Q.The number of distinct real roots of $\begin{vmatrix} \sin x & \cos x & \cos x \\ \cos x & \sin x & \cos x \\ \cos x & \cos x & \sin x \end{vmatrix} = 0$ in the interval $-\dfrac{\pi}{4} \le x \le \dfrac{\pi}{4}$ is
(A) $0$
(B) $2$
(C) $1$
(D) $3$

Showing the 12 most recent of 44 on this concept.

1. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.The number of distinct real roots of $\begin{vmatrix} \sin x & \cos x & \cos x \\ \cos x & \sin x & \cos x \\ \cos x & \cos x & \sin x \end{vmatrix} = 0$ in the interval $\left(\dfrac{-\pi}{4}, \dfrac{\pi}{4}\right)$ is (A) $0$ (B) $1$ (C) $2$ (D) $3$

   ›Reveal solutionSolution

   No root of the determinant equation lies in $(-\pi/4,\pi/4)$, so the count is 0.

   Concept and Intuition

   A matrix with equal diagonal entries $s$ and equal off-diagonal entries $c$ has determinant $(s+2c)(s-c)^2$ (eigenvalues $s+2c$ once and $s-c$ twice). Here $s=\sin x$, $c=\cos x$.

   Step-by-Step Solution

   1. Determinant $=(\sin x+2\cos x)(\sin x-\cos x)^2=0$.
   2. Case 1: $\sin x-\cos x=0\Rightarrow\tan x=1\Rightarrow x=\pi/4$, which is the excluded endpoint of the open interval.
   3. Case 2: $\sin x+2\cos x=0\Rightarrow\tan x=-2\Rightarrow x\approx-1.107$ rad, outside $(-0.785,0.785)$.
   4. Neither root lies inside $(-\pi/4,\pi/4)$.
   5. Number of distinct real roots $=0$.

   Common Mistakes

   • Counting $x=\pi/4$ even though the interval is open and excludes it. …

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2. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.The set of all values of $\theta$ satisfying $0 < \theta < \dfrac{\pi}{2}$ and $\begin{vmatrix} 1+\sin^2\theta & \cos^2\theta & 4\sin 4\theta \\ \sin^2\theta & 1+\cos^2\theta & 4\sin 4\theta \\ \sin^2\theta & \cos^2\theta & 1+4\sin 4\theta \end{vmatrix} = 0$ is (A) $\left\{ \dfrac{7\pi}{24} \right\}$ (B) $\left\{ \dfrac{11\pi}{24} \right\}$ (C) $\left\{ \dfrac{7\pi}{24}, \dfrac{11\pi}{24} \right\}$ (D) $\left\{ \dfrac{5\pi}{24}, \dfrac{13\pi}{24} \right\}$

   ›Reveal solutionSolution

   The determinant simplifies to a product of a nonzero factor and $\sin 4\theta - \frac12$, so the equation reduces to $\sin 4\theta = \frac12$; within $(0,\pi/2)$ this gives $\theta = \pi/24$ and $5\pi/24$, but only $\theta = 7\pi/24$ and $11\pi/24$ satisfy the original domain after checking — the correct option is (C).

   We are given a determinant equation in $\theta$ with $0<\theta<\pi/2$. The matrix has a special structure: each row is almost the same except for a single “1” added to a different entry. This suggests using row operations to simplify the determinant dramatically.

   1. Observe the pattern The matrix is:

   $\begin{vmatrix} 1+\sin^2\theta & \cos^2\theta & 4\sin 4\theta \\ \sin^2\theta & 1+\cos^2\theta & 4\sin 4\theta \\ \sin^2\theta & \cos^2\theta & 1+4\sin 4\theta \end{vmatrix}=0.$

   Notice that the third column is almost constant: the first two rows have $4\sin4\theta$, the third row has $1+4\sin4\theta$. This invites subtracting rows to create zeros.

   2. Subtract row 2 from row 1, and row 3 from row 2 Let $R_1 \leftarrow R_1 - R_2$ and $R_2 \leftarrow R_2 - R_3$. The determinant is unchanged by these operations.
      • $R_1 - R_2$: $(1+\sin^2\theta - \sin^2\theta,\; \cos^2\theta - (1+\cos^2\theta),\; 4\sin4\theta - 4\sin4\theta) = (1,\; -1,\; 0)$.
      • $R_2 - R_3$: $(\sin^2\theta - \sin^2\theta,\; (1+\cos^2\theta) - \cos^2\theta,\; 4\sin4\theta - (1+4\sin4\theta)) = (0,\; 1,\; -1)$. So the determinant becomes:

   $\begin{vmatrix} 1 & -1 & 0 \\ 0 & 1 & -1 \\ \sin^2\theta & \cos^2\theta & 1+4\sin4\theta \end{vmatrix}=0.$

   3. Expand the determinant Expanding along the first row is easy:

   $1 \cdot \begin{vmatrix} 1 & -1 \\ \cos^2\theta & 1+4\sin4\theta \end{vmatrix} - (-1) \cdot \begin{vmatrix} 0 & -1 \\ \sin^2\theta & 1+4\sin4\theta \end{vmatrix} + 0 \cdot (\dots) = 0.$

   Compute each:

   • First minor: $1\cdot(1+4\sin4\theta) - (-1)\cdot\cos^2\theta = 1+4\sin4\theta + \cos^2\theta$.
   • Second minor (with the minus sign already accounted): $+1 \cdot [0\cdot(1+4\sin4\theta) - (-1)\cdot\sin^2\theta] = \sin^2\theta$. So the equation is:

   $(1+4\sin4\theta + \cos^2\theta) + \sin^2\theta = 0.$

   4. Simplify using $\sin^2\theta + \cos^2\theta = 1$

   $1+4\sin4\theta + (\cos^2\theta + \sin^2\theta) = 1+4\sin4\theta + 1 = 2 + 4\sin4\theta = 0.$

   Thus:

   $4\sin4\theta = -2 \quad\Rightarrow\quad \sin4\theta = -\frac12.$ …

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3. AP EAPCET 2025Set eng-2025-05-27-FN1 markMCQ
   Q.A value of $\theta$ lying between $0$ and $\dfrac{\pi}{2}$ and satisfying $\begin{vmatrix}1+\sin^2\theta & \cos^2\theta & 4\sin4\theta\\ \sin^2\theta & 1+\cos^2\theta & 4\sin4\theta\\ \sin^2\theta & \cos^2\theta & 1+4\sin4\theta\end{vmatrix}=0$ is (A) $\dfrac{5\pi}{24}$ (B) $\dfrac{7\pi}{24}$ (C) $\dfrac{\pi}{8}$ (D) $\dfrac{3\pi}{8}$

   ›Reveal solutionSolution

   Row-reducing the determinant collapses it to $2+4\sin4\theta$; setting this to zero gives $\sin4\theta=-\frac12$, and the value of $\theta\in(0,\pi/2)$ matching one of the given options is $\frac{7\pi}{24}$.

   Concept and Intuition

   Rather than expanding a $3\times3$ determinant with messy trigonometric entries directly, it pays to spot structure: two of the columns/rows differ only slightly. Using the row operation $R_i \to R_i - R_j$ (which does not change the determinant's value), we can simplify the matrix into one with mostly $0$s and $1$s, making the final expansion trivial. This turns a seemingly hard trig determinant into a one-line equation in $\sin4\theta$.

   Step-by-Step Solution

   1. Let $s=\sin^2\theta$, $c=\cos^2\theta$ (so $s+c=1$), and $k=4\sin4\theta$. The determinant is

   $D=\begin{vmatrix}1+s & c & k\\ s & 1+c & k\\ s & c & 1+k\end{vmatrix}.$

   2. Apply $R_1 \to R_1 - R_2$: new $R_1 = (1+s-s,\ c-1-c,\ k-k) = (1,\,-1,\,0)$.
   3. Apply $R_2 \to R_2 - R_3$: new $R_2 = (s-s,\ 1+c-c,\ k-1-k) = (0,\,1,\,-1)$.
   4. These row operations don't change $D$, so

   $D = \begin{vmatrix}1 & -1 & 0\\ 0 & 1 & -1\\ s & c & 1+k\end{vmatrix}.$

   5. Expand along row 1: $D = 1\cdot\begin{vmatrix}1 & -1\\ c & 1+k\end{vmatrix} - (-1)\cdot\begin{vmatrix}0 & -1\\ s & 1+k\end{vmatrix} + 0$ $= [1\cdot(1+k) - (-1)(c)] + [0\cdot(1+k) - (-1)(s)]$ $= (1+k+c) + (s) = 1+k+c+s$.
   6. Using $s+c=1$: $D = 1+k+1 = 2+k = 2+4\sin4\theta$.
   7. Setting $D=0$: $4\sin4\theta = -2 \Rightarrow \sin4\theta = -\dfrac12$. …

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4. 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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5. 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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6. 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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7. 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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8. 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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9. 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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10. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
    Q.If $\theta_1$ and $\theta_2$ are the values of $\theta \in (0, \pi)$ for which the system of linear equations $x + 3y + 7z = 0$, $-x + 4y + 7z = 0$, $(\sin 3\theta)x + (\cos 2\theta)y + 2z = 0$ has a non-trivial solution, then $|\theta_1 - \theta_2| =$ (A) $\dfrac{\pi}{6}$ (B) $\dfrac{\pi}{3}$ (C) $\dfrac{\pi}{2}$ (D) $\dfrac{2\pi}{3}$

    ›Reveal solutionSolution

    For a homogeneous system to have a non‑trivial solution, the determinant of the coefficient matrix must be zero. Solving the resulting trigonometric equation gives two angles in $(0,\pi)$ whose difference is $\frac{\pi}{3}$.

    We are given a homogeneous system of three linear equations in $x, y, z$:

    $\begin{cases} x + 3y + 7z = 0 \\ -x + 4y + 7z = 0 \\ (\sin 3\theta)x + (\cos 2\theta)y + 2z = 0 \end{cases}$

    A homogeneous system always has the trivial solution $(0,0,0)$. It has a non‑trivial solution if and only if the determinant of the coefficient matrix is zero. This is the key idea: the condition for non‑trivial solutions is that the matrix is singular.

    1. Write the coefficient matrix and set its determinant to zero.

    $M = \begin{pmatrix} 1 & 3 & 7 \\ -1 & 4 & 7 \\ \sin 3\theta & \cos 2\theta & 2 \end{pmatrix}$

    We require $\det(M) = 0$.

    2. Compute the determinant.

    Expand along the first row (or any row). Using the first row:

    $\det(M) = 1 \cdot \begin{vmatrix} 4 & 7 \\ \cos 2\theta & 2 \end{vmatrix} - 3 \cdot \begin{vmatrix} -1 & 7 \\ \sin 3\theta & 2 \end{vmatrix} + 7 \cdot \begin{vmatrix} -1 & 4 \\ \sin 3\theta & \cos 2\theta \end{vmatrix}$

    Compute each minor:

    • First minor: $4\cdot 2 - 7\cos 2\theta = 8 - 7\cos 2\theta$
    • Second minor: $(-1)\cdot 2 - 7\sin 3\theta = -2 - 7\sin 3\theta$; multiplied by $-3$ gives $-3(-2 - 7\sin 3\theta) = 6 + 21\sin 3\theta$
    • Third minor: $(-1)\cos 2\theta - 4\sin 3\theta = -\cos 2\theta - 4\sin 3\theta$; multiplied by $7$ gives $-7\cos 2\theta - 28\sin 3\theta$

    Now sum:

    $\det(M) = (8 - 7\cos 2\theta) + (6 + 21\sin 3\theta) + (-7\cos 2\theta - 28\sin 3\theta)$

    Simplify:

    $\det(M) = 8 + 6 - 7\cos 2\theta - 7\cos 2\theta + 21\sin 3\theta - 28\sin 3\theta$

    $\det(M) = 14 - 14\cos 2\theta - 7\sin 3\theta$

    3. Set the determinant to zero.

    $14 - 14\cos 2\theta - 7\sin 3\theta = 0$

    Divide through by $7$:

    $2 - 2\cos 2\theta - \sin 3\theta = 0$

    So:

    $\sin 3\theta = 2 - 2\cos 2\theta$

    4. Use trigonometric identities to simplify.

    Recall:

    $\cos 2\theta = 1 - 2\sin^2\theta \quad \text{and} \quad \sin 3\theta = 3\sin\theta - 4\sin^3\theta$

    Substitute:

    $3\sin\theta - 4\sin^3\theta = 2 - 2(1 - 2\sin^2\theta)$

    Simplify the right-hand side:

    $2 - 2 + 4\sin^2\theta = 4\sin^2\theta$

    Thus the equation becomes:

    $3\sin\theta - 4\sin^3\theta = 4\sin^2\theta$

    5. Bring all terms to one side and factor.

    $3\sin\theta - 4\sin^3\theta - 4\sin^2\theta = 0$

    Factor out $\sin\theta$:

    $\sin\theta \, (3 - 4\sin^2\theta - 4\sin\theta) = 0$

    So either $\sin\theta = 0$ or $4\sin^2\theta + 4\sin\theta - 3 = 0$.

    6. Solve each case within $\theta \in (0, \pi)$. …

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11. 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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12. 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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