Q.If $A = \begin{bmatrix} 1 & 1 & -2 \\ 2 & 1 & -3 \\ 5 & 4 & -9 \end{bmatrix}$, find $|A|$.

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

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

   🔒 Start your 14-day free trial to see the full solution →
3. 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. $f(-x)=2(-x-1)(-x-2)(-x-3)=2\cdot(-1)^3(x+1)(x+2)(x+3)=-2(x+1)(x+2)(x+3)$. …

   🔒 Start your 14-day free trial to see the full solution →
4. 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 →
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 $\tfrac13$: $b_n=\left(\tfrac13\right)^{n-1}$. …

   🔒 Start your 14-day free trial to see the full solution →
6. 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 →
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: $\det(sI+\mathbf1 v^T) = s^3 + s^2\,(v^T\mathbf1)$ …

   🔒 Start your 14-day free trial to see the full solution →
8. 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 →
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 $(n+3)(n+4)-(n+1)(n+2)=4n+10$: …

   🔒 Start your 14-day free trial to see the full solution →
10. 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 $x=1/5$. …

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

    🔒 Start your 14-day free trial to see the full solution →
12. AP EAPCET 2021Set eng-2021-08-19-FN1 markMCQ
    Q.If $a_1, a_2, \ldots a_9$ are in G.P, then $\begin{vmatrix} \log a_1 & \log a_2 & \log a_3 \\ \log a_4 & \log a_5 & \log a_6 \\ \log a_7 & \log a_8 & \log a_9 \end{vmatrix}$ is equal to (A) $\log(a_1, a_2, \ldots a_n)$ (B) $1$ (C) $(\log a_9)^9$ (D) $0$

    ›Reveal solutionSolution

    The logs of GP terms form an arithmetic sequence, which makes the three rows of the determinant linearly dependent ($R_1+R_3=2R_2$); hence the determinant is $0$.

    Concept and Intuition

    Taking logs of a GP converts multiplication into addition, turning the GP into an AP. When a $3\times3$ matrix's rows are built from consecutive terms of one long AP (each row a block of 3 consecutive AP terms), the rows themselves form an AP of row-vectors — meaning the middle row is the average of the outer two. Any such linear dependency among rows forces determinant $=0$.

    Step-by-Step Solution

    1. Since $a_1,\dots,a_9$ are in GP with common ratio $r$: $a_n = a_1 r^{n-1}$, so $\log a_n = \log a_1 + (n-1)\log r$.
    2. Let $L=\log a_1$, $d=\log r$. Then $\log a_n = L+(n-1)d$.
    3. Row 1: $(\log a_1,\log a_2,\log a_3) = (L, L+d, L+2d)$. Row 2: $(\log a_4,\log a_5,\log a_6) = (L+3d, L+4d, L+5d)$. Row 3: $(\log a_7,\log a_8,\log a_9) = (L+6d, L+7d, L+8d)$. …

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