) $1$

   (b) $-1$

   (c) $0$

   (d) $2$

   ›Reveal solutionSolution

   The determinant equals $0$ because one column is the sum of the other two.

   Inspect the columns of

   $\begin{vmatrix} 23 & 12 & 11 \\ 36 & 10 & 26 \\ 63 & 26 & 37 \end{vmatrix}.$

   Check: $12+11 = 23$, $10+26 = 36$, $26+37 = 63$.

   So $C_1 = C_2 + C_3$. When one column is a linear combination of the others, the columns are linearly dependent and the determinant is $0$.

   ✓Final answer

   (c) $0$.

4. CBSE 2026Set A1 markMCQ
   Q.$\begin{vmatrix} \cos 15^\circ & \sin 15^\circ \\ \sin 75^\circ & \cos 75^\circ \end{vmatrix} =$

   (a) $1$

   (b) $0$

   (c) $-1$

   (d) $\frac{1}{2}$

   ›Reveal solutionSolution

   The determinant equals $\cos 90^\circ = 0$.

   Expand:

   $\begin{vmatrix} \cos 15^\circ & \sin 15^\circ \\ \sin 75^\circ & \cos 75^\circ \end{vmatrix} = \cos 15^\circ\cos 75^\circ - \sin 15^\circ\sin 75^\circ.$

   By the cosine addition formula $\cos A\cos B - \sin A\sin B = \cos(A+B)$:

   $= \cos(15^\circ + 75^\circ) = \cos 90^\circ = 0.$

   ✓Final answer

   (b) $0$.

5. CBSE 2026Set A1 markMCQ
   Q.$\begin{vmatrix} a+ib & c+id \\ -c+id & a-ib \end{vmatrix} =$

   (a) $a^2 + b^2 + c^2 + d^2$

   (b) $a^2 - b^2 - c^2 - d^2$

   (c) $a^2 - b^2 + c^2 + d^2$

   (d) $a^2 + b^2 + c^2 - d^2$

   ›Reveal solutionSolution

   Expand the $2\times2$ determinant and simplify the complex products.

   $\begin{vmatrix} a+ib & c+id \\ -c+id & a-ib \end{vmatrix} = (a+ib)(a-ib) - (c+id)(-c+id).$

   First term: $(a+ib)(a-ib) = a^2 - (ib)^2 = a^2 + b^2$.

   Second term: $(c+id)(-c+id) = -c^2 + icd - icd + (id)^2 = -c^2 - d^2$.

   So the determinant $= (a^2 + b^2) - (-c^2 - d^2) = a^2 + b^2 + c^2 + d^2$.

   ✓Final answer

   (a) $a^2 + b^2 + c^2 + d^2$.

6. CBSE 2026Set ANNUAL1 markMCQ
   Q.Value of $\begin{vmatrix}x^2-x+1 & x-1\\ x+1 & x+1\end{vmatrix}$ will be

   (a) $x^2-x+2$

   (b) $x^3+x^2-2$

   (c) $x^3-x^2+2$

   (d) $x^3+x^2+4$

   ›Reveal solutionSolution

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

   $\Delta = (x^2-x+1)(x+1) - (x-1)(x+1)$

   $(x^2-x+1)(x+1) = x^3+1$ (the middle terms cancel).

   $(x-1)(x+1) = x^2-1$.

   $\Delta = (x^3+1)-(x^2-1) = x^3-x^2+2$.

   ✓Final answer

   The correct option is (c) $x^3-x^2+2$.

7. CBSE 2026Set ANNUAL1 mark
   Q.The value of determinant $\Delta = \begin{vmatrix}1 & 2 & 4\\ -1 & 3 & 0\\ 4 & 1 & 0\end{vmatrix}$ is __________.

   ›Reveal solutionSolution

   Expand the $3\times3$ determinant along the first row.

   $\Delta=\begin{vmatrix}1&2&4\\-1&3&0\\4&1&0\end{vmatrix}$

   Expanding along row 1:

   $\Delta = 1(3\cdot0-0\cdot1) - 2(-1\cdot0-0\cdot4) + 4(-1\cdot1-3\cdot4)$

   $= 1(0) - 2(0) + 4(-1-12) = 4(-13) = -52$.

   ✓Final answer

   $\Delta = -52$.

8. CBSE 2026Set ANNUAL1 mark
   Q.Find the value of determinant $\Delta = \begin{vmatrix}0 & \sin\alpha & -\cos\alpha\\ -\sin\alpha & 0 & \sin\beta\\ \cos\alpha & -\sin\beta & 0\end{vmatrix}$.

   ›Reveal solutionSolution

   The matrix is skew-symmetric (each $a_{ij}=-a_{ji}$) and every odd-order skew-symmetric matrix has determinant $0$.

   Check: $a_{12}=\sin\alpha=-a_{21}$, $a_{13}=-\cos\alpha=-a_{31}$, $a_{23}=\sin\beta=-a_{32}$, and all diagonal entries are $0$ — so the matrix is skew-symmetric.

   For a skew-symmetric matrix $A$ of odd order $n$, $\det A = \det A^T = \det(-A) = (-1)^n\det A = -\det A$, so $\det A=0$.

   ✓Final answer

   $\Delta = 0$.

9. CBSE 2026Set ANNUAL1 markMCQ
   Q.$\begin{vmatrix} \cos 30^\circ & \sin 30^\circ \\ \sin 30^\circ & \cos 30^\circ \end{vmatrix} =$

   (a) $\dfrac{1}{2}$

   (b) $\dfrac{\sqrt{3}}{2}$

   (c) 0

   (d) None of these

   ›Reveal solutionSolution

   This determinant has the form $\cos^2\theta - \sin^2\theta = \cos 2\theta$.

   $\begin{vmatrix} \cos 30^\circ & \sin 30^\circ \\ \sin 30^\circ & \cos 30^\circ \end{vmatrix} = \cos 30^\circ \cdot \cos 30^\circ - \sin 30^\circ \cdot \sin 30^\circ = \cos^2 30^\circ - \sin^2 30^\circ$

   Using $\cos^2\theta - \sin^2\theta = \cos 2\theta$: this equals $\cos 60^\circ = \dfrac{1}{2}$.

   ✓Final answer

   (a) $\dfrac{1}{2}$.

10. CBSE 2026Set ANNUAL1 mark
    Q.Evaluate the determinant \Delta = \begin{vmatrix}1 & 2 & 4\ -1 & 3 & 0\ 4 & 1 & 0\end{vmatrix}.

    ›Reveal solutionSolution

    Expand the $3\times3$ determinant along the first row (or any row/column) using cofactors.

    Working: Expanding along Row 1:

    $\Delta = \begin{vmatrix}1 & 2 & 4\\ -1 & 3 & 0\\ 4 & 1 & 0\end{vmatrix}$

    $= 1\begin{vmatrix}3 & 0\\ 1 & 0\end{vmatrix} - 2\begin{vmatrix}-1 & 0\\ 4 & 0\end{vmatrix} + 4\begin{vmatrix}-1 & 3\\ 4 & 1\end{vmatrix}$

    $= 1(3\cdot0 - 0\cdot1) - 2(-1\cdot0 - 0\cdot4) + 4(-1\cdot1 - 3\cdot4)$

    $= 1(0) - 2(0) + 4(-1-12) = 4(-13) = -52$

    ✓Final answer

    $\Delta = -52$.

11. CBSE 2025Set 65/4/11 markMCQ
    Q.If M and N are square matrices of order 3 such that $\det(M) = m$ and $MN = mI$, then $\det(N)$ is equal to : (A) $-1$ (B) $1$ (C) $-m^2$ (D) $m^2$

    ›Reveal solutionSolution

    The key idea is that $MN = mI$ implies $N = m M^{-1}$, so $\det(N) = m^3 \det(M^{-1}) = m^3 \cdot \frac{1}{m} = m^2$. The correct option is (D).

    The problem gives us two square matrices $M$ and $N$ of order 3, with $\det(M) = m$ and $MN = mI$, where $I$ is the $3 \times 3$ identity matrix. We need $\det(N)$.

    The central concept here is the relationship between matrix multiplication and determinants. When two matrices multiply to give a scalar times the identity, that scalar is intimately connected to the determinant of the first matrix. The equation $MN = mI$ is not just a product — it tells us that $N$ is essentially a scaled inverse of $M$.

    Why? Because if $MN = mI$, then multiplying both sides on the left by $M^{-1}$ (assuming $M$ is invertible) gives $N = m M^{-1}$. But we must first check: is $M$ invertible? Yes — since $\det(M) = m \neq 0$ (the problem doesn't state $m \neq 0$ explicitly, but if $m = 0$, then $MN = 0$, which would make $N$ singular and the answer ambiguous; in standard exam contexts, $m$ is taken as a non-zero scalar, often a real number, and the options suggest $m \neq 0$). So $M^{-1}$ exists.

    Now, the determinant of a scalar multiple of a matrix: for an $n \times n$ matrix $A$, $\det(kA) = k^n \det(A)$. Here $n = 3$, so $\det(m M^{-1}) = m^3 \det(M^{-1})$.

    And we know $\det(M^{-1}) = \frac{1}{\det(M)} = \frac{1}{m}$.

    Putting it together:

    1. From $MN = mI$, take determinant on both sides: $\det(MN) = \det(mI)$.
    2. $\det(MN) = \det(M) \cdot \det(N) = m \cdot \det(N)$.
    3. $\det(mI)$: $mI$ is a diagonal matrix with all diagonal entries $m$, so its determinant is $m^3$ (since it's $3 \times 3$).
    4. So $m \cdot \det(N) = m^3$.
    5. Divide both sides by $m$ (non-zero): $\det(N) = m^2$.

    Tip

    A faster route: from $MN = mI$, multiply both sides on left by $M^{-1}$ to get $N = m M^{-1}$. Then $\det(N) = \det(m M^{-1}) = m^3 \cdot \frac{1}{m} = m^2$. This avoids the determinant-of-product step, but both are equivalent.

    Watch out

    A common mistake is to forget the exponent on $m$ when taking $\det(mI)$. Since $I$ is $3 \times 3$, $\det(mI) = m^3$, not $m$. Also, do not confuse $MN = mI$ with $MN = I$ — the scalar $m$ changes the scaling factor.

    Thus, the determinant of $N$ is $m^2$.

    ✓Final answer

    The value is $\boxed{m^2}$, which corresponds to option (D).

12. CBSE 2025Set E1 markMCQ
    Q.$\begin{vmatrix} 21 & 11 & 10 \\ 25 & 15 & 10 \\ 64 & 27 & 37 \end{vmatrix} =$

    (a) $1190$

    (b) $841$

    (c) $0$

    (d) $1$

    ›Reveal solutionSolution

    A column that is the sum of the other two makes the determinant zero.

    Examine the columns of

    $\begin{vmatrix} 21 & 11 & 10 \\ 25 & 15 & 10 \\ 64 & 27 & 37 \end{vmatrix}.$

    Check $C_2+C_3$ against $C_1$:

    $11+10=21,\quad 15+10=25,\quad 27+37=64.$

    So $C_1=C_2+C_3$, i.e. the columns are linearly dependent. A determinant with linearly dependent columns is $0$ (apply $C_1\to C_1-C_2-C_3$ to get a zero column).

    ✓Final answer

    (C) $0$.