Q.Value of the determinant |cos 67𝑜 sin 67𝑜 sin 23𝑜 cos 23𝑜| is
(A) 0
(B) 1 2
(C) √3 2
(D) 1

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 $\begin{bmatrix}-1&1&-1\\1&-1&0\\-1&0&-1\end{bmatrix}$. …

   🔒 Start your 14-day free trial to see the full solution →
2. 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. $\det(B+sI) = (6+s)\cdot s\cdot s = s^2(6+s)$. …

   🔒 Start your 14-day free trial to see the full solution →
3. AP EAPCET 2021Set eng-2021-08-23-FN1 markMCQ
   Q.$\begin{bmatrix} 1 & 2 & 3 \\ -1 & 1 & 2 \\ 3 & 0 & 2 \end{bmatrix} \begin{vmatrix} 2022 & 2024 \\ 2021 & 2023 \end{vmatrix} =$ (A) $\begin{bmatrix} 8 & 4 & 11 \\ 4 & -1 & 3 \\ 9 & 6 & 13 \end{bmatrix}$ (B) $\begin{bmatrix} 8 & 4 & 13 \\ 4 & -1 & 3 \\ 9 & 6 & 12 \end{bmatrix}$ (C) $\begin{bmatrix} 8 & 4 & 13 \\ 4 & -1 & 3 \\ 9 & 6 & 13 \end{bmatrix}$ (D) $\begin{bmatrix} 8 & 4 & 11 \\ 4 & 1 & 13 \\ 9 & 6 & 13 \end{bmatrix}$

   ›Reveal solutionSolution

   Multiplying the given $3\times3$ matrix by itself, entry by entry using the row-times-column rule, produces $\begin{bmatrix}8&4&13\\4&-1&3\\9&6&13\end{bmatrix}$, matching option (C).

   Concept and Intuition

   Matrix multiplication of a square matrix $A$ with itself ($A\times A = A^2$) is computed entry by entry: the $(i,j)$ entry of the product is the dot product of row $i$ of the first matrix with column $j$ of the second matrix. Carrying this out carefully for all 9 entries of a $3\times3$ matrix gives the resulting product matrix.

   Step-by-Step Solution

   Let $A=\begin{bmatrix}1&2&3\\-1&1&2\\3&0&2\end{bmatrix}$. Compute $A^2 = A\cdot A$ entry by entry (row $i$ of $A$ dotted with column $j$ of $A$):

   1. $(1,1)$: $1(1)+2(-1)+3(3) = 1-2+9 = 8$
   2. $(1,2)$: $1(2)+2(1)+3(0) = 2+2+0 = 4$
   3. $(1,3)$: $1(3)+2(2)+3(2) = 3+4+6 = 13$
   4. $(2,1)$: $-1(1)+1(-1)+2(3) = -1-1+6 = 4$
   5. $(2,2)$: $-1(2)+1(1)+2(0) = -2+1+0 = -1$
   6. $(2,3)$: $-1(3)+1(2)+2(2) = -3+2+4 = 3$
   7. $(3,1)$: $3(1)+0(-1)+2(3) = 3+0+6 = 9$
   8. $(3,2)$: $3(2)+0(1)+2(0) = 6+0+0 = 6$
   9. $(3,3)$: $3(3)+0(2)+2(2) = 9+0+4 = 13$

   Assembling these gives $A^2 = \begin{bmatrix}8&4&13\\4&-1&3\\9&6&13\end{bmatrix}$, which is an exact, entry-by-entry match with option (C). …

   🔒 Start your 14-day free trial to see the full solution →
4. AP EAPCET 2021Set eng-2021-08-23-AN1 markMCQ
   Q.If $A(\theta)=\begin{bmatrix} i\sin\theta & \cos\theta \\ \cos\theta & i\sin\theta \end{bmatrix}$ is a matrix where $i=\sqrt{-1}$, then which of the following is not true (A) $\det A(\pi+\theta)=\det A(-\theta)$ (B) $\det A(-\theta)=\det A(\theta)$ (C) $\det[A(\theta)]^{-1}=1$ (D) $\det A(-\theta)=-1$

   ›Reveal solutionSolution

   The determinant of $A(\theta)$ is identically $-1$, independent of $\theta$; this makes (A), (B), (D) trivially true and exposes (C) as false.

   Concept and Intuition

   Whenever a matrix's determinant simplifies to a constant using $\sin^2\theta+\cos^2\theta=1$, every statement that only compares $\det A$ at different arguments becomes trivial — the real test is whether the algebra of determinants (like $\det(M^{-1})=1/\det M$) is applied correctly.

   Step-by-Step Solution

   1. Compute $\det A(\theta) = (i\sin\theta)(i\sin\theta) - (\cos\theta)(\cos\theta) = i^2\sin^2\theta - \cos^2\theta$.
   2. Since $i^2=-1$: $\det A(\theta) = -\sin^2\theta - \cos^2\theta = -(\sin^2\theta+\cos^2\theta) = -1$.
   3. This value is the SAME for every $\theta$ (it never even used the sign of the argument), so:
      • (A) $\det A(\pi+\theta)=\det A(-\theta)$: both sides are $-1$. True.
      • (B) $\det A(-\theta)=\det A(\theta)$: both sides are $-1$. True.
      • (D) $\det A(-\theta) = -1$: True. …

   🔒 Start your 14-day free trial to see the full solution →
5. 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: $-a\begin{vmatrix}-b&c\\b&-c\end{vmatrix} - b\begin{vmatrix}a&c\\a&-c\end{vmatrix} + c\begin{vmatrix}a&-b\\a&b\end{vmatrix}$ …

   🔒 Start your 14-day free trial to see the full solution →
6. 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: $\det = 5(4\cdot3-2\cdot3) - 1(2\cdot3-2\cdot3) + 1(2\cdot3-4\cdot3) = 5(12-6) - 1(0) + 1(6-12) = 30 - 0 - 6 = 24$. …

   🔒 Start your 14-day free trial to see the full solution →
7. AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
   Q.$\begin{vmatrix} 2 & 1 \\ 3 & 1 \end{vmatrix} + \begin{vmatrix} 1 & 1/3 \\ 3 & 1 \end{vmatrix} + \begin{vmatrix} 1/2 & 1/9 \\ 3 & 1 \end{vmatrix} + \begin{vmatrix} 1/4 & 1/27 \\ 3 & 1 \end{vmatrix} + \ldots \infty =$ (A) $0$ (B) $\dfrac{1}{2}$ (C) $-\dfrac{1}{2}$ (D) $-1$

   ›Reveal solutionSolution

   Expand each $2\times2$ determinant, recognise two independent geometric series in the result, and sum them separately. Answer: (C).

   Concept and Intuition

   Every term in the sum is a $2\times2$ determinant with a fixed bottom row $\begin{pmatrix}3&1\end{pmatrix}$, so $\begin{vmatrix}a&b\\3&1\end{vmatrix}=a-3b$ splits linearly into an $a$-part and a $b$-part. Once each part is recognised as its own geometric progression, the infinite sum is just the difference of two standard geometric series sums, $\dfrac{\text{first term}}{1-r}$.

   Step-by-Step Solution

   1. General term: $\begin{vmatrix}a_n & b_n\\3&1\end{vmatrix} = a_n\cdot1 - b_n\cdot3 = a_n - 3b_n$.
   2. First-column values across the given terms: $2, 1, \tfrac12, \tfrac14,\dots$ — a geometric sequence with first term $2$ and ratio $\tfrac12$: $a_n = 2\left(\tfrac12\right)^{n-1}$.
   3. Second-column values: $1, \tfrac13, \tfrac19, \tfrac1{27},\dots$ — geometric with first term $1$ and ratio $\tfrac13$: $b_n=\left(\tfrac13\right)^{n-1}$. …

   🔒 Start your 14-day free trial to see the full solution →
8. 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 …

   🔒 Start your 14-day free trial to see the full solution →
9. AP EAPCET 2024Set eng-2024-05-18-FN1 markMCQ
   Q.If the determinant of a $3^{rd}$ order matrix A is K, then the sum of the determinants of the matrices $(AA^T)$ and $(A-A^T)$ is (A) 2K (B) 0 (C) $K^2$ (D) K

   ›Reveal solutionSolution

   This tests two standard determinant facts: $\det(AA^T) = (\det A)^2$, and any odd-order skew-symmetric matrix (such as $A-A^T$ for a $3\times3$ matrix $A$) has determinant zero — giving the sum $K^2 + 0 = K^2$.

   Concept and Intuition

   Two separate determinant identities combine here:

   1. For any square matrix $A$, $\det(A^T) = \det(A)$, so $\det(AA^T) = \det(A)\cdot\det(A^T) = (\det A)^2$.
   2. The matrix $A - A^T$ is always skew-symmetric, since $(A - A^T)^T = A^T - A = -(A - A^T)$. For a skew-symmetric matrix $S$ of odd order $n$, we always have $\det S = 0$. This is because $\det(S^T) = \det(S)$ always, but also $S^T = -S$ gives $\det(S^T) = \det(-S) = (-1)^n \det(S)$. For odd $n$, $(-1)^n = -1$, so $\det(S) = -\det(S)$, forcing $\det(S) = 0$.

   Step-by-Step Solution

   1. Given: $A$ is a $3\times3$ matrix with $\det(A) = K$.
   2. Compute $\det(AA^T)$: using $\det(AA^T) = \det(A)\det(A^T)$ and $\det(A^T) = \det(A) = K$, we get $\det(AA^T) = K \cdot K = K^2$.
   3. Compute $\det(A - A^T)$: let $S = A - A^T$. Then $S^T = A^T - A = -S$, so $S$ is skew-symmetric.
   4. Since $S$ is a $3\times3$ (odd-order) skew-symmetric matrix: $\det(S) = \det(S^T) = \det(-S) = (-1)^3\det(S) = -\det(S)$. This gives $2\det(S) = 0 \Rightarrow \det(S) = 0$. …

   🔒 Start your 14-day free trial to see the full solution →
10. 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: $\det(sI+\mathbf1 v^T) = s^3 + s^2\,(v^T\mathbf1)$ …

    🔒 Start your 14-day free trial to see the full solution →
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 $D=22$. …

    🔒 Start your 14-day free trial to see the full solution →
12. 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 $R_3 = (a+4b-(a+2b),\ a+5b-(a+3b),\ a+6b-(a+4b)) = (2b,\ 2b,\ 2b)$. …

    🔒 Start your 14-day free trial to see the full solution →