Q.Find the value of the following:

Showing the 12 most recent of 18 on this concept.

1. KEAM 2026Set eng-2026-04184 marksMCQ
   Q.Let $f(x)=\begin{vmatrix}-1 & x & 3\\ 0 & 1 & 2x\\ 1 & -1 & 1\end{vmatrix}$. Then the value of $f(-1)$ is equal to (A) 6 (B) -4 (C) -2 (D) 2 (E) 0

   ›Reveal solutionSolution

   At $x=-1$ the determinant evaluates to $0$.

   At $x=-1$ the matrix is $\begin{pmatrix}-1&-1&3\\0&1&-2\\1&-1&1\end{pmatrix}$. Expanding along the first column:

   $f(-1)=-1\big[(1)(1)-(-2)(-1)\big]-0+1\big[(-1)(-2)-(3)(1)\big].$

   $=-1(1-2)+1(2-3)=-1(-1)+1(-1)=1-1=0.$

   ✓Final answer

   The correct option is (E).

2. KEAM 2025Set eng-2025-04284 marksMCQ
   Q.Let $\Delta = \begin{vmatrix} x & y & 1 \\ x+y & y+1 & x+1 \\ 1 & x & y \end{vmatrix}$. If $x+y=-1$, then the value of $\Delta$ is equal to (A) 3 (B) 2 (C) 1 (D) 0 (E) -3

   ›Reveal solutionSolution

   Expanding $\Delta$ and substituting $x+y=-1$, every term reduces to $0$ because it carries a factor $x+y+1=0$.

   Expanding along the first row:

   $\Delta=x\big[(y+1)y-(x+1)x\big]-y\big[(x+y)y-(x+1)\big]+\big[(x+y)x-(y+1)\big].$

   With $x+y=-1$:

   • First term: $x\big[(y^2-x^2)+(y-x)\big]=x(y-x)(x+y+1)=x(y-x)(0)=0$.
   • Second term: $-y\big[(-1)y-x-1\big]=-y\big[-(x+y)-1\big]=-y(1-1)=0$.
   • Third term: $(-1)x-y-1=-(x+y)-1=1-1=0$.

   Hence $\Delta=0$.

   ✓Final answer

   The correct option is (D).

3. KEAM 2026Set eng-2026-04224 marksMCQ
   Q.The value of $\begin{vmatrix} \sin 30^\circ & \cos 30^\circ & \sin(30^\circ + 75^\circ) \\ \sin 45^\circ & \cos 45^\circ & \sin(45^\circ + 75^\circ) \\ \sin 60^\circ & \cos 60^\circ & \sin(60^\circ + 75^\circ) \end{vmatrix}$ is equal to (A) $-2$ (B) $-1$ (C) $0$ (D) $1$ (E) $2$

   ›Reveal solutionSolution

   The third column is a fixed linear combination of the first two, forcing a zero determinant.

   Each entry of column 3 is $\sin(\theta+75^\circ)=\sin\theta\cos75^\circ+\cos\theta\sin75^\circ$.

   So $C_3=\cos75^\circ\,C_1+\sin75^\circ\,C_2$, i.e. column 3 is linearly dependent on columns 1 and 2.

   A determinant with linearly dependent columns is $0$.

   ✓Final answer

   The correct option is (C).

4. KEAM 2022Set eng-2022-P2-B14 marksMCQ
   Q.$\begin{vmatrix} \sin\alpha & \cos(\alpha+\theta) & \cos\alpha \\ \sin\beta & \cos(\beta+\theta) & \cos\beta \\ \sin\gamma & \cos(\gamma+\theta) & \cos\gamma \end{vmatrix} =$ (A) $-1$ (B) 1 (C) 2 (D) 4 (E) 0

   ›Reveal solutionSolution

   The determinant equals $0$.

   Concept and Intuition

   A determinant is zero when one column is a linear combination of the others.

   Step-by-Step Solution

   1. Expand $\cos(\phi+\theta)=\cos\theta\cos\phi-\sin\theta\sin\phi$ for each row's middle entry.
   2. So column 2 $=\cos\theta\,(\text{column of }\cos\phi)-\sin\theta\,(\text{column of }\sin\phi)$.
   3. Column of $\cos\phi$ is column 3, column of $\sin\phi$ is column 1.
   4. Thus $C_2=\cos\theta\,C_3-\sin\theta\,C_1$: columns are linearly dependent.
   5. Determinant $=0$.

   Common Mistakes

   • Trying a brute-force cofactor expansion instead of spotting dependence.
   • Sign error in the cosine addition formula.

   ✓Final answer

   The correct option is (E) — $0$.

   ANSWER: E

5. KEAM 2024Set eng-2024-06064 marksMCQ
   Q.Let $A=\begin{pmatrix}2 & -1 & 1\\ -1 & 0 & 2\\ 1 & -2 & -1\end{pmatrix}$ and let $B=\dfrac{1}{|A|}A$. Then the value of $|B|$ is equal to (A) $\dfrac{1}{9}$ (B) $\dfrac{1}{11}$ (C) $\dfrac{1}{81}$ (D) $\dfrac{1}{121}$ (E) 1

   ›Reveal solutionSolution

   For a $3\times3$ matrix, scaling by $1/|A|$ scales the determinant by $(1/|A|)^3$.

   $|A|=2(0+4)+1(1-2)+1(2-0)=8-1+2=9$. Then $|B|=\left(\dfrac{1}{|A|}\right)^3|A|=\dfrac{1}{|A|^2}=\dfrac{1}{81}$.

   ✓Final answer

   The correct option is (C).

6. KEAM 2026Set eng-2026-04194 marksMCQ
   Q.$\begin{vmatrix} 11 & 1 & 1 \\ 1 & 21 & 1 \\ 1 & 1 & 31 \end{vmatrix}$ is equal to (A) 7100 (B) 6800 (C) 7300 (D) 6900 (E) 6700

   ›Reveal solutionSolution

   Cofactor expansion along the first row gives $7100$.

   Expansion. For $\begin{vmatrix}11&1&1\\1&21&1\\1&1&31\end{vmatrix}$:

   $=11(21\cdot31-1\cdot1)-1(1\cdot31-1\cdot1)+1(1\cdot1-21\cdot1)$

   $=11(651-1)-1(31-1)+1(1-21)=11\cdot650-30-20.$

   $=7150-50=7100.$

   ✓Final answer

   The correct option is (A).

7. KEAM 2025Set eng-2025-04294 marksMCQ
   Q.If $\alpha + \beta + \gamma = 0$, then $\begin{vmatrix} e^{\alpha} & e^{2\alpha} & e^{3\alpha} - 1 \\ e^{\beta} & e^{2\beta} & e^{3\beta} - 1 \\ e^{\gamma} & e^{2\gamma} & e^{3\gamma} - 1 \end{vmatrix} =$ (A) $e^{-1}$ (B) $e$ (C) $e^2$ (D) $e^3$ (E) 0

   ›Reveal solutionSolution

   Let $a=e^{\alpha},b=e^{\beta},c=e^{\gamma}$, so $abc=e^{\alpha+\beta+\gamma}=1$. Split the third column $a^3-1=a^3+(-1)$; the two resulting Vandermonde determinants are equal and cancel, giving $0$.

   Write the rows as $(a,a^2,a^3-1)$ etc. By column-linearity the determinant splits as $D_1-D_2$ where

   $D_1=\begin{vmatrix}a&a^2&a^3\\\\ b&b^2&b^3\\\\ c&c^2&c^3\end{vmatrix}=abc\begin{vmatrix}1&a&a^2\\\\1&b&b^2\\\\1&c&c^2\end{vmatrix}=abc\\,V,$

   and $D_2=\begin{vmatrix}a&a^2&1\\\\ b&b^2&1\\\\ c&c^2&1\end{vmatrix}$. A cyclic (even) column permutation turns $D_2$ into the same Vandermonde $V$. Since $abc=1$, $D_1=V$ and $D_2=V$, so the value is $V-V=0$.

   ✓Final answer

   The correct option is (E).

8. KEAM 2025Set eng-2025-04254 marksMCQ
   Q.The numbers $a_1, a_2, a_3, a_4, a_5$ and $a_6$ are in G.P. If $a_1 = 2$ and the common ratio $r = \dfrac{1}{2}$, then the value of $\begin{vmatrix} a_1 & a_2 & 1 \\ a_3 & a_4 & 1 \\ a_5 & a_6 & 1 \end{vmatrix}$ is equal to (A) 1 (B) 2 (C) $\dfrac{1}{2}$ (D) 4 (E) 0

   ›Reveal solutionSolution

   Two columns are proportional, forcing the determinant to zero.

   With $a_1 = 2, r = \tfrac12$: the terms are $2, 1, \tfrac12, \tfrac14, \tfrac18, \tfrac{1}{16}$.

   In the determinant

   $\begin{vmatrix} a_1 & a_2 & 1 \\ a_3 & a_4 & 1 \\ a_5 & a_6 & 1 \end{vmatrix},$

   each second-column entry is $r$ times the first-column entry ($a_2 = r a_1$, $a_4 = r a_3$, $a_6 = r a_5$). Columns 1 and 2 are therefore linearly dependent, so the determinant equals $0$.

   ✓Final answer

   The correct option is (E).

9. KEAM 2021Set eng-2021-P2-B14 marksMCQ
   Q.The value of the determinant $\begin{vmatrix}4 & 4^2 & 4^3\\ 3 & 3^2 & 3^3\\ 2 & 2^2 & 2^3\end{vmatrix}$ is (A) $52$ (B) $-24$ (C) $24$ (D) $48$ (E) $-48$

   ›Reveal solutionSolution

   The determinant equals $-48$.

   Concept and Intuition

   Each row $(k, k^2, k^3)$ has a common factor $k$, which can be pulled out of the determinant. The remaining determinant is a small $3\times 3$ evaluation.

   Step-by-Step Solution

   1. Factor $4,3,2$ from rows 1,2,3: value $=4\cdot 3\cdot 2\,\begin{vmatrix}1 & 4 & 16\\ 1 & 3 & 9\\ 1 & 2 & 4\end{vmatrix}=24\,D$.
   2. Expand $D=1(3\cdot4-9\cdot2)-4(1\cdot4-9\cdot1)+16(1\cdot2-3\cdot1)$.
   3. $D=(-6)-4(-5)+16(-1)=-6+20-16=-2$.
   4. Value $=24\times(-2)=-48$.

   Common Mistakes

   • Forgetting the row factors, or sign slips in the cofactor expansion.

   ✓Final answer

   The correct option is (E) — $-48$.

   ANSWER: E

10. KEAM 2023Set eng-2023-P2-B24 marksMCQ
    Q.The determinant of the matrix $\begin{bmatrix} 1 & 4 & 8 \\ 1 & 9 & 27 \\ 1 & 16 & 64 \end{bmatrix}$ is (A) $13$ (B) $208$ (C) $104$ (D) $26$ (E) $52$

    ›Reveal solutionSolution

    Expanding along the first row gives determinant 52.

    Concept and Intuition

    A direct cofactor expansion along the first row is quickest for a 3×3 determinant.

    Step-by-Step Solution

    1. det = 1(9·64 − 27·16) − 4(1·64 − 27·1) + 8(1·16 − 9·1).
    2. = 1(576 − 432) − 4(64 − 27) + 8(16 − 9).
    3. = 144 − 4·37 + 8·7 = 144 − 148 + 56.
    4. = 52.

    Common Mistakes

    • Arithmetic slips in the 2×2 minors.

    ✓Final answer

    The correct option is (E) — 52.

    ANSWER: E

11. KEAM 2024Set eng-2024-06094 marksMCQ
    Q.Let $A=(a_{ij})$ be a square matrix of order 3 and let $M_{ij}$ be the minors of $a_{ij}$. If $M_{11}=-40, M_{12}=-10, M_{13}=35$ and $a_{11}=1, a_{12}=3, a_{13}=-2$ then the value of $|A|$ is equal to (A) -100 (B) -80 (C) 0 (D) 60 (E) 80

    ›Reveal solutionSolution

    Convert minors to cofactors with alternating signs, then expand: $|A|=-80$.

    Cofactors: $C_{11}=+M_{11}=-40$, $C_{12}=-M_{12}=10$, $C_{13}=+M_{13}=35$.

    Expanding along the first row:

    $|A|=a_{11}C_{11}+a_{12}C_{12}+a_{13}C_{13}=1(-40)+3(10)+(-2)(35)=-40+30-70=-80.$

    ✓Final answer

    The correct option is (B).

12. KEAM 2024Set eng-2024-06074 marksMCQ
    Q.Let $A=\begin{pmatrix}a_1&b_1&c_1\\a_2&b_2&c_2\\a_3&b_3&c_3\end{pmatrix}$ and $B=\begin{pmatrix}a_1&2b_1&4c_1\\2a_2&4b_2&8c_2\\4a_3&8b_3&16c_3\end{pmatrix}$. If $|B|=16$, then the value of $|A|$ is equal to (A) 4 (B) $\dfrac{1}{4}$ (C) 8 (D) $\dfrac{1}{8}$ (E) 16

    ›Reveal solutionSolution

    Pull common factors out of each row and each column of $B$.

    Rows of $B$ are $(a_1,2b_1,4c_1)$, $2(a_2,2b_2,4c_2)$, $4(a_3,2b_3,4c_3)$, giving a row factor $1\cdot2\cdot4=8$. The remaining matrix has columns with factors $1,2,4$, giving a column factor $1\cdot2\cdot4=8$, and what remains is $A$. So $|B|=8\cdot8\cdot|A|=64|A|$. With $|B|=16$, $|A|=\dfrac{16}{64}=\dfrac14$.

    ✓Final answer

    The correct option is (B).