Q.If $A = \begin{pmatrix} 2 & \lambda & -3 \\ 0 & 2 & 5 \\ 1 & 1 & 3 \end{pmatrix}$, then $A^{-1}$ exists if
(A) $\lambda = 2$
(B) $\lambda \neq 2$
(C) $\lambda \neq -2$
(D) None of these

Showing the 12 most recent of 32 on this concept.

1. 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 $x=1/5$. …

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2. AP EAPCET 2021Set eng-2021-08-23-AN1 markMCQ
   Q.If $A(\theta)=\begin{bmatrix} i\sin\theta & \cos\theta \\ \cos\theta & i\sin\theta \end{bmatrix}$ is a matrix where $i=\sqrt{-1}$, then which of the following is not true (A) $\det A(\pi+\theta)=\det A(-\theta)$ (B) $\det A(-\theta)=\det A(\theta)$ (C) $\det[A(\theta)]^{-1}=1$ (D) $\det A(-\theta)=-1$

   ›Reveal solutionSolution

   The determinant of $A(\theta)$ is identically $-1$, independent of $\theta$; this makes (A), (B), (D) trivially true and exposes (C) as false.

   Concept and Intuition

   Whenever a matrix's determinant simplifies to a constant using $\sin^2\theta+\cos^2\theta=1$, every statement that only compares $\det A$ at different arguments becomes trivial — the real test is whether the algebra of determinants (like $\det(M^{-1})=1/\det M$) is applied correctly.

   Step-by-Step Solution

   1. Compute $\det A(\theta) = (i\sin\theta)(i\sin\theta) - (\cos\theta)(\cos\theta) = i^2\sin^2\theta - \cos^2\theta$.
   2. Since $i^2=-1$: $\det A(\theta) = -\sin^2\theta - \cos^2\theta = -(\sin^2\theta+\cos^2\theta) = -1$.
   3. This value is the SAME for every $\theta$ (it never even used the sign of the argument), so:
      • (A) $\det A(\pi+\theta)=\det A(-\theta)$: both sides are $-1$. True.
      • (B) $\det A(-\theta)=\det A(\theta)$: both sides are $-1$. True.
      • (D) $\det A(-\theta) = -1$: True. …

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3. AP EAPCET 2022Set eng-2022-07-05-AN1 markMCQ
   Q.If b and c are non zero real numbers, $A = \begin{bmatrix} 1 & b & c \\ b & 2 & 3 \\ c & 3 & 4 \end{bmatrix}$ and $B = \begin{bmatrix} 0 & b & c \\ -b & 0 & 2 \\ -c & -2 & 0 \end{bmatrix}$, then $\det(A+B) =$ (A) 3 (B) 1 (C) -1 (D) 0

   ›Reveal solutionSolution

   Adding the symmetric $A$ and skew-symmetric $B$ entrywise cancels the off-diagonal $b,c$ terms below the diagonal, leaving a matrix whose determinant is a fixed number independent of $b,c$: 3.

   Concept and Intuition

   $A$ is symmetric and $B$ is skew-symmetric (its transpose is its negative, with zero diagonal). Adding them entrywise, the upper-triangular parts of $A+B$ pick up $b+b=2b$ and $c+c=2c$, while the lower-triangular parts get $b+(-b)=0$ and $c+(-c)=0$ — so $A+B$ becomes upper triangular in its first column, letting the determinant be computed by a clean cofactor expansion that never involves $b$ or $c$.

   Step-by-Step Solution

   1. $A=\begin{bmatrix}1&b&c\\b&2&3\\c&3&4\end{bmatrix}$, $B=\begin{bmatrix}0&b&c\\-b&0&2\\-c&-2&0\end{bmatrix}$.
   2. Add entrywise: $(A+B)_{11}=1$, $(A+B)_{12}=2b$, $(A+B)_{13}=2c$; $(A+B)_{21}=b-b=0$, $(A+B)_{22}=2$, $(A+B)_{23}=3+2=5$; $(A+B)_{31}=c-c=0$, $(A+B)_{32}=3-2=1$, $(A+B)_{33}=4$.
   3. So $A+B=\begin{bmatrix}1&2b&2c\\0&2&5\\0&1&4\end{bmatrix}$. …

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4. AP EAPCET 2021Set eng-2021-08-24-AN1 markMCQ
   Q.Let $A$ be a $2 \times 2$ matrix with real entries. Let $I$ be the $2 \times 2$ identity matrix. $Tr(A)$ denotes the sum of diagonal entries of $A$. Assume that $A^2 = I$. Statement I: If $A \ne I$ and $A \ne -I$ then $\det A = -1$. Statement II: If $A \ne I$ and $A \ne -I$ then $Tr\ A \ne 0$ (A) Statement I is true, statement II is true; statement II is a correct explanation for statement I (B) Statement I is true, statement II is true; statement II is not a correct explanation for statement I (C) Statement I is true, statement II is false (D) Statement I is false, statement II is true

   ›Reveal solutionSolution

   For a real $2\times 2$ matrix with $A^2=I$, $A\ne \pm I$ forces eigenvalues $\{1,-1\}$,

   giving $\det A=-1$ (Statement I true) and $\mathrm{Tr}\,A = 0$ exactly (so Statement II,

   which claims $\mathrm{Tr}\,A \ne 0$, is false).

   Concept and Intuition

   $A^2=I$ says $A$ is an "involution" — applying it twice returns the identity. Its eigenvalues

   must satisfy $\lambda^2=1$, so $\lambda \in \{1,-1\}$. Since $x^2-1$ has two distinct

   roots, $A$ is always diagonalizable (no repeated-root Jordan-block subtlety). For a

   $2\times2$ matrix there are only three possible eigenvalue multisets: $\{1,1\}$ (forces

   $A=I$), $\{-1,-1\}$ (forces $A=-I$), or $\{1,-1\}$ — and the hypothesis $A\ne \pm I$

   eliminates the first two, pinning down the third exactly.

   Step-by-Step Solution

   1. $A^2 = I \Rightarrow$ eigenvalues $\lambda$ satisfy $\lambda^2=1 \Rightarrow \lambda = \pm 1$; since these roots are distinct, $A$ is diagonalizable over $\mathbb{R}$.
   2. Case eigenvalues $\{1,1\}$: a diagonalizable matrix with both eigenvalues equal to $1$ is similar to $I$, and similar-to-$I$ forces the matrix itself to equal $I$ (since $P I P^{-1} = I$ for any $P$). This case is excluded ($A \ne I$).
   3. Case eigenvalues $\{-1,-1\}$: analogously forces $A=-I$. Excluded ($A\ne -I$).
   4. Remaining case: eigenvalues $\{1,-1\}$. This is the only possibility left once $A\ne I, -I$ is assumed. …

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5. 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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6. AP EAPCET 2022Set eng-2022-07-04-FN1 markMCQ
   Q.If $A = \begin{bmatrix} \alpha^2 & 5 \\ 5 & -\alpha \end{bmatrix}$ and $\det(A^{10}) = 1024$, then $\alpha =$ (A) $-2$ (B) $-1$ (C) $-3$ (D) $0$

   ›Reveal solutionSolution

   Use $\det(A^{10})=(\det A)^{10}$ to pin down $\det A=\pm2$, then solve the resulting cubic in $\alpha$ using the actual $2\times2$ determinant. Answer: $\alpha=-3$.

   Concept and Intuition

   For any square matrix, $\det(A^n)=(\det A)^n$. So knowing $\det(A^{10})=1024=2^{10}$ tells us $(\det A)^{10}=2^{10}$, meaning $\det A$ must be a real 10th root of $1024$ — since 10 is even, this allows $\det A=2$ or $\det A=-2$ (both give $(\pm2)^{10}=1024$). We then compute $\det A$ directly from the given matrix in terms of $\alpha$ and match against the candidate integer roots offered.

   Step-by-Step Solution

   1. $\det(A^{10}) = (\det A)^{10} = 1024 = 2^{10} \Rightarrow \det A = 2$ or $\det A=-2$.
   2. Compute $\det A$ directly: $A=\begin{bmatrix}\alpha^2 & 5\\ 5 & -\alpha\end{bmatrix}$, so $\det A = \alpha^2(-\alpha) - (5)(5) = -\alpha^3-25$.
   3. Case $\det A=2$: $-\alpha^3-25=2 \Rightarrow \alpha^3=-27 \Rightarrow \alpha=-3$.
   4. Case $\det A=-2$: $-\alpha^3-25=-2 \Rightarrow \alpha^3=-23 \Rightarrow \alpha=-\sqrt[3]{23}$, not an integer, and not among the answer choices. …

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7. 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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8. AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQ
   Q.If $\alpha$ is a real root of the equation $x^3+6x^2+5x-42=0$, then the determinant of the matrix $\begin{bmatrix}\alpha-1 & \alpha+1 & \alpha+2\\ \alpha-2 & \alpha+3 & \alpha-3\\ \alpha+4 & \alpha-4 & \alpha+5\end{bmatrix}$ is (A) 90 (B) 120 (C) -105 (D) -135

   ›Reveal solutionSolution

   The cubic has a unique real root $\alpha=2$ (found by testing small integers and confirming the remaining quadratic has no real roots); substituting into the determinant and expanding gives $-105$.

   Concept and Intuition

   When a cubic is asked to have "a real root," it's often meant to be found by the rational root theorem (testing small divisors of the constant term) — here the constant is $-42$, and $2$ divides it. Once found, factor it out and check the resulting quadratic's discriminant to confirm no other real roots exist, so $\alpha$ is unambiguous.

   Step-by-Step Solution

   1. Test $x=2$ in $x^3+6x^2+5x-42$: $8+24+10-42=0$. So $x=2$ is a root.
   2. Factor: $x^3+6x^2+5x-42=(x-2)(x^2+8x+21)$.
   3. Discriminant of $x^2+8x+21$ is $64-84=-20<0$, so this factor has no real roots — confirming $\alpha=2$ is the unique real root.
   4. Substitute $\alpha=2$ into the matrix:

   $\begin{bmatrix}1&3&4\\0&5&-1\\6&-2&7\end{bmatrix}$

   5. Expand along row 1: …

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9. 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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10. 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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11. 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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12. 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. $f(-x)=2(-x-1)(-x-2)(-x-3)=2\cdot(-1)^3(x+1)(x+2)(x+3)=-2(x+1)(x+2)(x+3)$. …

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