Q.If $f(x) = \begin{vmatrix} (1 + x)^{17} & (1 + x)^{19} & (1 + x)^{23} \\ (1 + x)^{23} & (1 + x)^{29} & (1 + x)^{34} \\ (1 + x)^{41} & (1 + x)^{43} & (1 + x)^{47} \end{vmatrix} = A + Bx + Cx^2 + \ldots$, then $A =$ ________ .

Showing the 12 most recent of 44 on this concept.

1. AP EAPCET 2022Set eng-2022-07-04-AN1 markMCQ
   Q.Let $A = \begin{bmatrix} -2 & x & 1 \\ x & 1 & 1 \\ 2 & 3 & -1 \end{bmatrix}$. If the roots of the equation $\det A = 0$ are $l, m$ then $l^3 - m^3 =$ (A) $35$ (B) $-35$ (C) $19$ (D) $-19$

   ›Reveal solutionSolution

   $\det A = x^2+5x+6=0$ has roots $-2,-3$; with $l=-2,\ m=-3$, we get $l^3-m^3=19$.

   Concept and Intuition

   The unknown $x$ appears in the matrix; setting $\det A=0$ turns the determinant into a quadratic equation in $x$ whose two roots are called $l$ and $m$. Once the roots are found, evaluating $l^3-m^3$ is direct arithmetic — the only subtlety is being consistent about which root is called $l$ and which $m$ (standard convention: the root from the "+" sign of the quadratic formula is taken first).

   Step-by-Step Solution

   1. Expand $\det A$ for $A=\begin{bmatrix}-2&x&1\\x&1&1\\2&3&-1\end{bmatrix}$ along the first row: $\det A = -2\big(1(-1)-1(3)\big) - x\big(x(-1)-1(2)\big) + 1\big(x(3)-1(2)\big)$ $= -2(-1-3) - x(-x-2) + (3x-2)$ $= 8 + x^2+2x+3x-2 = x^2+5x+6$.
   2. Set $\det A = 0$: $x^2+5x+6=0 \Rightarrow (x+2)(x+3)=0 \Rightarrow x=-2 \text{ or } x=-3$. …

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2. AP EAPCET 2021Set eng-2021-08-20-AN1 markMCQ
   Q.If $\begin{vmatrix} x & x^2 & 1+x^3 \\ y & y^2 & 1+y^3 \\ z & z^2 & 1+z^3 \end{vmatrix} = 0$ and $x, y, z$ are all distinct, then $xyz =$ (A) $-1$ (B) $1$ (C) $0$ (D) $3$

   ›Reveal solutionSolution

   Splitting the determinant by linearity in the third column reduces it to a Vandermonde factor

   times $(1+xyz)$; since $x,y,z$ are distinct the Vandermonde factor is nonzero, forcing

   $xyz=-1$.

   Concept and Intuition

   Whenever a determinant's column is a sum (here, $1+x^3$), linearity lets us split it into two

   separate determinants. Recognising the classic Vandermonde pattern ($1,x,x^2$-type columns) lets us

   avoid brute-force expansion.

   Step-by-Step Solution

   1. Write the third column as $1+x^3$ etc. and split by linearity: $\begin{vmatrix}x&x^2&1+x^3\\y&y^2&1+y^3\\z&z^2&1+z^3\end{vmatrix} = \begin{vmatrix}x&x^2&1\\y&y^2&1\\z&z^2&1\end{vmatrix} + \begin{vmatrix}x&x^2&x^3\\y&y^2&y^3\\z&z^2&z^3\end{vmatrix}$.
   2. The first determinant, with columns $(x,x^2,1)$, is a cyclic rearrangement of the standard Vandermonde columns $(1,x,x^2)$ — a 3-cycle of columns is an even permutation, so its value equals the Vandermonde determinant $V=(y-x)(z-x)(z-y)$ unchanged in sign.
   3. The second determinant factors out $x,y,z$ from each row: it equals $xyz\begin{vmatrix}1&x&x^2\\1&y&y^2\\1&z&z^2\end{vmatrix} = xyz\cdot V$. …

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3. AP EAPCET 2021Set eng-2021-08-25-AN1 markMCQ
   Q.Sum of the roots of the equation $\begin{vmatrix} x & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & x & 0 & 0 \\ 2 & 0 & x-1 & 0 \end{vmatrix} - \begin{vmatrix} 0 & x & 0 \\ 0 & 0 & x-1 \\ 2 & 2 & 0 \end{vmatrix} = 0$ is (A) 2 (B) 3 (C) 1 (D) 5

   ›Reveal solutionSolution

   Expand each determinant by cofactors along the sparsest row/column, subtract, factor the resulting cubic, and sum its roots. Answer: (B).

   Concept and Intuition

   Both determinants are sparse (mostly zeros), so expanding along the row/column with the fewest non-zero entries makes the computation short. After simplifying, the equation reduces to a simple cubic in $x$ whose roots can be read off by factoring.

   Step-by-Step Solution

   1. First (4×4) determinant: expanding along row 1 (only entry $x$ at position (1,1) is nonzero) reduces it to $x$ times a 3×3 minor $\begin{vmatrix}0&0&1\\x&0&0\\0&x-1&0\end{vmatrix}$, which itself (expanding along its first row) equals $x(x-1)$. So the 4×4 determinant $= x\cdot x(x-1) = x^2(x-1)$. …

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4. AP EAPCET 2021Set eng-2021-08-19-AN1 markMCQ
   Q.Let $a, b, c$ be such that $b + c \neq 0$ and $\begin{vmatrix} a & a+1 & a-1 \\ -b & b+1 & b-1 \\ c & c-1 & c+1 \end{vmatrix} + \begin{vmatrix} a+1 & b+1 & c-1 \\ a-1 & b-1 & c+1 \\ (-1)^{n+2}a & (-1)^{n-1}b & (-1)^{n}c \end{vmatrix} = 0$, then the value of '$n$' is (A) Zero (B) Any even integer (C) Any odd integer (D) Any integer

   ›Reveal solutionSolution

   Both determinants evaluate to the same expression $4b(a+c)$ (one scaled by $(-1)^n$), so the equation reduces to $4b(a+c)(1+(-1)^n)=0$, forcing $(-1)^n=-1$ — true exactly for odd $n$.

   Concept and Intuition

   When a sum of two determinants is set to zero and one of them carries a parity-dependent sign factor like $(-1)^{n+2}$, the standard approach is: (1) simplify each determinant symbolically or verify its value with a numeric substitution, (2) express the $(-1)^{\text{power}}$ terms in terms of a single variable $k=(-1)^n$ using the fact that shifting the exponent by an even number doesn't change the sign, while shifting by an odd number flips it, and (3) solve the resulting simple equation for which parity of $n$ makes it true.

   Step-by-Step Solution

   1. Call the first determinant $D_1 = \begin{vmatrix} a & a+1 & a-1 \\ -b & b+1 & b-1 \\ c & c-1 & c+1 \end{vmatrix}$. Expanding (or checking numerically, e.g. with $a=1,b=2,c=3$ giving $D_1=32$) shows $D_1 = 4ab+4bc = 4b(a+c)$ (numeric check: $4\times2\times(1+3)=32$ ✓).
   2. In the second determinant, note the exponents: $(-1)^{n+2} = (-1)^n\cdot(-1)^2 = (-1)^n$ (shifting by 2, an even number, preserves sign), and $(-1)^{n-1} = (-1)^n\cdot(-1)^{-1} = -(-1)^n$ (shifting by an odd number flips sign). Let $k=(-1)^n$; the third row becomes $(ka, -kb, kc) = k(a,-b,c)$.
   3. Factor $k$ out of the third row: $D_2 = k\cdot\begin{vmatrix} a+1 & b+1 & c-1 \\ a-1 & b-1 & c+1 \\ a & -b & c \end{vmatrix}$.
   4. Expanding this remaining determinant (or verifying numerically with the same $a=1,b=2,c=3$) gives the SAME value, $4b(a+c) = 32$ — call it $D_3=D_1$. …

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5. AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQ
   Q.If $A = \begin{bmatrix}2 & 2 & 1\\1 & 3 & 1\\1 & 2 & 2\end{bmatrix}$ and $\alpha, \beta, \gamma$ are the roots of the equation represented by $|A - xI| = 0$, then $\alpha^2+\beta^2+\gamma^2 =$ (A) $50$ (B) $29$ (C) $17$ (D) $27$

   ›Reveal solutionSolution

   Using the trace (sum of roots) and the sum of principal $2\times2$ minors (sum of pairwise products of roots), $\alpha^2+\beta^2+\gamma^2 = 49-22=27$ — no need to solve the cubic explicitly.

   Concept and Intuition

   For the characteristic equation $|A-xI|=0$ of a $3\times3$ matrix, the coefficients are directly the elementary symmetric functions of the eigenvalues: their sum equals the trace of $A$, and the sum of their pairwise products equals the sum of the three principal $2\times2$ minors of $A$ (the determinants obtained by deleting one matching row and column). This avoids expanding and solving the actual cubic.

   Step-by-Step Solution

   1. Sum of roots: $\alpha+\beta+\gamma = \operatorname{tr}(A) = 2+3+2 = 7$.
   2. Sum of pairwise products = sum of principal minors:
      • Delete row1,col1: $\begin{vmatrix}3&1\\2&2\end{vmatrix}=6-2=4$
      • Delete row2,col2: $\begin{vmatrix}2&1\\1&2\end{vmatrix}=4-1=3$
      • Delete row3,col3: $\begin{vmatrix}2&2\\1&3\end{vmatrix}=6-2=4$
      • Sum: $4+3+4=11=\alpha\beta+\beta\gamma+\gamma\alpha$. …

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6. AP EAPCET 2022Set eng-2022-07-07-FN1 markMCQ
   Q.If $A = \begin{bmatrix} x & 2 & 1 \\ 2 & x & 1 \\ 2 & 1 & 0 \end{bmatrix}$ and $\det(A^3) = 125$, then $x =$ (A) $1/3$ (B) $3$ (C) $-1/3$ (D) $-3$

   ›Reveal solutionSolution

   Using $\det(A^3)=(\det A)^3$, the real solution requires $\det A=5$; expanding the $3\times3$ determinant and equating to 5 gives $x=1/3$.

   Concept and Intuition

   For any square matrix, $\det(A^n)=(\det A)^n$. Here $\det(A^3)=125=5^3$, and since $x$ (and hence $\det A$, a polynomial in the real variable $x$) is real, the only real value satisfying $(\det A)^3=125$ is $\det A=5$ (the other two cube roots of 125 are complex).

   Step-by-Step Solution

   1. Expand $\det(A)$ along the first row for $A=\begin{bmatrix}x&2&1\\2&x&1\\2&1&0\end{bmatrix}$: $\det(A) = x(x\cdot0-1\cdot1) - 2(2\cdot0-1\cdot2) + 1(2\cdot1-x\cdot2)$
   2. $= x(-1) - 2(-2) + 1(2-2x) = -x+4+2-2x = 6-3x$. …

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7. AP EAPCET 2024Set eng-2024-05-20-FN1 markMCQ
   Q.Sum of the positive roots of the equation $\begin{vmatrix} x^2+2x & x+2 & 1 \\ 2x+1 & x-1 & 1 \\ x+2 & -1 & 1 \end{vmatrix} = 0$ is (A) $\dfrac{1+\sqrt{13}}{2}$ (B) $1$ (C) $\dfrac{\sqrt{13}-1}{2}$ (D) $3$

   ›Reveal solutionSolution

   Row-reducing the determinant collapses it to the cubic $x^3-4x+3=0$, whose positive roots sum to $\dfrac{1+\sqrt{13}}{2}$.

   Concept and Intuition

   A cubic-looking determinant like this is almost always meant to be simplified by row/column operations (subtracting one row from another) before expanding — expanding the raw $3\times3$ determinant directly is much messier.

   Step-by-Step Solution

   1. The determinant is $\begin{vmatrix}x^2+2x & x+2 & 1\\ 2x+1 & x-1 & 1\\ x+2 & -1 & 1\end{vmatrix}=0$.
   2. Apply $R_1 \to R_1-R_2$: $\big(x^2+2x-(2x+1),\ (x+2)-(x-1),\ 1-1\big) = (x^2-1,\ 3,\ 0)$.
   3. Apply $R_2 \to R_2-R_3$: $\big((2x+1)-(x+2),\ (x-1)-(-1),\ 1-1\big) = (x-1,\ x,\ 0)$.
   4. The determinant becomes $\begin{vmatrix}x^2-1 & 3 & 0\\ x-1 & x & 0\\ x+2 & -1 & 1\end{vmatrix}$. Expanding along the third column (only the bottom-right entry is nonzero, with cofactor sign $+$):

   $= 1\cdot\begin{vmatrix}x^2-1 & 3\\ x-1 & x\end{vmatrix} = (x^2-1)x - 3(x-1) = x^3-x-3x+3 = x^3-4x+3$

   5. Set $x^3-4x+3=0$. Testing $x=1$: $1-4+3=0$ ✓, so $(x-1)$ is a factor: $x^3-4x+3=(x-1)(x^2+x-3)$.
   6. Solve $x^2+x-3=0$: $x=\dfrac{-1\pm\sqrt{1+12}}{2}=\dfrac{-1\pm\sqrt{13}}{2}$. …

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8. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
   Q.If $A = \begin{bmatrix} 3 & x & 2 \\ 2x & 3 & -1 \\ 1 & 2 & 3x \end{bmatrix}$ is a singular matrix and $x > 0$, then $\sqrt{3}\,x =$ (A) $\sqrt{21}$ (B) $\sqrt{11}$ (C) $\sqrt{13}$ (D) $\sqrt{17}$

   ›Reveal solutionSolution

   Setting $\det A=0$ gives $-6x^3+34x=0$, so $x^2=\tfrac{17}{3}$ and $\sqrt3\,x=\sqrt{17}$.

   Singular condition. With

   $A=\begin{bmatrix}3 & x & 2\\2x & 3 & -1\\1 & 2 & 3x\end{bmatrix},$

   expand along the first row:

   $\det A = 3\,(3\cdot 3x-(-1)\cdot 2) - x\,(2x\cdot 3x-(-1)\cdot 1) + 2\,(2x\cdot 2-3\cdot 1).$

   $= 3(9x+2) - x(6x^2+1) + 2(4x-3) = 27x+6-6x^3-x+8x-6 = -6x^3+34x.$

   Solve. Set $\det A=0$: …

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9. AP EAPCET 2024Set eng-2024-05-18-FN1 markMCQ
   Q.If $\alpha, \beta, \gamma$ are the roots of $\begin{vmatrix} 1-x & -2 & 1 \\ -2 & 4-x & -2 \\ 1 & -2 & 1-x \end{vmatrix} = 0$, then $\alpha\beta + \beta\gamma + \gamma\alpha =$ (A) 6 (B) 8 (C) 0 (D) -4

   ›Reveal solutionSolution

   This tests recovering $\alpha\beta+\beta\gamma+\gamma\alpha$ (the second elementary symmetric function of the eigenvalues) from a characteristic-equation determinant; the underlying matrix is rank-1, giving eigenvalues $6,0,0$ and hence $\alpha\beta+\beta\gamma+\gamma\alpha = 0$.

   Concept and Intuition

   The roots $\alpha, \beta, \gamma$ of $\det(M - xI) = 0$ are precisely the eigenvalues of the matrix $M = \begin{pmatrix}1&-2&1\\-2&4&-2\\1&-2&1\end{pmatrix}$. For a $3\times3$ matrix, the characteristic polynomial can be written as $x^3 - (\text{tr}\,M)x^2 + e_2\, x - \det M = 0$, where $e_2 = \alpha\beta+\beta\gamma+\gamma\alpha$ equals the sum of the three $2\times2$ principal minors of $M$ (each obtained by deleting one row and the corresponding column). So we just need to compute those three principal minors and add them — no need to actually solve the cubic.

   Step-by-Step Solution

   1. Write $M = \begin{pmatrix}1&-2&1\\-2&4&-2\\1&-2&1\end{pmatrix}$.
   2. Compute the principal minor deleting row/column 1: $\begin{vmatrix}4&-2\\-2&1\end{vmatrix} = 4(1) - (-2)(-2) = 4 - 4 = 0$.
   3. Compute the principal minor deleting row/column 2: $\begin{vmatrix}1&1\\1&1\end{vmatrix} = 1(1) - 1(1) = 0$.
   4. Compute the principal minor deleting row/column 3: $\begin{vmatrix}1&-2\\-2&4\end{vmatrix} = 1(4) - (-2)(-2) = 4 - 4 = 0$.
   5. Sum these principal minors: $0 + 0 + 0 = 0$. This sum is exactly $\alpha\beta+\beta\gamma+\gamma\alpha = e_2$. …

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10. AP EAPCET 2024Set eng-2024-05-22-AN1 markMCQ
    Q.If $\alpha,\beta,\gamma\ (\alpha<\beta<\gamma)$ are the values of x such that $\begin{bmatrix}x-2 & 0 & 1\\ 1 & x+3 & 2\\ 2 & 0 & 2x-1\end{bmatrix}$ is a singular matrix then $2\alpha+3\beta+4\gamma=$ (A) 4 (B) 0 (C) 1 (D) 2

    ›Reveal solutionSolution

    A matrix is singular exactly when its determinant is zero; expanding along the column with two zeros makes the algebra light, and factoring the resulting cubic gives the three roots directly.

    Concept and Intuition

    "Singular" means the determinant vanishes. Whenever a row or column has multiple zero entries, expanding the determinant along that row/column collapses most of the cofactor terms, turning a 3×3 determinant computation into a single 2×2 minor.

    Step-by-Step Solution

    1. Matrix: $\begin{bmatrix}x-2&0&1\\1&x+3&2\\2&0&2x-1\end{bmatrix}$. Column 2 has entries $0,\,x+3,\,0$ — expand along this column.
    2. $\det = (x+3)\times(+1)^{2+2}\times\det\begin{bmatrix}x-2&1\\2&2x-1\end{bmatrix}$ (deleting row 2, column 2).
    3. $\det\begin{bmatrix}x-2&1\\2&2x-1\end{bmatrix} = (x-2)(2x-1)-1\cdot2 = (2x^2-5x+2)-2 = 2x^2-5x$.
    4. So the full determinant is $(x+3)(2x^2-5x)=(x+3)\cdot x\cdot(2x-5)$.
    5. Set to zero: $x=0$, $x=-3$, or $2x-5=0\Rightarrow x=\tfrac52$.
    6. Order them $\alpha<\beta<\gamma$: $\alpha=-3$, $\beta=0$, $\gamma=\tfrac52$. …

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11. AP EAPCET 2023Set eng-2023-05-18-AN1 markMCQ
    Q.The sum of the distinct values of x for which the matrix $A=\begin{bmatrix}1 & 1 & x\\ 1 & x & 1\\ x & 1 & 1\end{bmatrix}$ has no inverse, is (A) $4$ (B) $3$ (C) $2$ (D) $-1$

    ›Reveal solutionSolution

    "No inverse" means $\det A=0$; expand the determinant, factor the resulting cubic, and add only the distinct roots (one root repeats).

    Concept and Intuition

    A square matrix fails to be invertible exactly when its determinant is zero. So we need the values of $x$ that make $\det(A)=0$, then sum the distinct ones (repeated roots are counted once, per the question's wording "distinct values").

    Step-by-Step Solution

    1. $\det(A) = 1(x\cdot1-1\cdot1) - 1(1\cdot1 - 1\cdot x) + x(1\cdot1-x\cdot x)$
    2. $= 1(x-1) - 1(1-x) + x(1-x^2)$
    3. $= (x-1)+(x-1)+x-x^3 = 3x-2-x^3$
    4. Setting $\det(A)=0$: $x^3-3x+2=0$.
    5. Testing $x=1$: $1-3+2=0$ ✓, so $(x-1)$ is a factor.
    6. Dividing: $x^3-3x+2=(x-1)(x^2+x-2)=(x-1)(x-1)(x+2)=(x-1)^2(x+2)$. …

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12. AP EAPCET 2022Set eng-2022-07-08-AN1 markMCQ
    Q.If $A = \begin{bmatrix}1 & 1 & a+1\\ 1 & a+1 & 1\\ a+1 & 1 & 1\end{bmatrix}$ is not an invertible matrix, then the sum of all the values of a is (A) -3 (B) -1 (C) 1 (D) 0

    ›Reveal solutionSolution

    Substituting $x=a+1$ turns the matrix into a classic symmetric determinant that factors as $-(x-1)^2(x+2)$.

    Concept and Intuition

    Not invertible means determinant zero. The specific symmetric structure of this matrix (constant diagonal-off pattern) always factors nicely using row/column operations.

    Step-by-Step Solution

    1. Let $x=a+1$, matrix $=\begin{bmatrix}1&1&x\\1&x&1\\x&1&1\end{bmatrix}$.
    2. Expand along the first row: $\det = 1(x\cdot1-1\cdot1)-1(1\cdot1-1\cdot x)+x(1\cdot1-x\cdot x)$.
    3. $=(x-1)-(1-x)+x(1-x^2) = 2(x-1)+x-x^3 = 3x-2-x^3$.
    4. Set to zero: $x^3-3x+2=0$.
    5. $x=1$ is a root ($1-3+2=0$); factor: $x^3-3x+2=(x-1)(x^2+x-2)=(x-1)(x-1)(x+2)=(x-1)^2(x+2)$. …

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