Q.Find the value of the following:

Showing the 12 most recent of 32 on this concept.

1. 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: $\det = 5(4\cdot3-2\cdot3) - 1(2\cdot3-2\cdot3) + 1(2\cdot3-4\cdot3) = 5(12-6) - 1(0) + 1(6-12) = 30 - 0 - 6 = 24$.
   5. Compare with each option at $a=1,b=2,c=3$: (A) $abc=6$ — no. (B) $(a+b)(b+c)(c+a)=3\cdot5\cdot4=60$ — no. (C) $4abc = 4\cdot6=24$ — matches. (D) $(a-b)(b-c)(c-a)=(-1)(-1)(2)=2$ — no.
   6. Only option (C), $4abc$, matches the computed value.

   Common Mistakes

   • Attempting a fully symbolic cofactor expansion without organising the algebra carefully, leading to sign errors — a numeric check with simple distinct values is a fast, robust way to identify the correct closed form among given options.
   • Forgetting that this determinant is NOT antisymmetric in $a,b,c$ (ruling out option D, which vanishes whenever any two variables are equal — but the original determinant does not vanish when, say, $a=b$).

   ✓Final answer

   The correct option is (C) — $4abc$.

   ANSWER: C

2. 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).

   Common Mistakes

   • Multiplying corresponding entries directly (element-wise/"Hadamard" product) instead of using the proper row-times-column matrix multiplication rule.
   • Sign slip in row 2 (which contains a $-1$), which is where options (A) and (D) diverge from the correct value (they show entry $(2,2)$ or other entries inconsistent with the correct computation) — careful arithmetic through each dot product avoids this.

   ✓Final answer

   The correct option is (C) — $\begin{bmatrix} 8 & 4 & 13 \\ 4 & -1 & 3 \\ 9 & 6 & 13 \end{bmatrix}$.

   ANSWER: C

3. 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

   • Trying to compute $\det(A)$ first and squaring it via $\det(AA^T)=(\det A)^2$ — correct in principle, but the direct recognition that $AA^T=I$ is far faster and avoids sign ambiguities in $\det A$.
   • Misreading the orthogonality condition as applying to columns instead of rows — here it is explicitly stated in terms of the row-indexed $a_i,b_i,c_i$.

   ✓Final answer

   The correct option is (B) — $1$.

   ANSWER: B

4. 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 $x=1/5$.
   5. Since $D=0$ always, the matching option is the expression that equals $0$ at $x=\dfrac15$: $5x-1 = 5\left(\dfrac15\right)-1 = 1-1=0$. ✓ (Checking the rest: $\dfrac{x}{5}=\dfrac1{25}$, $x-5=-\dfrac{24}{5}$, $x+5=\dfrac{26}{5}$ — all nonzero.)

   Common Mistakes

   • Brute-force expanding the $3\times3$ determinant instead of spotting the AP-row structure, wasting time and inviting slips.
   • Forgetting that "$D=0$ identically" means the answer must be matched numerically against $x=1/5$, not simplified as a formula.
   • Mixing up which matrix is $A$ and which is the given inverse when extracting $x$.

   ✓Final answer

   The correct option is (C) — $5x-1$.

   ANSWER: C

5. 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 $\tfrac13$: $b_n=\left(\tfrac13\right)^{n-1}$.
   4. Sum $=\displaystyle\sum_{n=1}^\infty a_n - 3\sum_{n=1}^\infty b_n = \left[2\cdot\frac{1}{1-\frac12}\right] - 3\left[\frac{1}{1-\frac13}\right] = 2\cdot2 - 3\cdot\frac32 = 4 - 4.5 = -\frac12$.
   5. So the infinite sum equals $-\dfrac12$.

   Common Mistakes

   • Trying to expand the determinant as a whole (e.g. cross-multiplying $a_n\cdot1$ and $b_n\cdot 3$ inconsistently) instead of separating the two geometric parts.
   • Misidentifying the common ratio of either sequence (it's easy to mix up $\tfrac12$ and $\tfrac13$ between the two columns).

   ✓Final answer

   The correct option is (C).

   ANSWER: C

6. AP EAPCET 2023Set eng-2023-05-15-AN1 markMCQ
   Q.If the cofactors of the elements 3, 7 and 6 of the matrix $\begin{pmatrix} 1 & 2 & 3 \\ 4 & -1 & 7 \\ 2 & 4 & 6 \end{pmatrix}$ are $a, b$ and $c$ respectively, then $[a\ b\ c]\begin{pmatrix}1\\4\\2\end{pmatrix} + [a\ b\ c]\begin{pmatrix}3\\7\\6\end{pmatrix} =$ (A) $-1$ (B) $1$ (C) $0$ (D) $3$

   ›Reveal solutionSolution

   $a,b,c$ turn out to be the cofactors of the matrix's third column, and the two

   vectors being dotted with $[a\ b\ c]$ are exactly its first and third columns. The

   Laplace-expansion identity makes both dot products come out to (a multiple of) the

   matrix's determinant, which here is $0$ — so the sum is $0$.

   Concept and Intuition

   For a $3\times3$ matrix $A$, the cofactor expansion identity says: if you take the

   entries of any column of $A$ and dot them with the cofactors of a matching column,

   you get $\det(A)$; if you dot them with the cofactors of a different column, you get

   $0$ (this is the algebraic content behind $A\cdot\text{adj}(A)=\det(A)\,I$). Recognising

   that the two given vectors, $(1,4,2)^T$ and $(3,7,6)^T$, are precisely columns 1 and 3

   of the matrix — and that $a,b,c$ are the cofactors of column 3 — turns this into a

   one-line application of that identity instead of brute-force computation (though brute

   force also works and is done below to double-check).

   Step-by-Step Solution

   1. The matrix is $M=\begin{pmatrix}1&2&3\\4&-1&7\\2&4&6\end{pmatrix}$. The entries $3,7,6$ sit at positions $(1,3),(2,3),(3,3)$ — all in column 3. So $a=C_{13}$, $b=C_{23}$, $c=C_{33}$.
   2. Compute $C_{13}$ (delete row 1, col 3; sign $(+1)^{1+3}=+$):

   $C_{13} = +\begin{vmatrix}4 & -1\\ 2 & 4\end{vmatrix} = 4(4)-(-1)(2) = 16+2 = 18.$

   3. Compute $C_{23}$ (delete row 2, col 3; sign $(-1)^{2+3}=-$):

   $C_{23} = -\begin{vmatrix}1 & 2\\ 2 & 4\end{vmatrix} = -(1\cdot4-2\cdot2) = -(0) = 0.$

   4. Compute $C_{33}$ (delete row 3, col 3; sign $(+1)^{3+3}=+$):

   $C_{33} = +\begin{vmatrix}1 & 2\\ 4 & -1\end{vmatrix} = 1(-1)-2(4) = -1-8=-9.$

   5. So $[a\ b\ c] = [18\ \ 0\ \ -9]$.
   6. First dot product: $[a\ b\ c]\begin{pmatrix}1\\4\\2\end{pmatrix} = 18(1)+0(4)+(-9)(2) = 18-18 = 0$.
   7. Second dot product: $[a\ b\ c]\begin{pmatrix}3\\7\\6\end{pmatrix} = 18(3)+0(7)+(-9)(6) = 54-54 = 0$. (This equals $\det(M)$ by the same-column expansion rule; direct expansion of $M$ along column 3 confirms $\det(M)=3(18)+7(0)+6(-9)=54-54=0$, i.e. $M$ is singular.)
   8. Sum of both dot products: $0+0 = 0$.

   Common Mistakes

   • Sign errors in the cofactor signs $(-1)^{i+j}$ — easy to drop the minus sign on $C_{23}$ (row+col = odd).
   • Not noticing the two given column vectors are literally columns of $M$, leading to unnecessary/error-prone brute-force arithmetic instead of the quick identity check.
   • Forgetting to add the two dot products together (the question asks for their sum, not just one of them).

   ✓Final answer

   The correct option is (C) — $0$.

   ANSWER: C

7. 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:

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

   4. Compute each $2\times2$ minor: $\begin{vmatrix}-b&c\\b&-c\end{vmatrix} = (-b)(-c)-c(b) = bc-bc=0$; $\begin{vmatrix}a&c\\a&-c\end{vmatrix} = a(-c)-c(a) = -2ac$; $\begin{vmatrix}a&-b\\a&b\end{vmatrix} = ab-(-b)(a) = 2ab$.
   5. Substitute: $-a(0) - b(-2ac) + c(2ab) = 0 + 2abc + 2abc = 4abc$.
   6. So the reduced determinant equals $4abc$, and the original determinant is $abc \times 4abc = 4a^2b^2c^2$.

   Common Mistakes

   • Sign errors when expanding the $2\times2$ minors of the reduced matrix (the $-,+,-$ cofactor pattern is easy to mis-apply).
   • Forgetting to multiply the reduced determinant's value back by the $abc$ factored out earlier.

   ✓Final answer

   The correct option is (D) — $4a^2b^2c^2$.

   ANSWER: D

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 $R_3 = (a+4b-(a+2b),\ a+5b-(a+3b),\ a+6b-(a+4b)) = (2b,\ 2b,\ 2b)$.
   4. The matrix now has rows $(a+b, a+2b, a+3b)$, $(b,b,b)$, $(2b,2b,2b)$ — and row 3 is exactly $2\times$ row 2, i.e. the rows are linearly dependent.
   5. A determinant with two proportional rows is always $0$.

   Common Mistakes

   • Trying to expand the $3\times3$ determinant directly by cofactors without first noticing the arithmetic-progression row structure — far more error-prone than the row-operation shortcut.
   • Forgetting that row operations of the type $R_i \to R_i - R_j$ don't change the determinant's value, only simplify it.

   ✓Final answer

   The correct option is (C) — $0$.

   ANSWER: C

9. 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 $(n+3)(n+4)-(n+1)(n+2)=4n+10$:

   $\begin{vmatrix} 1 & n+1 & (n+1)(n+2) \\0 & 1 & 2(n+2) \\0 & 2 & 2(2n+5) \end{vmatrix}$

   Expanding along the first column:

   $1\cdot\big[\,1\cdot 2(2n+5) - 2(n+2)\cdot 2\,\big] = (4n+10) - (4n+8) = 2.$

   ✓Final answer

   $\dfrac{D}{n!\,(n+1)!\,(n+2)!} = 2$ — option (C).

   ANSWER: C

10. 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. $f(-x)=2(-x-1)(-x-2)(-x-3)=2\cdot(-1)^3(x+1)(x+2)(x+3)=-2(x+1)(x+2)(x+3)$.
    5. $\dfrac{f(x)}{f(-x)}=\dfrac{2(x-1)(x-2)(x-3)}{-2(x+1)(x+2)(x+3)}=\dfrac{-(x-1)(x-2)(x-3)}{(x+1)(x+2)(x+3)}$.
    6. As $x\to1$: numerator $\to-(0)(-1)(-2)=0$; denominator $\to(2)(3)(4)=24\ne0$. So the limit is $0/24=0$.

    Common Mistakes

    • Expanding the determinant the "long way" (all 6 terms) instead of noticing the trivial first-column expansion, inviting arithmetic slips.
    • Forgetting to fully factor before substituting, and instead trying to plug $x=1$ into unfactored polynomials (masking the zero).

    ✓Final answer

    The correct option is (C) — $0$.

    ANSWER: C

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: $R_2\to R_2-3R_1 = (0,\,1,\,2+\omega)$; $R_3\to R_3-6R_1=(0,\,3,\,5+3\omega)$.
    6. Determinant $=1\cdot\begin{vmatrix}1 & 2+\omega\\ 3 & 5+3\omega\end{vmatrix} = (5+3\omega)-3(2+\omega) = 5+3\omega-6-3\omega=-1$.

    Common Mistakes

    • Forgetting $\omega^2=-1-\omega$ and trying to expand the $3\times3$ determinant directly with raw $\omega$ powers, which is error-prone.
    • Sign errors in the row-reduction step.

    ✓Final answer

    The correct option is (B) — $-1$.

    ANSWER: B

12. 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 $A+B=\begin{bmatrix}1&2b&2c\\0&2&5\\0&1&4\end{bmatrix}$.
    4. Expand the determinant along the first column (only the $(1,1)$ entry is nonzero there): $\det(A+B) = 1\cdot\det\begin{bmatrix}2&5\\1&4\end{bmatrix} = 1\cdot(2\cdot4-5\cdot1) = 8-5=3$.

    Common Mistakes

    • Not recognising $B$ as skew-symmetric and instead trying to compute the full $3\times3$ determinant symbolically in $b,c$ (much more work, and error-prone).
    • Sign slip when subtracting $b-b$ or $c-c$ in the lower-triangular entries.

    ✓Final answer

    The correct option is (A) — 3.

    ANSWER: A