Q.Using properties of determinants, prove that $\begin{vmatrix} 1 & 1 & 1+3x \\ 1+3y & 1 & 1 \\ 1 & 1+3z & 1 \end{vmatrix} = 9(3xyz + xy + yz + zx)$

Showing the 12 most recent of 32 on this concept.

1. 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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2. 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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3. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.$A = \begin{bmatrix} b+c & a & a \\ b & c+a & b \\ c & c & a+b \end{bmatrix}$ is a matrix such that trace of $A = 18$ and $\det(A) = 96$. If $a, b, c \in N$ and $ab = 6$, then $ab + bc + ca =$ (A) $36$ (B) $26$ (C) $48$ (D) $24$

   ›Reveal solutionSolution

   Using trace(A) = 2(a+b+c) and the identity det(A) = 4abc, we get a+b+c=9 and abc=24; with ab=6 this pins down c=4, {a,b}={2,3}, and ab+bc+ca=26.

   Concept and Intuition

   The matrix

   $A=\begin{bmatrix} b+c & a & a \\ b & c+a & b \\ c & c & a+b \end{bmatrix}$

   has a very clean structure: each row sums to $a+b+c$, and there is a well-known identity that its determinant simplifies to $4abc$ (this can be shown by row/column operations: subtracting appropriately or direct cofactor expansion collapses most cross terms). Recognising this identity turns a messy 3×3 determinant into a one-line relation between $a,b,c$.

   Step-by-Step Solution

   1. Trace: $(b+c) + (c+a) + (a+b) = 2(a+b+c) = 18 \Rightarrow a+b+c = 9$.
   2. Determinant identity: expand $\det(A)$ directly — $\det(A) = (b+c)[(c+a)(a+b)-bc] - a[b(a+b)-bc] + a[bc - c(c+a)]$. Simplifying each bracket:
      • $(c+a)(a+b) - bc = a^2+ab+ac+bc-bc = a(a+b+c)$
      • Combining all terms and cancelling cross terms (ab, ac, b², c² all cancel) leaves $4abc$. So $\det(A) = 4abc = 96 \Rightarrow abc = 24$.
   3. Given $ab = 6$: $c = abc/(ab) = 24/6 = 4$.
   4. From step 1: $a+b = 9 - c = 9-4 = 5$. …

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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. $\det(B+sI) = (6+s)\cdot s\cdot s = s^2(6+s)$. …

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5. 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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6. 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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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 $\tfrac13$: $b_n=\left(\tfrac13\right)^{n-1}$. …

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8. AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQ
   Q.If $\alpha$ is a real root of the equation $x^3+6x^2+5x-42=0$, then the determinant of the matrix $\begin{bmatrix}\alpha-1 & \alpha+1 & \alpha+2\\ \alpha-2 & \alpha+3 & \alpha-3\\ \alpha+4 & \alpha-4 & \alpha+5\end{bmatrix}$ is (A) 90 (B) 120 (C) -105 (D) -135

   ›Reveal solutionSolution

   The cubic has a unique real root $\alpha=2$ (found by testing small integers and confirming the remaining quadratic has no real roots); substituting into the determinant and expanding gives $-105$.

   Concept and Intuition

   When a cubic is asked to have "a real root," it's often meant to be found by the rational root theorem (testing small divisors of the constant term) — here the constant is $-42$, and $2$ divides it. Once found, factor it out and check the resulting quadratic's discriminant to confirm no other real roots exist, so $\alpha$ is unambiguous.

   Step-by-Step Solution

   1. Test $x=2$ in $x^3+6x^2+5x-42$: $8+24+10-42=0$. So $x=2$ is a root.
   2. Factor: $x^3+6x^2+5x-42=(x-2)(x^2+8x+21)$.
   3. Discriminant of $x^2+8x+21$ is $64-84=-20<0$, so this factor has no real roots — confirming $\alpha=2$ is the unique real root.
   4. Substitute $\alpha=2$ into the matrix:

   $\begin{bmatrix}1&3&4\\0&5&-1\\6&-2&7\end{bmatrix}$

   5. Expand along row 1: …

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9. 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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10. 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 $2\det(S) = 0 \Rightarrow \det(S) = 0$. …

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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 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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