Q.Find minors and cofactors of the elements $a_{11}$, $a_{21}$ in the determinant $\Delta = \begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{vmatrix}$.

Showing the 12 most recent of 32 on this concept.

1. 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: $[a\ b\ c]\begin{pmatrix}1\\4\\2\end{pmatrix} = 18(1)+0(4)+(-9)(2) = 18-18 = 0$. …

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2. 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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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). …

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4. 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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5. 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: $\det = 5(4\cdot3-2\cdot3) - 1(2\cdot3-2\cdot3) + 1(2\cdot3-4\cdot3) = 5(12-6) - 1(0) + 1(6-12) = 30 - 0 - 6 = 24$. …

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6. 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). …

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7. AP EAPCET 2021Set eng-2021-08-23-AN1 markMCQ
   Q.If $\begin{vmatrix} x^3+2x^2+3x-2 & x^2+2x+4 \\ x^3-x^2-2x-1 & 3x^3-2x^2+4x-2 \end{vmatrix} = ax^6+bx^5+cx^4+dx^3+ex^2+fx+g$, then $a+b+c+d+e+f=$ (A) 23 (B) 25 (C) 21 (D) 20

   ›Reveal solutionSolution

   Evaluating the determinant-polynomial at $x=1$ gives the sum of ALL coefficients (including the constant $g$); evaluating at $x=0$ isolates $g$ alone. Subtracting removes $g$, leaving $a+b+c+d+e+f=25$.

   Concept and Intuition

   If $P(x)=ax^6+bx^5+cx^4+dx^3+ex^2+fx+g$, a classic trick to get the sum of coefficients excluding the constant term is: $P(1) = a+b+c+d+e+f+g$ gives the sum of all coefficients (since every power of $1$ is $1$), while $P(0)=g$ isolates just the constant term. So $P(1)-P(0) = a+b+c+d+e+f$. Here $P(x)$ is defined as the given $2\times2$ determinant, so we just need to evaluate that determinant at $x=1$ and $x=0$ directly — no need to expand the full degree-6 polynomial.

   Step-by-Step Solution

   1. The determinant is $\begin{vmatrix} x^3+2x^2+3x-2 & x^2+2x+4 \\ x^3-x^2-2x-1 & 3x^3-2x^2+4x-2 \end{vmatrix}$.
   2. At $x=1$: top-left $=1+2+3-2=4$; top-right $=1+2+4=7$; bottom-left $=1-1-2-1=-3$; bottom-right $=3-2+4-2=3$. Determinant $=4(3)-7(-3)=12+21=33 = P(1) = a+b+c+d+e+f+g$. …

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8. 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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9. AP EAPCET 2021Set eng-2021-08-20-AN1 markMCQ
   Q.If $k \in R$ and $\det A = \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} = k$ then $\det B = \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 + 2a_1 & b_2 + 2b_1 & c_2 + 2c_1 \\ a_3 & b_3 & c_3 \end{vmatrix}$ is equal to (A) $0$ (B) $2K$ (C) $K$ (D) $K^2$

   ›Reveal solutionSolution

   $\det B$ is obtained from $\det A$ by the elementary row operation $R_2 \to R_2 + 2R_1$, which

   leaves a determinant unchanged; so $\det B = \det A = k$.

   Concept and Intuition

   A core determinant property: adding a scalar multiple of one row to another row does not change the determinant's value. This is because splitting the determinant using linearity in

   row 2 gives the original determinant plus $2\times$ a determinant with two identical rows (rows 1

   and the added copy of row 1), and any determinant with two identical rows is exactly zero.

   Step-by-Step Solution

   1. $\det B = \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2+2a_1 & b_2+2b_1 & c_2+2c_1 \\ a_3 & b_3 & c_3 \end{vmatrix}$.
   2. By linearity in row 2, split this into two determinants: $\det B = \begin{vmatrix} a_1&b_1&c_1\\a_2&b_2&c_2\\a_3&b_3&c_3\end{vmatrix} + 2\begin{vmatrix} a_1&b_1&c_1\\a_1&b_1&c_1\\a_3&b_3&c_3\end{vmatrix}$. …

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10. AP EAPCET 2022Set eng-2022-07-04-AN1 markMCQ
    Q.If a, b, c are respectively the $5^{th}$, $8^{th}$, $13^{th}$ terms of an arithmetic progression, then $\begin{vmatrix} a & 5 & 1 \\ b & 8 & 1 \\ c & 13 & 1 \end{vmatrix} =$ (A) $0$ (B) $1$ (C) $abc$ (D) $520$

    ›Reveal solutionSolution

    Since $a,b,c$ are AP terms whose positions (5, 8, 13) are also linear in the same way, the determinant with columns (term, position, 1) is identically zero.

    Concept and Intuition

    If $a,b,c$ are terms of an AP at positions $p,q,r$ respectively, then $a=A+(p-1)d,\ b=A+(q-1)d,\ c=A+(r-1)d$ — i.e. each term is an affine (linear) function of its position. A determinant of the form $\begin{vmatrix}\text{term} & \text{position} & 1\end{vmatrix}$ over three such rows is always zero, because the "term" column is a linear combination of the "position" column and the "1" column (row reduction makes a column of zeros).

    Step-by-Step Solution

    1. Let the AP have first term $A$ and common difference $d$. Then $a = A+4d$ (5th term), $b=A+7d$ (8th term), $c=A+12d$ (13th term).
    2. Expand the determinant along the first row: $\Delta = a(8\cdot1-1\cdot13) - 5(b\cdot1-1\cdot c) + 1(b\cdot13-8\cdot c)$ $= a(-5) -5(b-c) + (13b-8c) = -5a -5b+5c+13b-8c = -5a+8b-3c$.
    3. Substitute: $-5a = -5A-20d$, $8b=8A+56d$, $-3c=-3A-36d$. …

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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 …

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12. 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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