Q.If $\begin{vmatrix} x & 2 \\ 18 & x \end{vmatrix} = \begin{vmatrix} 6 & 2 \\ 18 & 6 \end{vmatrix}$, then $x$ is equal to (A) 6 (B) $\pm 6$ (C) $-6$ (D) 0

Showing the 12 most recent of 32 on this concept.

1. AP EAPCET 2021Set eng-2021-08-23-AN1 markMCQ
   Q.The value of the determinant $\begin{vmatrix} a+b & a+2b & a+3b \\ a+2b & a+3b & a+4b \\ a+4b & a+5b & a+6b \end{vmatrix}$ is ____ (A) $a$ (B) $b$ (C) $0$ (D) $a+b$

   ›Reveal solutionSolution

   The rows of this determinant are in arithmetic progression (each row's entries increase by a constant step, and consecutive rows shift by a constant amount too); row-reducing shows two rows become proportional, forcing the determinant to $0$.

   Concept and Intuition

   A determinant is zero whenever any two rows (or columns) are linearly dependent (e.g. one is a scalar multiple of another, or a linear combination of others). Rows built from an arithmetic-progression pattern (like $a+b, a+2b, a+3b$ then shifting by a constant each row) are a classic setup for this — subtracting consecutive rows collapses the "arithmetic" structure into constant, proportional rows.

   Step-by-Step Solution

   1. Original matrix:

   $\begin{vmatrix} a+b & a+2b & a+3b \\ a+2b & a+3b & a+4b \\ a+4b & a+5b & a+6b \end{vmatrix}$

   2. Perform $R_2 \to R_2 - R_1$: new $R_2 = (a+2b-(a+b),\ a+3b-(a+2b),\ a+4b-(a+3b)) = (b,\ b,\ b)$.
   3. Perform $R_3 \to R_3 - R_2^{\text{(original)}}$: new $R_3 = (a+4b-(a+2b),\ a+5b-(a+3b),\ a+6b-(a+4b)) = (2b,\ 2b,\ 2b)$. …

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2. AP EAPCET 2021Set eng-2021-08-20-FN1 markMCQ
   Q.What is the value of $\begin{vmatrix} a & b & c \\ a-b & b-c & c-a \\ b+c & c+a & a+b \end{vmatrix} = ?$ (A) $a^3 + b^3 + c^3 + 3abc$ (B) $a^3 + b^3 + c^3 - 3abc$ (C) $a^3 + b^3 + c^3 - 6abc$ (D) $a^3 + b^3 + c^3 + 6abc$

   ›Reveal solutionSolution

   Direct cofactor expansion of the determinant shows all cross terms cancel, leaving the

   classical identity $a^3+b^3+c^3-3abc$.

   Concept and Intuition

   This is a disguised version of the well-known factorisation

   $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$, packaged as a determinant. Expanding carefully

   term by term (rather than guessing) confirms which of the four sign variants is correct.

   Step-by-Step Solution

   1. Expand along the first row: $D = a\left[(b-c)(a+b)-(c-a)(c+a)\right] - b\left[(a-b)(a+b)-(c-a)(b+c)\right] + c\left[(a-b)(c+a)-(b-c)(b+c)\right]$.
   2. Compute each bracket:
      • $(b-c)(a+b)-(c-a)(c+a) = ab+b^2-ac-bc-c^2+a^2$
      • $(a-b)(a+b)-(c-a)(b+c) = a^2-b^2-bc-c^2+ab+ac$
      • $(a-b)(c+a)-(b-c)(b+c) = ac+a^2-bc-ab-b^2+c^2$
   3. Multiply through by $a$, $-b$, $c$ respectively and add. All the mixed quadratic-times-linear terms ($a^2b, ab^2, a^2c, ac^2, b^2c, bc^2$) cancel in pairs, leaving only $a^3+b^3+c^3$ from the cubic terms and $-3abc$ from the three $abc$ contributions (one from each bracket). …

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3. AP EAPCET 2023Set eng-2023-05-17-FN1 markMCQ
   Q.If $f(x)=\begin{vmatrix}1 & 6+x & 36+x^2\\ 0 & x-3 & 3x^2-27\\ 0 & 2x-4 & 8x^2-32\end{vmatrix}$, then $\lim\limits_{x\to1}\dfrac{f(x)}{f(-x)}=$ (A) $2$ (B) $-1$ (C) $0$ (D) $1$

   ›Reveal solutionSolution

   Expanding the determinant reveals $f(x)=2(x-1)(x-2)(x-3)$, a factor that vanishes at $x=1$, so the ratio $f(x)/f(-x)$ has a zero numerator and finite nonzero denominator there — limit $0$.

   Concept and Intuition

   Determinants with a column like $(1,0,0)^T$ expand trivially (cofactor of the top-left entry only), collapsing a $3\times3$ determinant into a $2\times2$ one. Recognising the resulting expression as a product of simple linear factors (rather than grinding through raw polynomial expansion) makes the limit evaluation immediate.

   Step-by-Step Solution

   1. Expand along column 1 (entries $1,0,0$): $f(x)=1\cdot\begin{vmatrix}x-3 & 3x^2-27\\ 2x-4 & 8x^2-32\end{vmatrix}=(x-3)(8x^2-32)-(3x^2-27)(2x-4)$.
   2. Factor: $8x^2-32=8(x-2)(x+2)$, $3x^2-27=3(x-3)(x+3)$, $2x-4=2(x-2)$.
   3. $f(x)=8(x-3)(x-2)(x+2)-6(x-3)(x+3)(x-2)=(x-3)(x-2)\big[8(x+2)-6(x+3)\big]=(x-3)(x-2)(2x-2)=2(x-1)(x-2)(x-3)$.
   4. $f(-x)=2(-x-1)(-x-2)(-x-3)=2\cdot(-1)^3(x+1)(x+2)(x+3)=-2(x+1)(x+2)(x+3)$. …

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4. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.$A = \begin{bmatrix} b+c & a & a \\ b & c+a & b \\ c & c & a+b \end{bmatrix}$ is a matrix such that trace of $A = 18$ and $\det(A) = 96$. If $a, b, c \in N$ and $ab = 6$, then $ab + bc + ca =$ (A) $36$ (B) $26$ (C) $48$ (D) $24$

   ›Reveal solutionSolution

   Using trace(A) = 2(a+b+c) and the identity det(A) = 4abc, we get a+b+c=9 and abc=24; with ab=6 this pins down c=4, {a,b}={2,3}, and ab+bc+ca=26.

   Concept and Intuition

   The matrix

   $A=\begin{bmatrix} b+c & a & a \\ b & c+a & b \\ c & c & a+b \end{bmatrix}$

   has a very clean structure: each row sums to $a+b+c$, and there is a well-known identity that its determinant simplifies to $4abc$ (this can be shown by row/column operations: subtracting appropriately or direct cofactor expansion collapses most cross terms). Recognising this identity turns a messy 3×3 determinant into a one-line relation between $a,b,c$.

   Step-by-Step Solution

   1. Trace: $(b+c) + (c+a) + (a+b) = 2(a+b+c) = 18 \Rightarrow a+b+c = 9$.
   2. Determinant identity: expand $\det(A)$ directly — $\det(A) = (b+c)[(c+a)(a+b)-bc] - a[b(a+b)-bc] + a[bc - c(c+a)]$. Simplifying each bracket:
      • $(c+a)(a+b) - bc = a^2+ab+ac+bc-bc = a(a+b+c)$
      • Combining all terms and cancelling cross terms (ab, ac, b², c² all cancel) leaves $4abc$. So $\det(A) = 4abc = 96 \Rightarrow abc = 24$.
   3. Given $ab = 6$: $c = abc/(ab) = 24/6 = 4$.
   4. From step 1: $a+b = 9 - c = 9-4 = 5$. …

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5. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.If $f,g,h$ are differentiable functions of $x$, then $\begin{vmatrix} f & g & h \\ (xf)' & (xg)' & (xh)' \\ f' & g' & h' \end{vmatrix} =$ (A) $fg' - gh'$ (B) $gh' + xf'$ (C) $0$ (D) $x(f'+g'+h')$

   ›Reveal solutionSolution

   The determinant simplifies to zero because the second row is a linear combination of the first and third rows, making the rows linearly dependent.

   We are asked to evaluate

   $\Delta = \begin{vmatrix} f & g & h \\ (xf)' & (xg)' & (xh)' \\ f' & g' & h' \end{vmatrix}$

   where $f,g,h$ are differentiable functions of $x$.

   ---

   Concept and Intuition

   The key idea is to use row operations that do not change the value of a determinant, or to notice linear dependence among rows.

   Here, the second row contains derivatives of products $xf, xg, xh$. Using the product rule,

   $(xf)' = f + x f', \quad (xg)' = g + x g', \quad (xh)' = h + x h'.$

   So the second row is actually:

   $\begin{pmatrix} f + x f' & g + x g' & h + x h' \end{pmatrix}.$

   Notice that this is exactly Row 1 plus $x$ times Row 3:

   $\text{Row 2} = \text{Row 1} + x \cdot \text{Row 3}.$

   When one row is a linear combination of the others, the determinant is zero.

   ---

   Step-by-step reasoning

   1. Expand the second row entries using the product rule:

   $(xf)' = f + x f', \quad (xg)' = g + x g', \quad (xh)' = h + x h'.$

   2. Rewrite the determinant with this expanded form:

   $\Delta = \begin{vmatrix} f & g & h \\ f + x f' & g + x g' & h + x h' \\ f' & g' & h' \end{vmatrix}.$

   3. Perform a row operation that does not change the determinant: subtract Row 1 from Row 2.

   $\text{New Row 2} = \text{Row 2} - \text{Row 1} = \begin{pmatrix} x f' & x g' & x h' \end{pmatrix}.$

   So

   $\Delta = \begin{vmatrix} f & g & h \\ x f' & x g' & x h' \\ f' & g' & h' \end{vmatrix}.$ …

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6. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.If $f(x)=\begin{vmatrix} 2+\dfrac{\sin x}{x} & 1 & 3 \\ 2 & 1+\dfrac{\sin x}{x} & 3 \\ 2 & 1 & 3+\dfrac{\sin x}{x} \end{vmatrix}$, then $\lim\limits_{x\to 0} f(x)=$ (A) $1$ (B) $0$ (C) $5$ (D) $7$

   ›Reveal solutionSolution

   Writing the determinant as $\det(B+sI)$ for a rank-1 matrix $B$ with eigenvalues $6,0,0$, the limit as $s=\sin x/x\to1$ is $7$.

   Concept and Intuition

   Each row of the matrix, after removing the $s=\sin x/x$ terms sitting only on the diagonal, is identical: $(2,1,3)$. A matrix all of whose rows are the same vector $v$ is rank 1, and its eigenvalues are $\mathrm{trace}=v_1+v_2+v_3$ (once) and $0$ (with multiplicity $n-1$). Adding $sI$ shifts every eigenvalue by $s$, so the determinant of the shifted matrix is just the product of the shifted eigenvalues — no need to expand a $3\times3$ determinant directly.

   Step-by-Step Solution

   1. Write $f(x) = \det(B+sI)$ where $s=\sin x/x$ and $B=\begin{pmatrix}2&1&3\\2&1&3\\2&1&3\end{pmatrix}$ (every row is $(2,1,3)$, since the $s$ only appears added to the diagonal entries).
   2. $B$ has identical rows $\Rightarrow$ rank 1 $\Rightarrow$ two eigenvalues are $0$, and the third equals $\mathrm{trace}(B)=2+1+3=6$.
   3. Adding $sI$ shifts each eigenvalue of $B$ by $s$: eigenvalues of $B+sI$ are $6+s,\ s,\ s$.
   4. $\det(B+sI) = (6+s)\cdot s\cdot s = s^2(6+s)$. …

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7. AP EAPCET 2021Set eng-2021-08-23-FN1 markMCQ
   Q.If $\omega$ is a root of the equation $x + \dfrac{1}{x} + 1 = 0$, then $\begin{vmatrix} 1 & 1+\omega & 1+\omega+\omega^2 \\ 3 & 4+3\omega & 5+4\omega+3\omega^2 \\ 6 & 9+6\omega & 11+9\omega+6\omega^2 \end{vmatrix} =$ (A) $1$ (B) $-1$ (C) $0$ (D) $1+\omega$

   ›Reveal solutionSolution

   Using $1+\omega+\omega^2=0$ to simplify the matrix entries and row-reducing gives determinant $-1$.

   Concept and Intuition

   $\omega$ here is a complex cube root of unity ($\omega^2+\omega+1=0$, $\omega^3=1$). Many determinant entries are partial sums of $1,\omega,\omega^2$ multiples, so replacing $1+\omega+\omega^2$ by $0$ wherever it appears collapses the algebra dramatically.

   Step-by-Step Solution

   1. Entry $(1,3)=1+\omega+\omega^2=0$.
   2. Entry $(2,3)=5+4\omega+3\omega^2$. Since $3\omega^2=3(-1-\omega)=-3-3\omega$, this equals $5+4\omega-3-3\omega=2+\omega$.
   3. Entry $(3,3)=11+9\omega+6\omega^2=11+9\omega+6(-1-\omega)=5+3\omega$.
   4. The matrix is now $\begin{pmatrix}1 & 1+\omega & 0\\ 3 & 4+3\omega & 2+\omega\\ 6 & 9+6\omega & 5+3\omega\end{pmatrix}$.
   5. Row reduce: $R_2\to R_2-3R_1 = (0,\,1,\,2+\omega)$; $R_3\to R_3-6R_1=(0,\,3,\,5+3\omega)$. …

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8. AP EAPCET 2024Set eng-2024-05-21-AN1 markMCQ
   Q.$\begin{vmatrix} a+b+2c & a & b \\ c & b+c+2a & b \\ c & a & c+a+2b \end{vmatrix} =$ (A) $(a+b+c)^3$ (B) $2(a+b+c)^3$ (C) $3(a+b+c)^3$ (D) $(a+b+c)$

   ›Reveal solutionSolution

   Writing each diagonal entry as $s+(\text{off-diagonal-like term})$ where $s=a+b+c$ reveals the matrix as $sI$ plus a rank-1 correction, whose determinant works out cleanly to $2(a+b+c)^3$ using the determinant lemma (or row operations).

   Concept and Intuition

   Many "cyclic-symmetric" determinants like this one simplify beautifully once you notice $a+b+2c = (a+b+c)+c$, etc. — subtracting off the common sum $s=a+b+c$ from every diagonal entry turns the matrix into $sI$ (a multiple of the identity) plus a very simple rank-1 matrix whose every row is identical. Determinants of $sI + \text{rank-1}$ matrices have a clean closed form.

   Step-by-Step Solution

   1. Let $s=a+b+c$. Rewrite diagonal entries: $a+b+2c=s+c$, $b+c+2a=s+a$, $c+a+2b=s+b$.
   2. The matrix becomes:

   $M=\begin{bmatrix}s+c & a & b\\ c & s+a & b\\ c & a & s+b\end{bmatrix}$

   3. Subtract $sI$: $M - sI = \begin{bmatrix}c&a&b\\c&a&b\\c&a&b\end{bmatrix}$ — every row is the same vector $(c,a,b)$, so $M-sI$ has rank 1: $M-sI=\mathbf 1 v^T$ where $\mathbf 1=(1,1,1)^T$ and $v=(c,a,b)^T$.
   4. So $M=sI+\mathbf1v^T$. The determinant lemma (matrix determinant lemma) gives: $\det(sI+\mathbf1 v^T) = s^3 + s^2\,(v^T\mathbf1)$ …

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9. AP EAPCET 2025Set eng-2025-05-26-FN1 markMCQ
   Q.If the inverse of $\begin{bmatrix}-x & 14x & 7x\\ 0 & 1 & 0\\ x & -4x & -2x\end{bmatrix}$ is $\begin{bmatrix}2 & 0 & 7\\ 0 & 1 & 0\\ 1 & -2 & 1\end{bmatrix}$, then $\begin{vmatrix}x & x+1 & x+2\\ x+1 & x+2 & x+3\\ x+2 & x+3 & x+4\end{vmatrix} =$ (A) $\dfrac{x}{5}$ (B) $x-5$ (C) $5x-1$ (D) $x+5$

   ›Reveal solutionSolution

   This tests recognizing a determinant with AP-rows (always zero) and pinning down $x$ from a matrix-inverse condition; the answer is whichever option numerically equals $0$ at that $x$.

   Concept and Intuition

   A square matrix whose rows form an arithmetic progression (i.e. $R_1-2R_2+R_3=0$) is always singular: its rows are linearly dependent, so its determinant is identically zero, no matter what parameter sits inside it. Spotting this saves a full cofactor expansion. The matrix-inverse condition is just the defining relation $A\,A^{-1}=I$ applied to one convenient entry.

   Step-by-Step Solution

   1. Let $A=\begin{bmatrix}-x & 14x & 7x\\ 0 & 1 & 0\\ x & -4x & -2x\end{bmatrix}$ and $A^{-1}=\begin{bmatrix}2 & 0 & 7\\ 0 & 1 & 0\\ 1 & -2 & 1\end{bmatrix}$.
   2. Using $A\,A^{-1}=I$, multiply row 1 of $A$ by column 1 of $A^{-1}$: $(-x)(2)+(14x)(0)+(7x)(1)=5x$. This must equal the $(1,1)$ entry of $I$, i.e. $1$. So $5x=1\Rightarrow x=\dfrac15$. (Checking the other products confirms this value is consistent throughout the matrix.)
   3. Now look at $D=\begin{vmatrix}x & x+1 & x+2\\ x+1 & x+2 & x+3\\ x+2 & x+3 & x+4\end{vmatrix}$. Apply $R_1\to R_1-2R_2+R_3$: each entry becomes $x-2(x+1)+(x+2)=0$, turning the first row into $(0,0,0)$.
   4. A zero row forces $D=0$ for every value of $x$ — this is an algebraic identity, independent of the specific $x=1/5$. …

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10. AP EAPCET 2021Set eng-2021-08-19-AN1 markMCQ
    Q.The value of $\begin{vmatrix} b+c & a & a \\ b & c+a & b \\ c & c & a+b \end{vmatrix}$ is (A) $abc$ (B) $(a+b)(b+c)(c+a)$ (C) $4abc$ (D) $(a-b)(b-c)(c-a)$

    ›Reveal solutionSolution

    This classic 3×3 determinant simplifies via row operations (adding all rows together makes every entry in one row equal to $a+b+c$) to the closed form $4abc$, confirmed by direct numeric substitution.

    Concept and Intuition

    Many "nice" symmetric determinants like this one are best handled by first performing a row or column operation that reveals a common factor (here, adding all three rows makes every entry in the new row equal, exposing an $(a+b+c)$ or similar factor), then simplifying the reduced $2\times2$ structure. When the algebra gets intricate, a quick numeric sanity check with simple values of $a,b,c$ is an efficient way to confirm which answer choice matches.

    Step-by-Step Solution

    1. The determinant is $\begin{vmatrix} b+c & a & a \\ b & c+a & b \\ c & c & a+b \end{vmatrix}$.
    2. Apply $R_1 \to R_1 + R_2 + R_3$: the new first row becomes $(b+c+b+c,\ a+c+a+c,\ a+b+a+b)$... more carefully, summing column-wise: column 1 sum $= (b+c)+b+c = b+2c$... to avoid an error-prone symbolic expansion, verify by direct numeric substitution instead (a clean, reliable check for this type of determinant).
    3. Numeric check: let $a=1, b=2, c=3$. The matrix becomes $\begin{pmatrix}5&1&1\\2&4&2\\3&3&3\end{pmatrix}$.
    4. Expand along Row 1: $\det = 5(4\cdot3-2\cdot3) - 1(2\cdot3-2\cdot3) + 1(2\cdot3-4\cdot3) = 5(12-6) - 1(0) + 1(6-12) = 30 - 0 - 6 = 24$. …

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11. AP EAPCET 2021Set eng-2021-08-25-AN1 markMCQ
    Q.If the determinant $\begin{vmatrix} \cos 2x & \sin^2 x & \cos 2x \\ \sin^2 x & \cos 2x & \cos^2 x \\ \cos 2x & \cos^2 x & \cos 2x \end{vmatrix}$ is expanded in powers of $\cos x$, then the constant term in the expansion is (A) 1 (B) $-1$ (C) 0 (D) 2

    ›Reveal solutionSolution

    The constant term of a polynomial in $\cos x$ is exactly the value obtained by setting $\cos x=0$. Answer: (A).

    Concept and Intuition

    If a determinant's entries are polynomials in $c=\cos x$, expanding it produces a polynomial in $c$. Its constant term (the $c^0$ coefficient) equals the value of the whole expression when $c=0$ — a shortcut that avoids expanding the full polynomial in $\cos x$.

    Step-by-Step Solution

    1. Write each entry in terms of $c=\cos x$: $\cos2x = 2c^2-1$, $\sin^2x = 1-c^2$, $\cos^2x = c^2$.
    2. Set $c=0$: $\cos2x \to -1$, $\sin^2x \to 1$, $\cos^2x \to 0$.
    3. The matrix becomes $\begin{bmatrix}-1&1&-1\\1&-1&0\\-1&0&-1\end{bmatrix}$. …

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12. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
    Q.$\begin{vmatrix} -a^2 & ab & ac \\ ab & -b^2 & bc \\ ac & bc & -c^2 \end{vmatrix} =$ (A) $a^2b^2c^2$ (B) $2a^2b^2c^2$ (C) $3a^2b^2c^2$ (D) $4a^2b^2c^2$

    ›Reveal solutionSolution

    Factoring $a$, $b$, $c$ out of the three rows reduces the determinant to a simpler $\pm1$-coefficient determinant that evaluates to $4abc$, giving a total of $4a^2b^2c^2$. Answer: (D).

    Concept and Intuition

    Each row of the given determinant has a common factor: row 1 is $a\cdot(-a,b,c)$, row 2 is $b\cdot(a,-b,c)$, row 3 is $c\cdot(a,b,-c)$. Pulling a common factor out of a row simply multiplies the determinant by that factor (a standard determinant property), so we can simplify before directly expanding the messier original $3\times3$ determinant.

    Step-by-Step Solution

    1. Original determinant: $\begin{vmatrix} -a^2 & ab & ac \\ ab & -b^2 & bc \\ ac & bc & -c^2 \end{vmatrix}$.
    2. Factor $a$ from row 1, $b$ from row 2, $c$ from row 3:

    $= abc\begin{vmatrix} -a & b & c \\ a & -b & c \\ a & b & -c \end{vmatrix}$

    3. Expand this reduced determinant along the first row: $-a\begin{vmatrix}-b&c\\b&-c\end{vmatrix} - b\begin{vmatrix}a&c\\a&-c\end{vmatrix} + c\begin{vmatrix}a&-b\\a&b\end{vmatrix}$ …

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