Q.Find the value of the following:

Showing the 12 most recent of 18 on this concept.

1. 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

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 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).

5. 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).

6. KEAM 2021Set eng-2021-P2-B14 marksMCQ
   Q.Let $\begin{vmatrix} x-1 & 2 & 1 \\ 2 & x-1 & 2 \\ 1 & x+2 & x-1 \end{vmatrix} = ax^3 + bx^2 + cx + d$, where $a, b, c$ and $d$ are constants. Then the value of $d$ is (A) $-8$ (B) $6$ (C) $0$ (D) $-6$ (E) $16$

   ›Reveal solutionSolution

   Setting $x=0$ gives the constant term $d=16$.

   Concept and Intuition

   Writing the determinant as $ax^3+bx^2+cx+d$, the constant $d$ equals the value at $x=0$, since all $x$-bearing terms vanish there.

   Step-by-Step Solution

   1. At $x=0$ the matrix is $\begin{bmatrix}-1 & 2 & 1\\ 2 & -1 & 2\\ 1 & 2 & -1\end{bmatrix}$.
   2. Expand: $-1((-1)(-1)-2\cdot2)-2(2\cdot(-1)-2\cdot1)+1(2\cdot2-(-1)\cdot1)$.
   3. $=-1(1-4)-2(-2-2)+1(4+1)=3+8+5=16$.
   4. So $d=16$.

   Common Mistakes

   • Trying to expand the full cubic in $x$ instead of just substituting $x=0$.

   ✓Final answer

   The correct option is (E) — $16$.

   ANSWER: E

7. KEAM 2023Set eng-2023-P2-B24 marksMCQ
   Q.Let A be $(2n+1)\times(2n+1)$ matrix with integer entries and positive determinant, where $n\in\mathbb{N}$. If $AA^T=I=A^TA$, then which of the following statements always holds? (A) $\det(A)=0$ (B) $\det(A+I)\neq 0$ (C) $\det(A+I)=0$ (D) $\det(A-I)=0$ (E) $\det(A-I)\neq 0$

   ›Reveal solutionSolution

   An odd-dimensional orthogonal matrix with det +1 always has eigenvalue 1, so det(A−I) = 0.

   Concept and Intuition

   AA^T = I means A is orthogonal, and positive integer determinant forces det(A) = +1. In odd dimension a real rotation must fix an axis (eigenvalue 1) because complex eigenvalues occur in conjugate pairs and the leftover real eigenvalue, together with det = +1, must be +1.

   Step-by-Step Solution

   1. AA^T = I ⇒ A orthogonal ⇒ eigenvalues have modulus 1 and det = ±1.
   2. Integer positive det ⇒ det(A) = +1.
   3. Size 2n+1 is odd; non-real eigenvalues pair as conjugates, leaving at least one real eigenvalue ±1.
   4. Product of all eigenvalues = +1 forces a +1 eigenvalue to exist.
   5. Eigenvalue 1 ⇒ det(A − I) = 0.

   Common Mistakes

   • Assuming det could be 0 — an orthogonal matrix is invertible.

   ✓Final answer

   The correct option is (D) — det(A − I) = 0.

   ANSWER: D

8. KEAM 2026Set eng-2026-04204 marksMCQ
   Q.The value of the determinant $\begin{vmatrix} (10^5 + 10^{-5})^2 & (10^5 - 10^{-5})^2 & 1 \\ (100^6 + 100^{-6})^2 & (100^6 - 100^{-6})^2 & 1 \\ (6^{100} + 6^{-100})^2 & (6^{100} - 6^{-100})^2 & 1 \end{vmatrix}$ is equal to (A) $100$ (B) $200$ (C) $0$ (D) $6000$ (E) $60600$

   ›Reveal solutionSolution

   In every row the two entries are $(P+Q)^2$ and $(P-Q)^2$ with $PQ=1$, so $C_1-C_2=4PQ=4$ for all rows. That makes column $C_1-C_2$ a multiple of the all-ones column $C_3$; two proportional columns force the determinant to be $0$.

   In each row the entries have the form $(P+Q)^2,\ (P-Q)^2,\ 1$, where:

   • Row 1: $P=10^5,\ Q=10^{-5}$, $PQ=1$.
   • Row 2: $P=100^6,\ Q=100^{-6}$, $PQ=1$.
   • Row 3: $P=6^{100},\ Q=6^{-100}$, $PQ=1$.

   Apply the column operation $C_1\to C_1-C_2$. For every row,

   $(P+Q)^2-(P-Q)^2=4PQ=4\cdot1=4.$

   So the new first column is $(4,4,4)^T=4\,(1,1,1)^T$, which is exactly $4$ times the third column $C_3=(1,1,1)^T$.

   A determinant with two proportional columns is $0$.

   ✓Final answer

   The correct option is (C).

9. 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).

10. 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).

11. 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).

12. KEAM 2026Set eng-2026-04184 marksMCQ
    Q.Let $A$ and $B$ be two square matrices each of order 3. If $|AB|=21$ and $|A^{-1}|=-7$, then the value of $|B|$ is equal to (A) 3 (B) -3 (C) 147 (D) -63 (E) -147

    ›Reveal solutionSolution

    $|A|=-1/7$, and $|B|=|AB|/|A|=21/(-1/7)=-147$.

    Since $|A^{-1}|=\dfrac{1}{|A|}=-7$, we have $|A|=-\dfrac17$.

    Using $|AB|=|A|\,|B|$:

    $|B|=\frac{|AB|}{|A|}=\frac{21}{-1/7}=21\times(-7)=-147.$

    ✓Final answer

    The correct option is (E).