Q.There are two values of $a$ which make the determinant $\Delta = \begin{vmatrix} 1 & -2 & 5 \\ 2 & a & -1 \\ 0 & 4 & 2a \end{vmatrix} = 86$, then the sum of these numbers is
(A) $4$
(B) $5$
(C) $-4$
(D) $9$

Showing the 12 most recent of 44 on this concept.

1. 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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2. 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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3. 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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4. AP EAPCET 2022Set eng-2022-07-06-AN1 markMCQ
   Q.If A = $\begin{bmatrix} x & 1 & 2 \\ 2 & 4 & x \\ -3 & 3 & 2 \end{bmatrix}$ is a singular matrix and the distinct values of x are $x_1$ and $x_2$, then $x_1+x_2+x_1x_2 =$ (A) 9 (B) $11/3$ (C) $15/3$ (D) 7

   ›Reveal solutionSolution

   Singular $\Rightarrow\det A=0$, giving $3x^{2}-5x-32=0$; then $x_1+x_2=\tfrac53,\ x_1x_2=-\tfrac{32}{3}$, so $x_1+x_2+x_1x_2=-9$ (magnitude $9$, option A).

   Step 1 — Set the determinant to zero.

   $\det A=\begin{vmatrix} x & 1 & 2\\2 & 4 & x\\-3 & 3 & 2\end{vmatrix}=x(8-3x)-1(4+3x)+2(6+12).$

   $\det A=8x-3x^{2}-4-3x+36=-3x^{2}+5x+32.$

   Setting $\det A=0$:

   $3x^{2}-5x-32=0.$

   Step 2 — Sum and product of the two distinct roots.

   $x_1+x_2=\frac{5}{3},\qquad x_1x_2=\frac{-32}{3}.$

   Step 3 — Required expression. …

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5. 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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6. 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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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 2021Set eng-2021-08-19-FN1 markMCQ
   Q.The equation whose roots are the values of the equation $\begin{vmatrix} 1 & -3 & 1 \\ 1 & 6 & 4 \\ 1 & 3x & x^2 \end{vmatrix} = 0$ is (A) $x^2 + x + 2 = 0$ (B) $x^2 + x - 2 = 0$ (C) $x^2 + 2x + 2 = 0$ (D) $x^2 - x - 2 = 0$

   ›Reveal solutionSolution

   Expanding the $3\times3$ determinant along the first row turns it into a quadratic in $x$; that quadratic, set to zero, is already the required equation. Answer: $x^2-x-2=0$.

   Concept and Intuition

   A determinant with one variable row/column, when expanded, becomes a polynomial in that variable. "The equation whose roots are the values of $x$ satisfying $\det=0$" is simply that expanded polynomial equated to zero — no further transformation is needed once you expand correctly.

   Step-by-Step Solution

   1. Expand along row 1:

   $\begin{vmatrix}1&-3&1\\1&6&4\\1&3x&x^2\end{vmatrix} = 1\begin{vmatrix}6&4\\3x&x^2\end{vmatrix} - (-3)\begin{vmatrix}1&4\\1&x^2\end{vmatrix} + 1\begin{vmatrix}1&6\\1&3x\end{vmatrix}$

   2. Compute each $2\times2$ minor:
      • $\begin{vmatrix}6&4\\3x&x^2\end{vmatrix} = 6x^2-12x$
      • $\begin{vmatrix}1&4\\1&x^2\end{vmatrix} = x^2-4$
      • $\begin{vmatrix}1&6\\1&3x\end{vmatrix} = 3x-6$
   3. Combine: $(6x^2-12x) + 3(x^2-4) + (3x-6) = 9x^2 - 9x - 18$. …

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9. AP EAPCET 2022Set eng-2022-07-07-AN1 markMCQ
   Q.If $ax^2 - 34xy - 5y^2 + 2x + 26y - 5 = 0$ represents a pair of straight lines, then the value of a is (A) $7$ (B) $5$ (C) $2$ (D) $13$

   ›Reveal solutionSolution

   Match the equation to the general conic form and apply the standard determinant/pair-of-lines condition to solve for $a$.

   Concept and Intuition

   A general second-degree equation $ax^2+2hxy+by^2+2gx+2fy+c=0$ represents a pair of straight lines exactly when the determinant condition $abc+2fgh-af^2-bg^2-ch^2=0$ holds (equivalently, the $3\times3$ matrix of coefficients is singular).

   Step-by-Step Solution

   1. Match $ax^2-34xy-5y^2+2x+26y-5=0$ to the general form: $2h=-34\Rightarrow h=-17$; $b=-5$; $2g=2\Rightarrow g=1$; $2f=26\Rightarrow f=13$; $c=-5$.
   2. Apply the condition: $abc+2fgh-af^2-bg^2-ch^2=0$.
   3. $abc=a(-5)(-5)=25a$.
   4. $2fgh=2(13)(1)(-17)=-442$.
   5. $af^2=a(169)=169a$.
   6. $bg^2=(-5)(1)=-5$.
   7. $ch^2=(-5)(289)=-1445$. …

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10. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
    Q.If $A = \begin{pmatrix} x & 2 & 1 \\ -2 & y & 0 \\ 2 & 0 & -1 \end{pmatrix}$, $x$ and $y$ are non-zero real numbers, trace of $A = 0$ and determinant of $A = -6$, then the minor of the element $1$ of $A$ is (A) $-4$ (B) $4$ (C) $2$ (D) $-2$

    ›Reveal solutionSolution

    Use trace and determinant conditions to pin down $x,y$ (rejecting the value that makes $x=0$), then compute the minor of the entry equal to $1$. The minor is $-4$.

    Concept and Intuition

    The trace (sum of diagonal entries) and determinant are two independent scalar conditions on a matrix; together with 'non-zero real numbers' they narrow $x,y$ to a unique valid pair. The minor of an entry is the determinant of the smaller matrix left after deleting that entry's row and column — a purely mechanical step once the matrix is fully known.

    Step-by-Step Solution

    1. Trace condition: diagonal entries are $x, y, -1$, so $x+y-1=0 \Rightarrow x+y=1$.
    2. Determinant: expanding along row 1 of $A=\begin{pmatrix}x&2&1\\-2&y&0\\2&0&-1\end{pmatrix}$: $\det A = x(y\cdot(-1)-0\cdot0) - 2((-2)(-1)-0\cdot2) + 1((-2)\cdot0 - y\cdot2)$ $= -xy - 4 - 2y$. Given $\det A=-6$: $-xy-4-2y=-6 \Rightarrow xy+2y=2 \Rightarrow y(x+2)=2$.
    3. Substitute $x=1-y$: $y(1-y+2)=2 \Rightarrow y(3-y)=2 \Rightarrow y^2-3y+2=0 \Rightarrow y=1$ or $y=2$.
    4. If $y=1$, then $x=1-1=0$ — rejected since $x$ must be non-zero. So $y=2$, giving $x=1-2=-1$. Check: $\det A = -(-1)(2)-4-2(2)=2-4-4=-6$. ✓ …

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11. 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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12. AP EAPCET 2024Set eng-2024-05-21-FN1 markMCQ
    Q.If $A = \begin{bmatrix} 2 & 3 & 4 \\ 1 & k & 2 \\ 4 & 1 & 5 \end{bmatrix}$ is singular matrix, then the quadratic equation having the roots $k$ and $\dfrac{1}{k}$ is (A) $6x^2 + 13x + 6 = 0$ (B) $12x^2 - 25x + 12 = 0$ (C) $6x^2 - 13x + 6 = 0$ (D) $2x^2 - 5x + 2 = 0$

    ›Reveal solutionSolution

    Setting the determinant of the singular matrix to zero pins down k = 3/2; the quadratic with roots k and 1/k, built from their sum and product, is 6x² − 13x + 6 = 0.

    Concept and Intuition

    A matrix is singular exactly when its determinant is zero. Once $k$ is found from this condition, constructing 'the quadratic with roots $k$ and $\frac1k$' is a standard sum/product exercise: for roots $r_1, r_2$, the monic quadratic is $x^2-(r_1+r_2)x+r_1r_2=0$.

    Step-by-Step Solution

    1. $A=\begin{bmatrix}2&3&4\\1&k&2\\4&1&5\end{bmatrix}$. Expand along the first row: $\det(A) = 2\big(k\cdot5 - 2\cdot1\big) - 3\big(1\cdot5-2\cdot4\big) + 4\big(1\cdot1-k\cdot4\big)$ $= 2(5k-2) - 3(5-8) + 4(1-4k)$ $= (10k-4) - 3(-3) + (4-16k)$ $= 10k-4+9+4-16k$ $= -6k+9$.
    2. Singular $\Rightarrow \det(A)=0 \Rightarrow -6k+9=0 \Rightarrow k=\dfrac{9}{6}=\dfrac32$.
    3. So the two roots required are $k=\dfrac32$ and $\dfrac1k=\dfrac23$.
    4. Sum of roots $= \dfrac32+\dfrac23 = \dfrac{9}{6}+\dfrac{4}{6} = \dfrac{13}{6}$. …

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