Q.Using the properties of determinants, evaluate: $\begin{vmatrix} x + 4 & x & x \\ x & x + 4 & x \\ x & x & x + 4 \end{vmatrix}$

Showing the 12 most recent of 32 on this concept.

1. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.For a fixed positive integer $n$, if $D = \begin{vmatrix} n! & (n+1)! & (n+2)! \\ (n+1)! & (n+2)! & (n+3)! \\ (n+2)! & (n+3)! & (n+4)! \end{vmatrix}$, then $\dfrac{D}{n!\,(n+1)!\,(n+2)!} =$ (A) $-4$ (B) $-2$ (C) $2$ (D) $4$

   ›Reveal solutionSolution

   Taking the factorials common from each row collapses the determinant to a constant: $\dfrac{D}{n!\,(n+1)!\,(n+2)!} = 2$ — option (C).

   Working. Factor $n!$, $(n+1)!$, $(n+2)!$ from rows 1, 2, 3 respectively:

   $D = n!\,(n+1)!\,(n+2)!\begin{vmatrix} 1 & n+1 & (n+1)(n+2) \\1 & n+2 & (n+2)(n+3) \\1 & n+3 & (n+3)(n+4) \end{vmatrix}$

   Hence $\dfrac{D}{n!\,(n+1)!\,(n+2)!}$ equals that $3\times3$ determinant. Apply $R_2\to R_2-R_1$ and $R_3\to R_3-R_1$, using $(n+2)(n+3)-(n+1)(n+2)=2(n+2)$ and $(n+3)(n+4)-(n+1)(n+2)=4n+10$: …

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2. 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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3. 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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4. 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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5. AP EAPCET 2021Set eng-2021-08-23-AN1 markMCQ
   Q.If $\begin{vmatrix} x^3+2x^2+3x-2 & x^2+2x+4 \\ x^3-x^2-2x-1 & 3x^3-2x^2+4x-2 \end{vmatrix} = ax^6+bx^5+cx^4+dx^3+ex^2+fx+g$, then $a+b+c+d+e+f=$ (A) 23 (B) 25 (C) 21 (D) 20

   ›Reveal solutionSolution

   Evaluating the determinant-polynomial at $x=1$ gives the sum of ALL coefficients (including the constant $g$); evaluating at $x=0$ isolates $g$ alone. Subtracting removes $g$, leaving $a+b+c+d+e+f=25$.

   Concept and Intuition

   If $P(x)=ax^6+bx^5+cx^4+dx^3+ex^2+fx+g$, a classic trick to get the sum of coefficients excluding the constant term is: $P(1) = a+b+c+d+e+f+g$ gives the sum of all coefficients (since every power of $1$ is $1$), while $P(0)=g$ isolates just the constant term. So $P(1)-P(0) = a+b+c+d+e+f$. Here $P(x)$ is defined as the given $2\times2$ determinant, so we just need to evaluate that determinant at $x=1$ and $x=0$ directly — no need to expand the full degree-6 polynomial.

   Step-by-Step Solution

   1. The determinant is $\begin{vmatrix} x^3+2x^2+3x-2 & x^2+2x+4 \\ x^3-x^2-2x-1 & 3x^3-2x^2+4x-2 \end{vmatrix}$.
   2. At $x=1$: top-left $=1+2+3-2=4$; top-right $=1+2+4=7$; bottom-left $=1-1-2-1=-3$; bottom-right $=3-2+4-2=3$. Determinant $=4(3)-7(-3)=12+21=33 = P(1) = a+b+c+d+e+f+g$. …

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6. 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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7. 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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8. 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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9. 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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10. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
    Q.If $A = \begin{bmatrix} a & 2b & 3c \\ 2b & c & 3a \\ 3c & 2a & b \end{bmatrix}$ and $\det(A) = pa^3 + qb^3 + rc^3 + s(abc)$, then $p+q+r+s =$ (A) $12$ (B) $20$ (C) $24$ (D) $30$

    ›Reveal solutionSolution

    Expanding the determinant gives $-6a^3-4b^3-9c^3+31abc$, so $p+q+r+s=-6-4-9+31=12$ — option (A).

    $\det(A)=\begin{vmatrix} a & 2b & 3c \\2b & c & 3a \\3c & 2a & b \end{vmatrix} = a(cb-6a^2)-2b(2b^2-9ac)+3c(4ab-3c^2).$

    Expand:

    $= abc-6a^3-4b^3+18abc+12abc-9c^3 = -6a^3-4b^3-9c^3+31abc.$ …

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11. AP EAPCET 2024Set eng-2024-05-19-AN1 markMCQ
    Q.$\begin{vmatrix} 1 & 1 & 1 \\ a^2 & b^2 & c^2 \\ a^3 & b^3 & c^3 \end{vmatrix} =$ (A) $(a-b)(b-c)(c-a)(a+b+c)$ (B) $(a-b)(b-c)(c-a)$ (C) $(a-b)(b-c)(a-c)(ab+bc+ca)$ (D) $(a-b)(b-c)(c-a)(ab+bc+ca)$

    ›Reveal solutionSolution

    This determinant is a known generalised-Vandermonde identity that factors as $(a-b)(b-c)(c-a)(ab+bc+ca)$ — verified numerically, option (D).

    Concept and Intuition

    Determinants with rows $1,x,x^2$ (Vandermonde) factor neatly as $(a-b)(b-c)(c-a)$. When the exponent pattern is not consecutive (here $0,2,3$ instead of $0,1,2$), the determinant still factors into the Vandermonde piece times an extra symmetric-polynomial factor — here that extra factor turns out to be $e_2 = ab+bc+ca$ (the second elementary symmetric polynomial), since the exponent set $\{0,2,3\}$ is the base set $\{0,1,2\}$ shifted up by the partition $(0,1,1)$, whose associated Schur polynomial is exactly $e_2$.

    Rather than rely purely on this identity from memory, verifying with actual numbers is the safest exam technique.

    Step-by-Step Solution

    1. Expand the determinant along the first row (all entries 1): $D = (b^2c^3-b^3c^2) - (a^2c^3-a^3c^2) + (a^2b^3-a^3b^2)$ $= b^2c^2(c-b) + a^2c^2(a-c) + a^2b^2(b-a)$.
    2. Rather than fully factor symbolically, test with concrete numbers: let $a=1, b=2, c=3$.
       • Direct determinant: rows $(1,1,1)$, $(1,4,9)$, $(1,8,27)$. $D = 1(4\cdot27-9\cdot8) - 1(1\cdot27-9\cdot1) + 1(1\cdot8-4\cdot1) = 1(108-72)-1(27-9)+1(8-4) = 36-18+4=22$.
    3. Test option (D): $(a-b)(b-c)(c-a)(ab+bc+ca)$ with these values: $(1-2)(2-3)(3-1) = (-1)(-1)(2) = 2$; $ab+bc+ca = 2+6+3=11$; product $=2\times11=22$. Matches $D=22$. …

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12. 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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