Q.If $A$ is an invertible matrix of order 2, then $\det(A^{-1})$ is equal to (A) $\det(A)$ (B) $\dfrac{1}{\det(A)}$ (C) $1$ (D) $0$

Showing the 12 most recent of 20 on this concept.

1. 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. …

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2. 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): …

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3. 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 $\det(A)\det(\operatorname{adj}A) = (\det A)^n$. …

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4. 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 $|A| = \pm\dfrac{1}{4}$. …

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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 $B=\begin{bmatrix}1&2&3\\4&-3&-1\\2&1&-4\end{bmatrix}$: $\det B = 1((-3)(-4)-(-1)(1)) - 2(4(-4)-(-1)(2)) + 3(4(1)-(-3)(2))$ …

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6. 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: $a=-8,\,b=-2,\,c=12,\,d=10,\,e=3,\,f=-15,\,l=6,\,m=2,\,n=-10$. …

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7. 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): …

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8. AP EAPCET 2024Set eng-2024-05-22-FN1 markMCQ
   Q.If $P$ and $Q$ are two $3 \times 3$ matrices such that $|PQ| = 1$ and $|P| = 9$, then the determinant of adjoint of the matrix $P \cdot Adj\, 3Q$ is (A) $9^4$ (B) $\dfrac{1}{9^4}$ (C) $9^2$ (D) $\dfrac{1}{9^2}$

   ›Reveal solutionSolution

   Using $|Q|=1/|P|\cdot|PQ|$, the scaling property $Adj(kA)=k^{n-1}AdjA$, and $|Adj\,M|=|M|^{n-1}$ for $3\times3$ matrices gives $9^4$.

   Concept and Intuition

   For an $n\times n$ matrix, $|kA|=k^n|A|$, $Adj(kA)=k^{n-1}Adj\,A$, and $|Adj\,A|=|A|^{n-1}$. Chaining these scaling rules for $n=3$ solves the problem without ever computing $P$ or $Q$ explicitly.

   Step-by-Step Solution

   1. $|PQ|=|P||Q|=1\Rightarrow |Q|=1/9$ (since $|P|=9$).
   2. $Adj(3Q)=3^{3-1}\,Adj\,Q=9\,Adj\,Q$.
   3. Let $M=P\cdot Adj(3Q)=9\,(P\cdot Adj\,Q)$.
   4. $|P\cdot Adj\,Q|=|P|\cdot|Adj\,Q|=|P|\cdot|Q|^{3-1}=9\cdot\left(\dfrac19\right)^2=\dfrac{9}{81}=\dfrac19$.
   5. $|M|=9^3\cdot|P\cdot Adj\,Q|=729\cdot\dfrac19=81=9^2$. …

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9. 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:

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

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

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

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

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

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

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

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

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

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10. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
    Q.If $A = \begin{bmatrix} 1 & 2 & 3 \\ 1 & 3 & 5 \\ 2 & 1 & 6 \end{bmatrix}$ and $|\mathrm{adj}(\mathrm{adj}\,A)|(\mathrm{adj}\,A)^{-1} = kA$, then $k =$ (A) 1296 (B) 216 (C) 36 (D) 432

    ›Reveal solutionSolution

    Using the standard adjugate identities for a $3\times3$ matrix, $k=|A|^3$; direct computation gives $|A|=6$, so $k=216$ — option (B).

    Concept and Intuition

    The adjugate (classical adjoint) of an $n\times n$ matrix satisfies two workhorse identities: $|\mathrm{adj}\,A| = |A|^{n-1}$, and (applying that twice) $|\mathrm{adj}(\mathrm{adj}\,A)| = |A|^{(n-1)^2}$. Also, since $A\cdot\mathrm{adj}\,A = |A|\,I$, we get $\mathrm{adj}\,A = |A|A^{-1}$, i.e. $(\mathrm{adj}\,A)^{-1} = \dfrac{A}{|A|}$. Combining these turns the whole expression into a scalar power of $|A|$ times $A$ — exactly the form $kA$ the question wants.

    Step-by-Step Solution

    1. For $n=3$: $|\mathrm{adj}\,A|=|A|^{n-1}=|A|^2$, so $|\mathrm{adj}(\mathrm{adj}\,A)| = |\mathrm{adj}\,A|^{n-1} = (|A|^2)^{2} = |A|^4$.
    2. $(\mathrm{adj}\,A)^{-1} = \dfrac{A}{|A|}$ (from $A\,\mathrm{adj}\,A=|A|I$).
    3. So $|\mathrm{adj}(\mathrm{adj}\,A)|\,(\mathrm{adj}\,A)^{-1} = |A|^4\cdot\dfrac{A}{|A|} = |A|^3 A$. Comparing to $kA$: $k=|A|^3$. …

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11. 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.

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

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

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12. 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 $A=\begin{pmatrix}1&-3&2\\-2&1&3\\3&2&-1\end{pmatrix}$: $\det A = 1(1\cdot(-1)-3\cdot2) -(-3)((-2)(-1)-3\cdot3) +2((-2)(2)-1\cdot3)$ $= 1(-7) +3(-7) +2(-7) = -7-21-14 = -42$. …

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