Q.If $\begin{vmatrix} 2x & 5 \\ 8 & x \end{vmatrix} = \begin{vmatrix} 6 & -2 \\ 7 & 3 \end{vmatrix}$, then value of $x$ is
(A) $3$
(B) $\pm 3$
(C) $\pm 6$
(D) $6$

Showing the 12 most recent of 44 on this concept.

1. 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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2. AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
   Q.If $A = \begin{bmatrix} 1 & 2 & x \\ 4 & -1 & 7 \\ 2 & 4 & -6 \end{bmatrix}$ and the rank of $A$ is 2, then the value of $x$ is equal to (A) $1$ (B) $0$ (C) $-3$ (D) $3$

   ›Reveal solutionSolution

   Rank $<3$ forces $\det A=0$; solving the resulting linear equation in $x$ gives $x=-3$, and a nonzero $2\times2$ minor confirms the rank really is 2. Answer: (C).

   Concept and Intuition

   For a $3\times3$ matrix, rank $2$ means the matrix is not invertible (determinant zero — since full rank 3 would be needed for that) but has at least one nonzero $2\times2$ minor (so the rank doesn't drop further to 1 or 0). So the strategy is: force $\det A=0$ to find the candidate value(s) of $x$, then verify a $2\times2$ minor is nonzero at that $x$ to confirm the rank is exactly 2 (not less).

   Step-by-Step Solution

   1. $A=\begin{bmatrix}1&2&x\\4&-1&7\\2&4&-6\end{bmatrix}$. Expand $\det A$ along row 1: $\det A = 1\big[(-1)(-6)-7(4)\big] - 2\big[4(-6)-7(2)\big] + x\big[4(4)-(-1)(2)\big]$.
   2. Compute each bracket: $(-1)(-6)-7(4)=6-28=-22$; $4(-6)-7(2)=-24-14=-38$; $4(4)-(-1)(2)=16+2=18$.
   3. $\det A = 1(-22) - 2(-38) + x(18) = -22+76+18x = 54+18x$.
   4. Rank $<3$ requires $\det A = 0$: $54+18x=0 \Rightarrow x=-3$. …

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3. 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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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 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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6. 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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7. 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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8. 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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9. AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
   Q.$A = \begin{bmatrix} 0 & k & k \\ k & -4 & -6 \\ k & -3 & -5 \end{bmatrix}$ is a singular matrix for (A) $k = 2$ only (B) $k = \pm 2$ only (C) no real value of $k$ (D) all real values of $k$

   ›Reveal solutionSolution

   Cofactor expansion shows $\det A \equiv 0$ for every real $k$ — the two $k$-dependent terms exactly cancel. Answer: (D).

   Concept and Intuition

   A matrix is singular exactly when its determinant is zero. Here the determinant, expanded in terms of $k$, should in general be some polynomial in $k$ whose roots give the singular values — but it's worth checking whether the polynomial is identically zero (which would make the matrix singular for literally every value of the parameter), since the structure of this particular matrix (rows 2 and 3 are related, first-row entries are $0,k,k$) hints strongly at that possibility.

   Step-by-Step Solution

   1. $A=\begin{bmatrix}0&k&k\\k&-4&-6\\k&-3&-5\end{bmatrix}$. Expand along row 1: $\det A = 0\cdot M_{11} - k\cdot M_{12} + k\cdot M_{13}$, where $M_{12}=\det\begin{bmatrix}k&-6\\k&-5\end{bmatrix}$ and $M_{13}=\det\begin{bmatrix}k&-4\\k&-3\end{bmatrix}$.
   2. $M_{12} = k(-5) - (-6)(k) = -5k+6k = k$.
   3. $M_{13} = k(-3) - (-4)(k) = -3k+4k = k$.
   4. $\det A = 0 - k\cdot k + k\cdot k = -k^2+k^2 = 0$. …

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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 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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12. 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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