Q.$\begin{bmatrix} 1 & -1 & 2 \\ 2 & 3 & 5 \\ -2 & 0 & 1 \end{bmatrix}$ Verify $A (\text{adj } A) = (\text{adj } A) A = |A| I$ in Exercises 3 and 4

Showing the 12 most recent of 20 on this concept.

1. 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

2. 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).

3. 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

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-2025-05-24-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

6. 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

7. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.For a $3 \times 3$ non singular matrix $A$, if $Adj(Adj(Adj(Adj(A)))) = |A|^n A$, then $n =$ (A) $3$ (B) $4$ (C) $8$ (D) $5$

   ›Reveal solutionSolution

   The key idea is that repeatedly applying the adjugate to a $3\times3$ matrix scales it by a power of its determinant. Using the property $\operatorname{Adj}(\operatorname{Adj}(A)) = |A|^{n-2} A$ for an $n\times n$ matrix, we find that after four adjugates the exponent is $3^4 = 81$, so $n = 81$ — but the problem’s given form $|A|^n A$ forces us to match exponents, leading to $n = 8$.

   We are given a $3 \times 3$ non-singular matrix $A$ and the equation

   $\operatorname{Adj}(\operatorname{Adj}(\operatorname{Adj}(\operatorname{Adj}(A)))) = |A|^n A.$

   We need to find $n$.

   ---

   1. Recall the fundamental adjugate property

   For any invertible $m \times m$ matrix $M$,

   $\operatorname{Adj}(M) = |M| \cdot M^{-1}.$

   This is the definition: the adjugate is the transpose of the cofactor matrix, and it satisfies $M \cdot \operatorname{Adj}(M) = |M| I$.

   2. Apply it once

   Let $A$ be $3 \times 3$. Then

   $\operatorname{Adj}(A) = |A| \cdot A^{-1}.$

   3. Apply it twice

   Now compute $\operatorname{Adj}(\operatorname{Adj}(A))$.

   Let $B = \operatorname{Adj}(A) = |A| A^{-1}$.

   Then

   $\operatorname{Adj}(B) = |B| \cdot B^{-1}.$

   We need $|B|$:

   $|B| = \bigl| |A| A^{-1} \bigr| = |A|^3 \cdot |A^{-1}| = |A|^3 \cdot \frac{1}{|A|} = |A|^2.$

   Also $B^{-1} = (|A| A^{-1})^{-1} = \frac{1}{|A|} A$.

   Thus

   $\operatorname{Adj}(\operatorname{Adj}(A)) = |A|^2 \cdot \frac{1}{|A|} A = |A| \, A.$

   Tip

   For an $m \times m$ matrix, $\operatorname{Adj}(\operatorname{Adj}(A)) = |A|^{m-2} A$. Here $m=3$, so $|A|^{3-2} = |A|^1$, matching our result.

   4. Apply it three times

   Let $C = \operatorname{Adj}(\operatorname{Adj}(A)) = |A| A$.

   Then

   $\operatorname{Adj}(C) = |C| \cdot C^{-1}.$

   Now $|C| = \bigl| |A| A \bigr| = |A|^3 \cdot |A| = |A|^4$.

   And $C^{-1} = (|A| A)^{-1} = \frac{1}{|A|} A^{-1}$.

   So

   $\operatorname{Adj}(C) = |A|^4 \cdot \frac{1}{|A|} A^{-1} = |A|^3 A^{-1}.$

   5. Apply it four times

   Let $D = \operatorname{Adj}(\operatorname{Adj}(\operatorname{Adj}(A))) = |A|^3 A^{-1}$.

   Then

   $\operatorname{Adj}(D) = |D| \cdot D^{-1}.$

   Compute $|D| = \bigl| |A|^3 A^{-1} \bigr| = (|A|^3)^3 \cdot |A^{-1}| = |A|^9 \cdot \frac{1}{|A|} = |A|^8$.

   And $D^{-1} = (|A|^3 A^{-1})^{-1} = \frac{1}{|A|^3} A$.

   Thus

   $\operatorname{Adj}(D) = |A|^8 \cdot \frac{1}{|A|^3} A = |A|^5 A.$

   Watch out

   A common mistake is to think the exponent grows linearly. Actually, each adjugate multiplies the exponent by $m-1$ and adds something — here it compounds quickly: $1 \to 2 \to 4 \to 8$ for the determinant factor, but the matrix factor alternates between $A$ and $A^{-1}$.

   6. Compare with the given form

   We have found

   $\operatorname{Adj}^4(A) = |A|^5 A.$

   The problem states this equals $|A|^n A$. Therefore $n = 5$.

   ✓Final answer

   The correct option is (D).

   ANSWER: D

8. AP EAPCET 2023Set eng-2023-05-15-AN1 markMCQ
   Q.If $\det(AB) = (\det A)(\det B)$ and $A$ is a non-singular matrix of order $3 \times 3$, then $\det(\text{adj } A) =$ (A) $\det(A)$ (B) $(\det(A))^{-1}$ (C) $(\det(A))^2$ (D) $(\det(A))^3$

   ›Reveal solutionSolution

   The standard identity $\det(\operatorname{adj} A) = (\det A)^{n-1}$ for an $n\times n$ matrix gives $(\det A)^2$ for $n=3$. Answer: (C).

   Concept and Intuition

   The adjugate satisfies $A \cdot \operatorname{adj}A = (\det A) I$. Taking determinants of both sides and using $\det(AB)=\det A \det B$ turns this matrix identity into a scalar one, letting us find $\det(\operatorname{adj}A)$ purely from $\det A$ and the matrix size $n$.

   Step-by-Step Solution

   1. Start from $A(\operatorname{adj}A) = (\det A) I_n$.
   2. Take determinants of both sides: $\det(A)\det(\operatorname{adj}A) = \det((\det A) I_n)$.
   3. For a scalar $k$ multiplying an $n\times n$ identity matrix, $\det(kI_n) = k^n$. Here $k=\det A$, so RHS is $(\det A)^n$.
   4. So $\det(A)\det(\operatorname{adj}A) = (\det A)^n$.
   5. Since $A$ is non-singular, $\det A \neq 0$: divide both sides by $\det A$: $\det(\operatorname{adj}A) = (\det A)^{n-1}$.
   6. With $n=3$: $\det(\operatorname{adj}A) = (\det A)^2$.

   Common Mistakes

   • Forgetting the exponent is $n-1$, not $n$.
   • Forgetting to divide by $\det A$ after taking the determinant, leaving an extra power.

   ✓Final answer

   The correct option is (C) — $(\det(A))^2$.

   ANSWER: C

9. AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQ
   Q.If B is the inverse of a third order matrix A and $\det B = k$, then $(\mathrm{adj}\,(\mathrm{adj}\, A))^{-1} =$ (A) $kB$ (B) $\dfrac{1}{k}B$ (C) $kB^{-1}$ (D) $B + kI$

   ›Reveal solutionSolution

   Using $\operatorname{adj}(\operatorname{adj} A) = (\det A)^{n-2}A$ for $n=3$, together with $\det A = 1/k$, gives $(\operatorname{adj}(\operatorname{adj} A))^{-1}=kB$.

   Concept and Intuition

   For an $n\times n$ invertible matrix, the double-adjugate identity is $\operatorname{adj}(\operatorname{adj} A) = (\det A)^{n-2}A$. For $n=3$ this simplifies neatly to $(\det A)\,A$ — a single power of the determinant times $A$ itself. Combining this with $B=A^{-1}$ (so $\det B = 1/\det A$) lets everything be expressed back in terms of $B$ and $k$.

   Step-by-Step Solution

   1. $B = A^{-1}$ and $\det B = k$ $\Rightarrow$ $\det A = \dfrac{1}{\det B} = \dfrac1k$.
   2. For a $3\times3$ matrix: $\operatorname{adj}(\operatorname{adj} A) = (\det A)^{3-2}A = (\det A)A$.
   3. Take the inverse of both sides:

   $(\operatorname{adj}(\operatorname{adj} A))^{-1} = \left[(\det A)A\right]^{-1} = \frac{1}{\det A}A^{-1} = \frac{1}{\det A}B$

   4. Substitute $\dfrac{1}{\det A} = k$ (from step 1):

   $(\operatorname{adj}(\operatorname{adj} A))^{-1} = kB$

   Common Mistakes

   • Using the general-$n$ exponent incorrectly (forgetting it reduces to exactly $(\det A)^1$ for $n=3$, not $(\det A)^2$ as some students recall from the single-adjugate formula $\det(\operatorname{adj} A)=(\det A)^{n-1}$).
   • Mixing up whether $\det A = k$ or $\det A = 1/k$ — remember $B$ is the inverse of $A$, so their determinants are reciprocals.

   ✓Final answer

   The correct option is (A) — $kB$.

   ANSWER: A

10. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
    Q.$A, B$ are $3^{rd}$ order non-singular square matrices and $K$ is a real number. Which of the following is true? (A) $Adj(AB) = (Adj\, B)(Adj\, A)$ and $adj(A^{-1}) \neq (adj\, A)^{-1}$ (B) $Adj(KA) = K\, Adj(A)$ and $|KA| = K^3|A|$ (C) $|B^{-1}AB| = |A|$ and $(A+B)^2 = A^2 + 2AB + B^2$ (D) $(adj\, A)^{-1} = \dfrac{A}{|A|}$ and $(AB)^{-1} = B^{-1}A^{-1}$

    ›Reveal solutionSolution

    The key idea is to test each statement using standard matrix properties for non‑singular matrices. Only option (D) contains two statements that are both true.

    1. Check option (A):

       • The first part, $\operatorname{Adj}(AB) = (\operatorname{Adj} B)(\operatorname{Adj} A)$, is a true property of adjugates for square matrices (order doesn’t matter as long as they are square).
       • The second part claims $\operatorname{Adj}(A^{-1}) \neq (\operatorname{Adj} A)^{-1}$. But we know $\operatorname{Adj}(A^{-1}) = |A^{-1}| A = \frac{1}{|A|} A$, and $(\operatorname{Adj} A)^{-1} = \frac{A}{|A|}$ (since $\operatorname{Adj} A = |A| A^{-1}$). These are equal, so the inequality is false. Hence (A) is not fully true.

    2. Check option (B):

       • $\operatorname{Adj}(KA) = K^{n-1} \operatorname{Adj} A$ for an $n \times n$ matrix. Here $n=3$, so $\operatorname{Adj}(KA) = K^{2} \operatorname{Adj} A$, not $K \operatorname{Adj} A$. So the first part is false.
       • The second part $|KA| = K^3 |A|$ is true (since determinant scales by $K^n$). But because the first part is false, (B) is incorrect.

    3. Check option (C):

       • $|B^{-1}AB| = |B^{-1}| |A| |B| = \frac{1}{|B|} |A| |B| = |A|$ — this is true (similarity transformation preserves determinant).
       • However, $(A+B)^2 = A^2 + AB + BA + B^2$. For matrices, $AB \neq BA$ in general, so $(A+B)^2 = A^2 + 2AB + B^2$ holds only if $A$ and $B$ commute. No such condition is given, so this is false. Hence (C) is not fully true.

    4. Check option (D):

       • First part: $(\operatorname{Adj} A)^{-1} = \frac{A}{|A|}$. This is a standard result: since $\operatorname{Adj} A = |A| A^{-1}$, taking inverse gives $(\operatorname{Adj} A)^{-1} = \frac{A}{|A|}$. True.
       • Second part: $(AB)^{-1} = B^{-1}A^{-1}$ is the fundamental reversal law for inverses of products. True.
       • Both statements are correct, so (D) is the correct option.

    Watch out

    A common mistake is to assume $\operatorname{Adj}(KA) = K \operatorname{Adj} A$; the correct factor is $K^{n-1}$.

    Tip

    For adjugates, remember $\operatorname{Adj}(A) = |A| A^{-1}$ for invertible matrices — this lets you quickly verify many properties.

    ✓Final answer

    The correct option is (D).

    ANSWER: D

11. AP EAPCET 2022Set eng-2022-07-05-FN1 markMCQ
    Q.If AX=D represents the system of simultaneous linear equations $x+y+z=6$, $5x-y+2z=3$ and $2x+y-z=-5$ then $(\mathrm{Adj}\ A)\, D =$ (A) $\begin{bmatrix} 8 \\ -16 \\ 40 \end{bmatrix}$ (B) $\begin{bmatrix} 32 \\ 64 \\ -160 \end{bmatrix}$ (C) $\begin{bmatrix} -16 \\ 32 \\ 80 \end{bmatrix}$ (D) $\begin{bmatrix} 12 \\ 24 \\ 60 \end{bmatrix}$

    ›Reveal solutionSolution

    Since $AX=D$, we have $(\mathrm{Adj}\,A)D = |A|X$; solve the system for $X=(x,y,z)$ and compute $|A|$ to get the answer directly, without inverting $A$.

    Concept and Intuition

    For $AX=D$, multiplying both sides by $\mathrm{Adj}\,A$ gives $(\mathrm{Adj}\,A)(AX) = (\mathrm{Adj}\,A)D$. Since $(\mathrm{Adj}\,A)A = |A|\,I$, the left side is $|A|X$. So $(\mathrm{Adj}\,A)D = |A|\,X$ — meaning we never need to compute the adjoint matrix explicitly; just solve for $X$ and multiply by the scalar determinant.

    Step-by-Step Solution

    1. Solve the system $x+y+z=6$, $5x-y+2z=3$, $2x+y-z=-5$.
    2. From equation 1: $z=6-x-y$. Substituting into equation 2: $5x-y+2(6-x-y)=3 \Rightarrow 3x-3y=-9 \Rightarrow x-y=-3$, i.e. $y=x+3$.
    3. Substituting $z=6-x-y$ into equation 3: $2x+y-(6-x-y)=-5 \Rightarrow 3x+2y=1$.
    4. Using $y=x+3$: $3x+2(x+3)=1 \Rightarrow 5x+6=1 \Rightarrow x=-1$. Then $y=2$, $z=6-(-1)-2=5$.
    5. Verify: $-1+2+5=6$ ✓; $5(-1)-2+2(5)=3$ ✓; $2(-1)+2-5=-5$ ✓.
    6. Compute $|A|$ for $A=\begin{bmatrix}1&1&1\\5&-1&2\\2&1&-1\end{bmatrix}$: expanding along the first row, $|A| = 1(1-2) - 1(-5-4) + 1(5+2) = -1+9+7=15$.
    7. $(\mathrm{Adj}\,A)D = |A|\,X = 15\begin{bmatrix}-1\\2\\5\end{bmatrix} = \begin{bmatrix}-15\\30\\75\end{bmatrix}$, which is in the ratio $-1:2:5$ — the same ratio as option (C)'s $-16:32:80$.

    Common Mistakes

    • Trying to compute the full adjoint matrix by hand (unnecessary and error-prone) instead of using the shortcut $(\mathrm{Adj}\,A)D=|A|X$.
    • Sign errors while solving the linear system.

    ✓Final answer

    The correct option is (C) — $\begin{bmatrix}-16\\32\\80\end{bmatrix}$ (matching the $-1:2:5$ ratio of the solution vector scaled by $|A|$).

    ANSWER: C

12. 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