Q.Find $\operatorname{adj} A$ for $A = \begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix}$.

Showing the 12 most recent of 20 on this concept.

1. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
   Q.If $A = \begin{bmatrix} 2 & -3 & 1 \\ -3 & 1 & -2 \\ 1 & -2 & 3 \end{bmatrix}$, then $Adj(A) =$ (A) $\begin{bmatrix} 1 & -7 & 5 \\ -7 & 5 & -1 \\ 5 & -1 & 7 \end{bmatrix}$ (B) $\begin{bmatrix} 1 & 7 & -5 \\ 7 & 5 & 1 \\ -5 & 1 & 7 \end{bmatrix}$ (C) $\begin{bmatrix} -1 & 7 & 5 \\ 7 & 5 & 1 \\ 5 & 1 & -7 \end{bmatrix}$ (D) $\begin{bmatrix} -1 & 7 & -5 \\ 7 & -5 & 1 \\ -5 & 1 & -7 \end{bmatrix}$

   ›Reveal solutionSolution

   $A$ is symmetric, so its adjoint equals its cofactor matrix; computing the cofactors gives $\begin{bmatrix}-1&7&5\\7&5&1\\5&1&-7\end{bmatrix}$.

   Setup. $\operatorname{Adj}(A)$ is the transpose of the cofactor matrix. Here

   $A=\begin{bmatrix}2 & -3 & 1\\-3 & 1 & -2\\1 & -2 & 3\end{bmatrix}$

   is symmetric, so the adjoint is also symmetric.

   Cofactors.

   $C_{11}=\begin{vmatrix}1&-2\\-2&3\end{vmatrix}=3-4=-1,\quad C_{12}=-\begin{vmatrix}-3&-2\\1&3\end{vmatrix}=-(-9+2)=7,\quad C_{13}=\begin{vmatrix}-3&1\\1&-2\end{vmatrix}=6-1=5,$

   $C_{22}=\begin{vmatrix}2&1\\1&3\end{vmatrix}=6-1=5,\quad C_{23}=-\begin{vmatrix}2&-3\\1&-2\end{vmatrix}=-(-4+3)=1,\quad C_{33}=\begin{vmatrix}2&-3\\-3&1\end{vmatrix}=2-9=-7.$

   By symmetry $C_{21}=7,\;C_{31}=5,\;C_{32}=1$.

   Adjoint. Transposing the cofactor matrix (which is symmetric here) gives

   $\operatorname{Adj}(A)=\begin{bmatrix}-1 & 7 & 5\\7 & 5 & 1\\5 & 1 & -7\end{bmatrix}.$

   ✓Final answer

   $\operatorname{Adj}(A)=\begin{bmatrix}-1&7&5\\7&5&1\\5&1&-7\end{bmatrix}$ — option (C).

2. AP EAPCET 2022Set eng-2022-07-07-AN1 markMCQ
   Q.If $A = \begin{bmatrix} 1 & -2 & 2 \\ 2 & -6 & 5 \\ 5 & 0 & 4 \end{bmatrix}$ then Adj A = (A) $\begin{bmatrix} -24 & 8 & 2 \\ 17 & -6 & -1 \\ 30 & -10 & 2 \end{bmatrix}$ (B) $\begin{bmatrix} -24 & 8 & 2 \\ 17 & -6 & 1 \\ -30 & 10 & -2 \end{bmatrix}$ (C) $\begin{bmatrix} -24 & 8 & 2 \\ 17 & -6 & -1 \\ 30 & -10 & -2 \end{bmatrix}$ (D) $\begin{bmatrix} 24 & -8 & 2 \\ -17 & -6 & 1 \\ 30 & -10 & -2 \end{bmatrix}$

   ›Reveal solutionSolution

   Computing every $2\times2$ cofactor of $A$ and transposing the resulting cofactor matrix gives the adjugate, which matches option (C) exactly.

   Concept and Intuition

   The adjugate (classical adjoint) of a $3\times3$ matrix is the transpose of its cofactor matrix: $\text{Adj}(A) = C^T$, where $C_{ij}=(-1)^{i+j}M_{ij}$ and $M_{ij}$ is the minor obtained by deleting row $i$ and column $j$. This requires carefully computing all nine $2\times2$ minors with correct alternating signs.

   Step-by-Step Solution

   For $A=\begin{bmatrix}1&-2&2\\2&-6&5\\5&0&4\end{bmatrix}$, compute each cofactor:

   $C_{11}=+\begin{vmatrix}-6&5\\0&4\end{vmatrix}=(-24-0)=-24$

   $C_{12}=-\begin{vmatrix}2&5\\5&4\end{vmatrix}=-(8-25)=17$

   $C_{13}=+\begin{vmatrix}2&-6\\5&0\end{vmatrix}=(0-(-30))=30$

   $C_{21}=-\begin{vmatrix}-2&2\\0&4\end{vmatrix}=-(-8-0)=8$

   $C_{22}=+\begin{vmatrix}1&2\\5&4\end{vmatrix}=(4-10)=-6$

   $C_{23}=-\begin{vmatrix}1&-2\\5&0\end{vmatrix}=-(0-(-10))=-10$

   $C_{31}=+\begin{vmatrix}-2&2\\-6&5\end{vmatrix}=(-10-(-12))=2$

   $C_{32}=-\begin{vmatrix}1&2\\2&5\end{vmatrix}=-(5-4)=-1$

   $C_{33}=+\begin{vmatrix}1&-2\\2&-6\end{vmatrix}=(-6-(-4))=-2$

   Cofactor matrix: $C=\begin{bmatrix}-24&17&30\\8&-6&-10\\2&-1&-2\end{bmatrix}$

   Transpose to get the adjugate:

   $\text{Adj }A = C^T = \begin{bmatrix}-24&8&2\\17&-6&-1\\30&-10&-2\end{bmatrix}$

   Common Mistakes

   • Forgetting to transpose the cofactor matrix (a very common slip — the adjugate is $C^T$, not $C$ itself).
   • Sign errors in the alternating $(-1)^{i+j}$ pattern, especially for $C_{12}, C_{21}, C_{23}, C_{32}$.

   ✓Final answer

   The correct option is (C) — $\begin{bmatrix}-24&8&2\\17&-6&-1\\30&-10&-2\end{bmatrix}$.

   ANSWER: C

3. AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQ
   Q.If $a$ is the determinant of the adjoint of the matrix $\begin{bmatrix} 1 & 1 & 2 \\ 1 & 2 & 3 \\ 2 & 3 & 3 \end{bmatrix}$ and $b$ is the determinant of the inverse of the matrix $\begin{bmatrix} 1 & 2 & 3 \\ 4 & -3 & -1 \\ 2 & 1 & -4 \end{bmatrix}$ then $\dfrac{b+1}{18b} =$ (A) $a$ (B) $10a$ (C) $2+a$ (D) $2a$

   ›Reveal solutionSolution

   This tests the identities $\det(\operatorname{adj}A)=(\det A)^{n-1}$ and $\det(A^{-1})=1/\det A$, then simplifying an algebraic expression in $b$ to match $a$.

   Concept and Intuition

   For an $n\times n$ matrix, $\det(\operatorname{adj}A) = (\det A)^{n-1}$; for $n=3$ this is $(\det A)^2$, always non-negative. Also, since $A\cdot A^{-1}=I$, taking determinants gives $\det(A^{-1}) = 1/\det A$. Both facts let us compute $a$ and $b$ purely from the determinants of the given matrices, then plug into the target expression.

   Step-by-Step Solution

   1. Compute $\det A$ for $A=\begin{bmatrix}1&1&2\\1&2&3\\2&3&3\end{bmatrix}$: $\det A = 1(2\cdot3-3\cdot3) - 1(1\cdot3-3\cdot2) + 2(1\cdot3-2\cdot2) = 1(-3)-1(-3)+2(-1) = -3+3-2=-2$.
   2. $a = \det(\operatorname{adj}A) = (\det A)^{2} = (-2)^2 = 4$.
   3. Compute $\det B$ for $B=\begin{bmatrix}1&2&3\\4&-3&-1\\2&1&-4\end{bmatrix}$: $\det B = 1((-3)(-4)-(-1)(1)) - 2(4(-4)-(-1)(2)) + 3(4(1)-(-3)(2))$ $= 1(12+1) - 2(-16+2) + 3(4+6) = 13 + 28 + 30 = 71$.
   4. $b = \det(B^{-1}) = \dfrac{1}{71}$.
   5. Now evaluate $\dfrac{b+1}{18b} = \dfrac{\frac{1}{71}+1}{18\cdot\frac1{71}} = \dfrac{\frac{72}{71}}{\frac{18}{71}} = \dfrac{72}{18} = 4$.
   6. Since $a=4$, we get $\dfrac{b+1}{18b} = a$.

   Common Mistakes

   • Using $\det(\operatorname{adj}A)=\det A$ instead of $(\det A)^{n-1}$.
   • Sign errors in the $3\times3$ cofactor expansion.

   ✓Final answer

   The correct option is (A) — $a$.

   ANSWER: A

4. AP EAPCET 2022Set eng-2022-07-08-FN1 markMCQ
   Q.If $A = \begin{pmatrix}1 & -3 & 2\\ -2 & 1 & 3\\ 3 & 2 & -1\end{pmatrix}$ then $A^2\, \mathrm{Adj}\, A =$ (A) $21A$ (B) $-42A$ (C) $7A^{-1}$ (D) $14(\mathrm{Adj}\,A)$

   ›Reveal solutionSolution

   This tests the identity $A\cdot\mathrm{Adj}(A) = (\det A)I$ applied cleverly to reduce $A^2\,\mathrm{Adj}\,A$.

   Concept and Intuition

   Rather than computing $\mathrm{Adj}\,A$ explicitly (tedious for a 3×3), use the fundamental relation $A\cdot\mathrm{Adj}\,A = (\det A)I$ to convert one factor of $A$ times $\mathrm{Adj}\,A$ directly into a scalar multiple of the identity, leaving a single $A$ behind.

   Step-by-Step Solution

   1. Compute $\det A$ for $A=\begin{pmatrix}1&-3&2\\-2&1&3\\3&2&-1\end{pmatrix}$: $\det A = 1(1\cdot(-1)-3\cdot2) -(-3)((-2)(-1)-3\cdot3) +2((-2)(2)-1\cdot3)$ $= 1(-7) +3(-7) +2(-7) = -7-21-14 = -42$.
   2. Use $A\cdot\mathrm{Adj}\,A = (\det A)I = -42I$.
   3. So $A^2\,\mathrm{Adj}\,A = A\,(A\,\mathrm{Adj}\,A) = A(-42I) = -42A$.

   Common Mistakes

   • Trying to compute the full adjugate matrix explicitly and multiply out — unnecessary and error-prone; the identity shortcut avoids that.

   ✓Final answer

   The correct option is (B) — $-42A$.

   ANSWER: B

5. AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQ
   Q.If $P=\begin{bmatrix}1 & \alpha & 3\\ 1 & 3 & 3\\ 2 & 4 & 4\end{bmatrix}$ is the adjoint of a matrix A and $\det A = 4$, then the value of $\alpha$ is (A) 3 (B) 22 (C) 11 (D) 4

   ›Reveal solutionSolution

   Uses $\det(\operatorname{adj}A)=(\det A)^{n-1}$ for an $n\times n$ matrix; with $n=3,\det A=4$, we need $\det P=16$, which pins down $\alpha=11$.

   Concept and Intuition

   The adjoint (adjugate) matrix of $A$ satisfies $A\cdot\operatorname{adj}A=(\det A)I$, and taking determinants of both sides gives the identity $\det(\operatorname{adj}A)=(\det A)^{n-1}$ for an $n\times n$ matrix. For $n=3$ this is $(\det A)^2$, always non-negative regardless of the sign of $\det A$ — so knowing $\det A=4$ immediately tells us $\det(\operatorname{adj}A)=16$ without needing $A$ itself.

   Step-by-Step Solution

   1. $P=\operatorname{adj}A$ is $3\times3$, so $\det P=(\det A)^{3-1}=(\det A)^2=4^2=16$.
   2. Expand $\det P$ along row 1:

   $\det P = 1\begin{vmatrix}3&3\\4&4\end{vmatrix}-\alpha\begin{vmatrix}1&3\\2&4\end{vmatrix}+3\begin{vmatrix}1&3\\2&4\end{vmatrix}$

   3. Compute the $2\times2$ determinants: $\begin{vmatrix}3&3\\4&4\end{vmatrix}=12-12=0$; $\begin{vmatrix}1&3\\2&4\end{vmatrix}=4-6=-2$.
   4. So $\det P = 1(0)-\alpha(-2)+3(-2) = 2\alpha-6$.
   5. Set $2\alpha-6=16 \implies 2\alpha=22 \implies \alpha=11$.

   Common Mistakes

   • Using $\det(\operatorname{adj}A)=\det A$ (wrong power) instead of $(\det A)^{n-1}$.
   • Sign errors in cofactor expansion, especially the alternating $+,-,+$ pattern along a row.

   ✓Final answer

   The correct option is (C) — $11$.

   ANSWER: C

6. AP EAPCET 2021Set eng-2021-10-05-FN1 markMCQ
   Q.If $A = \begin{bmatrix} -1 & -2 & -2 \\ 2 & 1 & -2 \\ 2 & -2 & 1 \end{bmatrix}$ then $adj(A) = ?$ (A) $2A^T$ (B) $A^T$ (C) $3A^T$ (D) $4A^T$

   ›Reveal solutionSolution

   $A$'s rows are mutually orthogonal with equal norm, so $AA^T=9I$; combined with $\det A=27$, this gives $\operatorname{adj}(A)=3A^T$ directly, without computing all nine cofactors.

   Concept and Intuition

   For any invertible square matrix, $\operatorname{adj}(A) = \det(A)\,A^{-1}$. If $A$ happens to have orthogonal rows of equal length (a scaled orthogonal matrix), $AA^T$ collapses to a scalar multiple of $I$, instantly giving $A^{-1}$ (and hence the adjugate) without a full cofactor expansion.

   Step-by-Step Solution

   1. $A=\begin{bmatrix}-1&-2&-2\\2&1&-2\\2&-2&1\end{bmatrix}$. Compute $\det A$ by cofactor expansion along row 1: $\det A = -1(1\cdot1-(-2)(-2)) -(-2)(2\cdot1-(-2)\cdot2) + (-2)(2(-2)-1\cdot2)$ $= -1(1-4) +2(2+4) -2(-4-2) = 3+12+12 = 27$.
   2. Check rows of $A$ for orthogonality: Row1$\cdot$Row1 $=1+4+4=9$; Row2$\cdot$Row2$=4+1+4=9$; Row3$\cdot$Row3$=4+4+1=9$. Row1$\cdot$Row2$=-2-2+4=0$; Row1$\cdot$Row3$=-2+4-2=0$; Row2$\cdot$Row3$=4-2-2=0$.
   3. So $AA^T = 9I$, i.e. $A^{-1} = \dfrac{A^T}{9}$.
   4. $\operatorname{adj}(A) = \det(A)\cdot A^{-1} = 27\cdot\dfrac{A^T}{9} = 3A^T$.

   Common Mistakes

   • Grinding through all nine cofactors instead of noticing the orthogonal-row shortcut, which is much faster and less error-prone.

   ✓Final answer

   The correct option is (C) — $3A^T$.

   ANSWER: C

7. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
   Q.If $A=\begin{pmatrix}a & b & c\\ d & e & f\\ l & m & n\end{pmatrix}$ is a matrix such that $|A|>0$ and $\mathrm{Adj}\,A=\begin{pmatrix}0 & 4 & -6\\ 10 & 8 & 0\\ 2 & 4 & -4\end{pmatrix}$, then $\dfrac{cd}{fb}+\dfrac{ln}{em}=$ (A) $2a$ (B) $a+m$ (C) $a+b$ (D) $a$

   ›Reveal solutionSolution

   Reconstructing $A$ from $\mathrm{adj}(A)=A^{-1}\det(A)$ via $A=\mathrm{adj}(\mathrm{adj}\,A)/\det(A)$ gives concrete entries, letting $\frac{cd}{fb}+\frac{log}{em}$ be evaluated directly and matched to $a+m$.

   Concept and Intuition

   Since $A\cdot\mathrm{adj}(A)=\det(A)\cdot I$, once $\mathrm{adj}(A)$ is fully known and $\det(A)$ is pinned down, $A$ itself is uniquely determined: $A=\det(A)\cdot[\mathrm{adj}(A)]^{-1}=\dfrac{\mathrm{adj}(\mathrm{adj}\,A)}{\det(A)}$ (using $\mathrm{adj}(\mathrm{adj}\,A)=\det(A)^{n-2}A$ for $n=3$). This turns an abstract cofactor question into a concrete numeric matrix, from which any algebraic combination of entries can just be computed.

   Step-by-Step Solution

   1. $\det(\mathrm{adj}\,A)=\det(A)^{2}$ for a $3\times3$ matrix. Computing $\det\begin{pmatrix}0&4&-6\\10&8&0\\2&4&-4\end{pmatrix}=16$, so $\det(A)^2=16\Rightarrow\det(A)=\pm4$; given $|A|>0$, $\det(A)=4$.
   2. Compute $\mathrm{adj}(\mathrm{adj}\,A)$ (cofactor-transpose of the given adjugate matrix): $\begin{pmatrix}-32&-8&48\\40&12&-60\\24&8&-40\end{pmatrix}$.
   3. $A=\dfrac{\mathrm{adj}(\mathrm{adj}\,A)}{\det A}=\dfrac14\begin{pmatrix}-32&-8&48\\40&12&-60\\24&8&-40\end{pmatrix}=\begin{pmatrix}-8&-2&12\\10&3&-15\\6&2&-10\end{pmatrix}$.
   4. Read off: $a=-8,\,b=-2,\,c=12,\,d=10,\,e=3,\,f=-15,\,l=6,\,m=2,\,n=-10$.
   5. Verify consistency: $cd=120=af=(-8)(-15)=120$ correct; $en=-30=fm=(-15)(2)=-30$ correct; and $\det(A)$ recomputes to $4$ correct.
   6. $\dfrac{cd}{fb}=\dfrac{120}{(-15)(-2)}=\dfrac{120}{30}=4$; $\dfrac{log}{em}=\dfrac{(6)(-10)}{(3)(2)}=\dfrac{-60}{6}=-10$.
   7. Sum $=4-10=-6$. Compare to options: $a+m=-8+2=-6$ — matches.

   Common Mistakes

   • Trying to guess the answer symbolically without reconstructing $A$ — the combination $\frac{cd}{fb}+\frac{log}{em}$ doesn't simplify to a clean symbolic identity without knowing actual entry values here.
   • Sign errors in the cofactor-of-cofactor computation (easy to mismatch which row/column to delete).

   ✓Final answer

   The correct option is (B) — $a+m$.

   ANSWER: B

8. AP EAPCET 2022Set eng-2022-07-08-AN1 markMCQ
   Q.If $\begin{pmatrix}5 & a & -7\\ b & -7 & c\\ -7 & d & -1\end{pmatrix}$ is the adjoint of the matrix $\begin{pmatrix}1 & 2 & 3\\ 2 & 3 & 1\\ 3 & 1 & 2\end{pmatrix}$, then a+b+c+d = (A) 8 (B) 10 (C) 0 (D) 2

   ›Reveal solutionSolution

   Direct cofactor computation of the adjoint matrix, then matching term-by-term against the given entries.

   Concept and Intuition

   The adjoint is the transpose of the cofactor matrix: $\operatorname{adj}(M)_{ij}=C_{ji}$, where $C_{ij}=(-1)^{i+j}\times$(minor deleting row $i$, column $j$).

   Step-by-Step Solution

   1. $C_{11}=\det\begin{pmatrix}3&1\\1&2\end{pmatrix}=5$.
   2. $C_{12}=-\det\begin{pmatrix}2&1\\3&2\end{pmatrix}=-1$.
   3. $C_{13}=\det\begin{pmatrix}2&3\\3&1\end{pmatrix}=-7$.
   4. $C_{21}=-\det\begin{pmatrix}2&3\\1&2\end{pmatrix}=-1$.
   5. $C_{22}=\det\begin{pmatrix}1&3\\3&2\end{pmatrix}=-7$.
   6. $C_{23}=-\det\begin{pmatrix}1&2\\3&1\end{pmatrix}=5$.
   7. $C_{31}=\det\begin{pmatrix}2&3\\3&1\end{pmatrix}=-7$.
   8. $C_{32}=-\det\begin{pmatrix}1&3\\2&1\end{pmatrix}=5$.
   9. $C_{33}=\det\begin{pmatrix}1&2\\2&3\end{pmatrix}=-1$.
   10. $\operatorname{adj}(M)=$ transpose of $[C_{ij}]=\begin{pmatrix}5&-1&-7\\-1&-7&5\\-7&5&-1\end{pmatrix}$.
   11. Compare with $\begin{pmatrix}5&a&-7\\b&-7&c\\-7&d&-1\end{pmatrix}$: $a=-1,\ b=-1,\ c=5,\ d=5$.
   12. $a+b+c+d=-1-1+5+5=8$.

   Common Mistakes

   • Forgetting to transpose the cofactor matrix to get the adjoint.
   • Sign slip on the checkerboard $(-1)^{i+j}$ pattern.

   ✓Final answer

   The correct option is (A) — 8.

   ANSWER: A

9. AP EAPCET 2023Set eng-2023-05-18-FN1 markMCQ
   Q.If A and B are non-singular matrices and $\det(AB)=(\det A)(\det B)$, then $\big((\det A)(\det B)\big)B^{-1}A^{-1} =$ (A) $\mathrm{Adj}(BA)$ (B) $\mathrm{Adj}(A)+\mathrm{Adj}(B)$ (C) $\mathrm{Adj}(AB)$ (D) $(\mathrm{Adj}\,B)(\mathrm{Adj}\,A)$

   ›Reveal solutionSolution

   Using the identity det(M)M⁻¹ = Adj(M) with M = AB (noting (AB)⁻¹ = B⁻¹A⁻¹) gives the answer directly as Adj(AB).

   Concept and Intuition

   For any invertible square matrix M, the adjugate satisfies Adj(M) = det(M)·M⁻¹. This is a standard identity coming from M·Adj(M) = det(M)·I.

   Step-by-Step Solution

   1. Recall (AB)⁻¹ = B⁻¹A⁻¹ (reverse order rule for inverses of a product).
   2. The given expression is [(det A)(det B)]·B⁻¹A⁻¹ = det(AB)·B⁻¹A⁻¹ (using the given det(AB)=(det A)(det B)).
   3. Rewrite B⁻¹A⁻¹ as (AB)⁻¹.
   4. So the expression equals det(AB)·(AB)⁻¹.
   5. By the identity Adj(M) = det(M)·M⁻¹ applied with M=AB, this is exactly Adj(AB).

   Common Mistakes

   • Forgetting the reversal in (AB)⁻¹ = B⁻¹A⁻¹ (not A⁻¹B⁻¹).
   • Trying to expand Adj(A) and Adj(B) separately instead of recognising the compact identity det(M)M⁻¹=Adj(M).

   ✓Final answer

   The correct option is (C) — Adj(AB).

   ANSWER: C

10. AP EAPCET 2021Set eng-2021-08-20-AN1 markMCQ
    Q.Let $A$ be a $n \times n$ matrix such that $A$ is upper-triangular. Then $adj(A) =$ (A) lower triangular matrix (B) upper triangular matrix (C) diagonal matrix (D) scalar matrix

    ›Reveal solutionSolution

    The adjoint (transpose of the cofactor matrix) of an upper-triangular matrix is again upper

    triangular — this is a standard structural fact, easily checked on a $2\times2$ or $3\times3$

    example.

    Concept and Intuition

    $adj(A) = [C_{ji}]$ (the transpose of the cofactor matrix). For triangular matrices, cofactors

    obtained by deleting a row/column keep enough triangular structure that the resulting adjoint stays

    triangular of the same type (upper stays upper, lower stays lower) — this is a well-known

    result, most easily verified by direct computation on a small case and generalised by induction on

    matrix size.

    Step-by-Step Solution

    1. Take the smallest instructive example, a $2\times2$ upper-triangular matrix $A=\begin{pmatrix}a&b\\0&d\end{pmatrix}$.
    2. Its adjoint (for $2\times2$, swap diagonal entries and negate off-diagonal) is $adj(A) = \begin{pmatrix}d&-b\\0&a\end{pmatrix}$ — still upper triangular.
    3. For a $3\times3$ upper-triangular $A$, each cofactor $C_{ij}$ involves deleting row $i$/column $j$; working through the six off-diagonal cofactors shows that $adj(A)_{ij}=C_{ji}$ is zero whenever $i>j$, and generally nonzero for $i\le j$ — i.e. $adj(A)$ is upper triangular.
    4. This pattern generalises to any $n\times n$ upper-triangular matrix: $adj(A)$ remains upper triangular (though it need not be diagonal or scalar in general, ruling out options (C)/(D); and it is not lower triangular in general, ruling out (A)).

    Common Mistakes

    • Confusing "upper triangular" with "diagonal" — the adjoint of an upper triangular matrix is not generally diagonal unless $A$ itself is diagonal.
    • Assuming taking the adjoint (which involves a transpose of cofactors) automatically flips upper to lower — it does not, because the cofactor computation itself already accounts for the transpose in a way that preserves the triangular "shape".

    ✓Final answer

    The correct option is (B) — upper triangular matrix.

    ANSWER: B

11. AP EAPCET 2024Set eng-2024-05-21-FN1 markMCQ
    Q.Let $A$ be a $4 \times 4$ matrix and $P$ be its adjoint matrix. If $|P| = \left|\dfrac{A}{2}\right|$, then $|A^{-1}| =$ (A) $\pm \dfrac{1}{4}$ (B) $\pm 8$ (C) $\pm 2$ (D) $\pm 4$

    ›Reveal solutionSolution

    Using |adj(A)| = |A|^(n−1) for n=4 and |kA| = k^n|A|, the given equation reduces to |A|² = 1/16, so |A| = ±1/4 and |A⁻¹| = ±4.

    Concept and Intuition

    Two standard determinant identities are needed here:

    • For an $n\times n$ matrix $A$, $\det(\text{adj}\,A) = (\det A)^{n-1}$ (this follows from $A\cdot\text{adj}(A) = |A|I$, taking determinants of both sides: $|A|\cdot|\text{adj}\,A| = |A|^n$, so if $|A|\neq0$, $|\text{adj}\,A|=|A|^{n-1}$).
    • For a scalar $k$ and $n\times n$ matrix $A$: $\det(kA) = k^n\det(A)$ (each of the $n$ rows contributes a factor of $k$).

    Step-by-Step Solution

    1. $A$ is $4\times4$, so $n=4$. Given $P=\text{adj}(A)$, we have $|P| = |A|^{4-1} = |A|^3$.
    2. $\left|\dfrac{A}{2}\right| = \left(\dfrac12\right)^4|A| = \dfrac{1}{16}|A|$.
    3. Given $|P| = \left|\dfrac{A}{2}\right|$: $|A|^3 = \dfrac{1}{16}|A|$.
    4. Since $A^{-1}$ is asked for, $A$ must be invertible, i.e. $|A|\neq 0$. Divide both sides by $|A|$ (valid since $|A|\neq0$): $|A|^2 = \dfrac{1}{16}$.
    5. So $|A| = \pm\dfrac{1}{4}$.
    6. $|A^{-1}| = \dfrac{1}{|A|} = \dfrac{1}{\pm1/4} = \pm4$.

    Common Mistakes

    • Using $|\text{adj}\,A|=|A|^n$ instead of the correct $|A|^{n-1}$.
    • Forgetting the $\left(\dfrac12\right)^4$ scaling factor (using $n=4$, not some other power) when computing $|A/2|$.
    • Dividing by $|A|$ without first confirming $|A|\neq0$ (though here it's guaranteed since $A^{-1}$ is asked for).

    ✓Final answer

    The correct option is (D) — $\pm 4$.

    ANSWER: D

12. AP EAPCET 2024Set eng-2024-05-22-FN1 markMCQ
    Q.If $A = \begin{bmatrix} a & 1 & 2 \\ 1 & 2 & b \\ c & 1 & 3 \end{bmatrix}$ and $Adj\, A = \begin{bmatrix} 7 & -1 & -5 \\ -3 & 9 & 5 \\ 1 & -3 & 5 \end{bmatrix}$ then $a^2 + b^2 + c^2 =$ (A) $10$ (B) $14$ (C) $11$ (D) $29$

    ›Reveal solutionSolution

    Each entry of $Adj\,A$ is the corresponding cofactor of $A$, transposed: $(Adj\,A)_{ij}=C_{ji}$. Writing out these equations pins down $a,b,c$ uniquely. Answer: $a^2+b^2+c^2=10$.

    Concept and Intuition

    $Adj\,A$ is the transpose of the cofactor matrix of $A$. So $(Adj\,A)_{ij}=C_{ji}(A)$ where $C_{ji}$ is the cofactor obtained by deleting row $j$ and column $i$ of $A$. Comparing entries of the given $Adj\,A$ to cofactors computed from the unknown entries $a,b,c$ of $A$ gives a solvable (over-determined but consistent) system.

    Step-by-Step Solution

    1. With $A=\begin{bmatrix}a&1&2\\1&2&b\\c&1&3\end{bmatrix}$, compute cofactor $C_{11}=\begin{vmatrix}2&b\\1&3\end{vmatrix}=6-b$. This equals $(Adj\,A)_{11}=7\Rightarrow b=-1$.
    2. Cofactor $C_{31}=\begin{vmatrix}1&2\\2&b\end{vmatrix}=b-4$. This equals $(Adj\,A)_{13}=-5\Rightarrow b-4=-5\Rightarrow b=-1$ (consistent).
    3. Cofactor $C_{12}=-\begin{vmatrix}1&b\\c&3\end{vmatrix}=bc-3$, equal to $(Adj\,A)_{21}=-3\Rightarrow bc=0$. With $b=-1$, get $c=0$.
    4. Cofactor $C_{22}=\begin{vmatrix}a&2\\c&3\end{vmatrix}=3a-2c$, equal to $(Adj\,A)_{22}=9\Rightarrow 3a-0=9\Rightarrow a=3$.
    5. Check consistency: $C_{33}=\begin{vmatrix}a&1\\1&2\end{vmatrix}=2a-1=5\Rightarrow a=3$ ✓; $C_{32}=-\begin{vmatrix}a&2\\1&b\end{vmatrix}=2-ab=2-3(-1)=5=(Adj\,A)_{23}$ ✓.
    6. So $a=3,b=-1,c=0\Rightarrow a^2+b^2+c^2=9+1+0=10$.

    Common Mistakes

    • Mixing up $(Adj\,A)_{ij}$ with the un-transposed cofactor $C_{ij}$ instead of $C_{ji}$.
    • Solving only one equation and not cross-checking with the others (the system is over-determined and must be fully consistent).

    ✓Final answer

    The correct option is (A) — $10$.

    ANSWER: A