Q.Prove that the determinant $\begin{vmatrix} x & \sin\theta & \cos\theta \\ -\sin\theta & -x & 1 \\ \cos\theta & 1 & x \end{vmatrix}$ is independent of $\theta$.

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 $\begin{bmatrix}-1&1&-1\\1&-1&0\\-1&0&-1\end{bmatrix}$.
   4. Expand: $-1[(-1)(-1)-(0)(0)] - 1[(1)(-1)-(0)(-1)] + (-1)[(1)(0)-(-1)(-1)]$ $= -1(1) - 1(-1) + (-1)(-1) = -1+1+1 = 1$.

   Common Mistakes

   • Trying to fully expand the determinant symbolically in $\cos x$ (needlessly long) instead of using the $c=0$ substitution shortcut for the constant term.

   ✓Final answer

   The correct option is (A) — 1.

   ANSWER: A

2. 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.
   4. For (C): $\det[A(\theta)]^{-1} = \dfrac{1}{\det A(\theta)} = \dfrac{1}{-1} = -1$. The statement claims this equals $1$, which is false.

   Common Mistakes

   • Assuming $\det(M^{-1})=\det M$ instead of the correct reciprocal relation $\det(M^{-1})=1/\det M$.
   • Forgetting $i^2=-1$ and treating the top-left/bottom-right entries as if they contributed $+\sin^2\theta$.

   ✓Final answer

   The correct option is (C) — $\det[A(\theta)]^{-1}=1$ is NOT true (it actually equals $-1$).

   ANSWER: C

3. 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).
   4. Verify with a quick numeric check ($a=1,b=0,c=0$): the matrix becomes $\begin{bmatrix}1&0&0\\1&0&-1\\0&1&1\end{bmatrix}$, whose determinant is $1$; and $1^3+0+0-0=1$ — matches.
   5. So $D = a^3+b^3+c^3-3abc$.

   Common Mistakes

   • Sign error on the $3abc$ term (getting $+3abc$ instead of $-3abc$, or $\pm 6abc$) from mis-tracking how many times the $abc$ term appears across the three bracket expansions.

   ✓Final answer

   The correct option is (B) — $a^3+b^3+c^3-3abc$.

   ANSWER: B

4. 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. $\det(B+sI) = (6+s)\cdot s\cdot s = s^2(6+s)$.
   5. Sanity check at $s=0$: $\det(B)=0$, matching that $B$'s rows are identical (determinant zero). ✓.
   6. As $x\to0$, $\sin x/x\to1$, so $s\to1$ and the determinant is a continuous (polynomial) function of $s$.
   7. $\lim_{x\to0} f(x) = 1^2(6+1)=7$.

   Common Mistakes

   • Trying to brute-force expand the $3\times3$ determinant symbolically in $s$ (doable but error-prone) instead of using the rank-1-shift trick.
   • Forgetting that $\lim_{x\to0}\sin x/x = 1$, not $0$.

   ✓Final answer

   The correct option is (D) — $7$.

   ANSWER: D

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:

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

   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

6. 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 $D=22$.
   4. Test option (A): $(a-b)(b-c)(c-a)(a+b+c) = 2\times6=12\ne22$ — rejected.
   5. Test option (C): $(a-b)(b-c)(a-c)(ab+bc+ca)$: note $(a-c)=1-3=-2$ (opposite sign convention to $(c-a)=2$), giving $(-1)(-1)(-2)(11)=-22\ne22$ — rejected (sign mismatch).
   6. Option (D) is confirmed as the correct general factorisation.

   Common Mistakes

   • Sign confusion between $(c-a)$ and $(a-c)$ when comparing to option (C) — a single sign flip changes the whole product's sign.
   • Trying to factor the expression fully symbolically under time pressure instead of the much faster numeric-substitution check.

   ✓Final answer

   The correct option is (D) — $(a-b)(b-c)(c-a)(ab+bc+ca)$.

   ANSWER: D

7. 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

8. 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

9. 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

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

   4. Factor $x$ from Row 2:

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

   5. Observe that Row 2 and Row 3 are now identical. A determinant with two equal rows is zero. Hence

   $\Delta = x \cdot 0 = 0.$

   Watch out

   A common mistake is to forget that the product rule gives $(xf)' = f + x f'$, not just $x f'$. If you miss the $f$ term, you won't see the linear dependence.

   Tip

   Whenever you see derivatives of products inside a determinant, expand them first. Often the structure reveals that rows are linear combinations, making the determinant vanish.

   ---

   ✓Final answer

   The correct option is (C).

   ANSWER: C

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:

    $\det(sI+\mathbf1 v^T) = s^3 + s^2\,(v^T\mathbf1)$

    (since $\det(sI)=s^3$ and $\operatorname{adj}(sI)=s^2 I$ for a $3\times3$ matrix).

    5. $v^T\mathbf1 = c+a+b = s$. So $\det(M) = s^3+s^2\cdot s = 2s^3$.

    6. Substituting back $s=a+b+c$: $\det = 2(a+b+c)^3$.

    Common Mistakes

    • Trying to expand the $3\times3$ determinant directly by cofactors without the $s$-substitution — it's algebraically messy and error-prone; the "subtract the common sum" trick is much cleaner and standard for this whole family of problems.
    • Forgetting the factor of $s^2$ multiplying $(v^T\mathbf1)$ and just adding $s^3+v^T\mathbf1$.

    ✓Final answer

    The correct option is (B) — $2(a+b+c)^3$.

    ANSWER: B

11. 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

12. 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