Q.Using properties of determinants, prove that $\begin{vmatrix} 1 & 1 & 1+3x \\ 1+3y & 1 & 1 \\ 1 & 1+3z & 1 \end{vmatrix} = 9(3xyz + xy + yz + zx)$

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1. 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 $|AB|=|A|\,|B|$: …

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2. 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: …

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

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

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4. 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, …

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5. 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 $\sin(\theta+75^\circ)=\sin\theta\cos75^\circ+\cos\theta\sin75^\circ$. …

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6. 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

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

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7. 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: $x\big[(y^2-x^2)+(y-x)\big]=x(y-x)(x+y+1)=x(y-x)(0)=0$. …

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8. 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

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

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9. 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 $(1/|A|)^3$. …

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10. 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 i …

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11. KEAM 2024Set eng-2024-06074 marksMCQ
    Q.If $A=\begin{pmatrix}-7&3\\3&-1\end{pmatrix}$, then $\det\left(A^5\right)$ is equal to (A) 81 (B) -81 (C) 243 (D) -243 (E) -32

    ›Reveal solutionSolution

    $\det(A^5)=(\det A)^5$.

    $\det A=(-7)(-1)-(3)(3)=7-9=-2$. Then $\det(A^5)=(\det A)^5=(-2)^5=-32$ …

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12. 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: …

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