Q.Show that the $\triangle ABC$ is an isosceles triangle if the determinant $\Delta = \begin{vmatrix} 1 & 1 & 1 \\ 1 + \cos A & 1 + \cos B & 1 + \cos C \\ \cos^2 A + \cos A & \cos^2 B + \cos B & \cos^2 C + \cos C \end{vmatrix} = 0$.

Showing the 12 most recent of 32 on this concept.

1. 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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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 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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4. 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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5. AP EAPCET 2021Set eng-2021-08-25-FN1 markMCQ
   Q.If $a, b, c$ are real numbers such that $a^2+b^2+c^2-ab-bc-ac \le 0$, then $\begin{vmatrix} (a-b+1)^5 & b^7-c^7 & c^9-a^9 \\ a^{11}-b^{11} & (b-c+2)^3 & c^{13}-a^{13} \\ a^{15}-b^{15} & b^{17}-c^{17} & (c-a+3)^1 \end{vmatrix} =$ (A) $2abc$ (B) $0$ (C) $24abc$ (D) $24$

   ›Reveal solutionSolution

   The inequality forces $a=b=c$; substituting collapses the matrix to a simple upper-triangular-looking form whose determinant is just the product of its surviving diagonal terms, $24$.

   Concept and Intuition

   The expression $a^2+b^2+c^2-ab-bc-ca$ is a sum of squares in disguise: $\tfrac12[(a-b)^2+(b-c)^2+(c-a)^2]$, which can never be negative. So the given condition "$\le 0$" combined with this built-in "$\ge0$" forces it to be exactly zero, which only happens when $a=b=c$. That single deduction massively simplifies every entry of the determinant, since almost every entry is a difference of equal powers of $a,b,c$ (which all vanish when $a=b=c$), leaving only three surviving diagonal-type terms.

   Step-by-Step Solution

   1. Rewrite the condition: $a^2+b^2+c^2-ab-bc-ca=\tfrac12[(a-b)^2+(b-c)^2+(c-a)^2]$, which is always $\ge0$.
   2. Given this quantity is also $\le0$, it must equal exactly $0$, forcing $(a-b)^2=(b-c)^2=(c-a)^2=0$, i.e. $a=b=c$.
   3. Substitute $a=b=c$ into the matrix: all terms of the form $b^7-c^7$, $c^9-a^9$, $a^{11}-b^{11}$, $c^{13}-a^{13}$, $a^{15}-b^{15}$, $b^{17}-c^{17}$ become $0$ (difference of equal quantities). …

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6. 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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7. 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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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 2022Set eng-2022-07-04-FN1 markMCQ
   Q.For $i = 1,2,3$ and $j = 1,2,3$. If $a_i^2+b_i^2+c_i^2 = 1$, $a_ia_j+b_ib_j+c_ic_j = 0$, $\forall i \neq j$ and $A = \begin{bmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{bmatrix}$ then $\det(AA^T) =$ (A) $0$ (B) $1$ (C) $-1$ (D) $3$

   ›Reveal solutionSolution

   The conditions given say the rows of A form an orthonormal set, which is precisely the condition $AA^T=I$; so its determinant is 1. Answer: 1.

   Concept and Intuition

   For a matrix A with rows $R_1, R_2, R_3$, the $(i,j)$ entry of $AA^T$ is exactly the dot product $R_i\cdot R_j$. The problem states $a_i^2+b_i^2+c_i^2=1$ (each row is a unit vector) and $a_ia_j+b_ib_j+c_ic_j=0$ for $i\ne j$ (distinct rows are perpendicular). Together these say the rows are orthonormal — which is the defining property of an orthogonal matrix, for which $AA^T = I$.

   Step-by-Step Solution

   1. Write $(AA^T)_{ij} = R_i \cdot R_j = a_ia_j + b_ib_j + c_ic_j$.
   2. For $i=j$: $(AA^T)_{ii} = a_i^2+b_i^2+c_i^2 = 1$ (given).
   3. For $i\ne j$: $(AA^T)_{ij} = a_ia_j+b_ib_j+c_ic_j = 0$ (given).
   4. So $AA^T = I_3$, the $3\times3$ identity matrix.
   5. $\det(AA^T) = \det(I_3) = 1$.

   Common Mistakes …

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10. AP EAPCET 2022Set eng-2022-07-05-AN1 markMCQ
    Q.If b and c are non zero real numbers, $A = \begin{bmatrix} 1 & b & c \\ b & 2 & 3 \\ c & 3 & 4 \end{bmatrix}$ and $B = \begin{bmatrix} 0 & b & c \\ -b & 0 & 2 \\ -c & -2 & 0 \end{bmatrix}$, then $\det(A+B) =$ (A) 3 (B) 1 (C) -1 (D) 0

    ›Reveal solutionSolution

    Adding the symmetric $A$ and skew-symmetric $B$ entrywise cancels the off-diagonal $b,c$ terms below the diagonal, leaving a matrix whose determinant is a fixed number independent of $b,c$: 3.

    Concept and Intuition

    $A$ is symmetric and $B$ is skew-symmetric (its transpose is its negative, with zero diagonal). Adding them entrywise, the upper-triangular parts of $A+B$ pick up $b+b=2b$ and $c+c=2c$, while the lower-triangular parts get $b+(-b)=0$ and $c+(-c)=0$ — so $A+B$ becomes upper triangular in its first column, letting the determinant be computed by a clean cofactor expansion that never involves $b$ or $c$.

    Step-by-Step Solution

    1. $A=\begin{bmatrix}1&b&c\\b&2&3\\c&3&4\end{bmatrix}$, $B=\begin{bmatrix}0&b&c\\-b&0&2\\-c&-2&0\end{bmatrix}$.
    2. Add entrywise: $(A+B)_{11}=1$, $(A+B)_{12}=2b$, $(A+B)_{13}=2c$; $(A+B)_{21}=b-b=0$, $(A+B)_{22}=2$, $(A+B)_{23}=3+2=5$; $(A+B)_{31}=c-c=0$, $(A+B)_{32}=3-2=1$, $(A+B)_{33}=4$.
    3. So $A+B=\begin{bmatrix}1&2b&2c\\0&2&5\\0&1&4\end{bmatrix}$. …

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11. 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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12. AP EAPCET 2022Set eng-2022-07-04-AN1 markMCQ
    Q.If a, b, c are respectively the $5^{th}$, $8^{th}$, $13^{th}$ terms of an arithmetic progression, then $\begin{vmatrix} a & 5 & 1 \\ b & 8 & 1 \\ c & 13 & 1 \end{vmatrix} =$ (A) $0$ (B) $1$ (C) $abc$ (D) $520$

    ›Reveal solutionSolution

    Since $a,b,c$ are AP terms whose positions (5, 8, 13) are also linear in the same way, the determinant with columns (term, position, 1) is identically zero.

    Concept and Intuition

    If $a,b,c$ are terms of an AP at positions $p,q,r$ respectively, then $a=A+(p-1)d,\ b=A+(q-1)d,\ c=A+(r-1)d$ — i.e. each term is an affine (linear) function of its position. A determinant of the form $\begin{vmatrix}\text{term} & \text{position} & 1\end{vmatrix}$ over three such rows is always zero, because the "term" column is a linear combination of the "position" column and the "1" column (row reduction makes a column of zeros).

    Step-by-Step Solution

    1. Let the AP have first term $A$ and common difference $d$. Then $a = A+4d$ (5th term), $b=A+7d$ (8th term), $c=A+12d$ (13th term).
    2. Expand the determinant along the first row: $\Delta = a(8\cdot1-1\cdot13) - 5(b\cdot1-1\cdot c) + 1(b\cdot13-8\cdot c)$ $= a(-5) -5(b-c) + (13b-8c) = -5a -5b+5c+13b-8c = -5a+8b-3c$.
    3. Substitute: $-5a = -5A-20d$, $8b=8A+56d$, $-3c=-3A-36d$. …

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