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

If A = 2 λ -3 \0 2 5 \1 1 3 , then A⁻¹ exists if (A) λ = 2 (B) λ ≠ 2 (C) λ ≠ -2 (D) None of these

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A = \begin{pmatrix} 2 & \lambda & -3 \\ 0 & 2 & 5 \\ 1 & 1 & 3 \end{pmatrix}

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Concept understanding — Determinant Evaluation Using Identities

Determinant Evaluation Using Identities

Expanding a $4\times4$ or $5\times5$ determinant term by term is painful and error-prone. The smarter route is to transform the determinant into an easy form using properties (the "identities") that change its value in a known, controlled way — then read the answer off a triangular matrix.

The geometric intuition

A determinant measures the signed "volume" of the box spanned by the rows in $n$ -dimensional space. Sliding one row parallel to another doesn't change that volume; swapping two rows flips its sign; scaling a row scales the volume. The algebraic identities are just these facts translated into rules.

The three row (or column) operations

\det \to -\det

The identical rules hold for columns. There is also row-wise linearity: if a row is a sum $R_i = R_i' + R_i''$ , the determinant splits into the sum of two determinants with all other rows fixed.

Watch out

Row-wise linearity is not $\det(A+B)=\det A+\det B$ — that is false. The splitting works one row at a time.

The strategy

Use operation 3 to create zeros in a row or column (value unchanged).
Factor out common factors with operation 2.
Swap rows if needed to reach upper-triangular form (track the sign change).
The determinant is then the product of the diagonal entries.

Worked example

\det\begin{pmatrix}1&2&3\\4&5&6\\7&8&10\end{pmatrix}.

Apply $R_2\to R_2-4R_1$ and $R_3\to R_3-7R_1$ (no change), then $R_3\to R_3-2R_2$ : …

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$\det A = 5\lambda + 8$ , which vanishes only at $\lambda = -\tfrac{8}{5}$ ; so $A^{-1}$ exists for $\lambda \neq -\tfrac{8}{5}$ , matching none of (A)–(C). Correct option: (D).

$A^{-1}$ exists precisely when $\det A \neq 0$ . Expand along the first column (which contains a zero):

\det A = 2\begin{vmatrix} 2 & 5 \\1 & 3 \end{vmatrix} + 1\cdot\begin{vmatrix} \lambda & -3 \\2 & 5 \end{vmatrix} = 2(6-5) + (5\lambda + 6) = 5\lambda + 8.

Check by expanding along the second row:

$2\begin{vmatrix}2&-3\\1&3\end{vmatrix} - 5\begin{vmatrix}2&\lambda\\1&1\end{vmatrix} = 2(9) - 5(2-\lambda) = 5\lambda + 8,$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 18 on this concept.

KEAM 2024Set eng-2024-06064 marksMCQ
$A=\begin{pmatrix}2 & -1 & 1\\ -1 & 0 & 2\\ 1 & -2 & -1\end{pmatrix}$
›Reveal solutionSolution

For a $3\times3$ matrix, scaling by $1/|A|$ scales the determinant by $(1/|A|)^3$ . …

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KEAM 2026Set eng-2026-04184 marksMCQ
$f(x)=\begin{vmatrix}-1 & x & 3\\ 0 & 1 & 2x\\ 1 & -1 & 1\end{vmatrix}$
›Reveal solutionSolution

At $x=-1$ the determinant evaluates to $0$ .

At $x=-1$ the matrix is $\begin{pmatrix}-1&-1&3\\0&1&-2\\1&-1&1\end{pmatrix}$ . Expanding along the first column: …
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KEAM 2026Set eng-2026-04184 marksMCQ
$A$
›Reveal solutionSolution

$|A|=-1/7$ , and $|B|=|AB|/|A|=21/(-1/7)=-147$ .

Since $|A^{-1}|=\dfrac{1}{|A|}=-7$ , we have $|A|=-\dfrac17$ .

Using $|AB|=|A|\,|B|$ : …
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KEAM 2024Set eng-2024-06094 marksMCQ
$A=(a_{ij})$
›Reveal solutionSolution

Convert minors to cofactors with alternating signs, then expand: $|A|=-80$ .

Cofactors: $C_{11}=+M_{11}=-40$ , $C_{12}=-M_{12}=10$ , $C_{13}=+M_{13}=35$ .

Expanding along the first row: …
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KEAM 2025Set eng-2025-04284 marksMCQ
$\Delta = \begin{vmatrix} x & y & 1 \\ x+y & y+1 & x+1 \\ 1 & x & y \end{vmatrix}$
›Reveal solutionSolution
Expanding $\Delta$ and substituting $x+y=-1$ , every term reduces to $0$ because it carries a factor $x+y+1=0$ .

Expanding along the first row:
$\Delta=x\big[(y+1)y-(x+1)x\big]-y\big[(x+y)y-(x+1)\big]+\big[(x+y)x-(y+1)\big].$
With $x+y=-1$ :

First term: $x\big[(y^2-x^2)+(y-x)\big]=x(y-x)(x+y+1)=x(y-x)(0)=0$ . …
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KEAM 2023Set eng-2023-P2-B24 marksMCQ
$(2n+1)\times(2n+1)$
›Reveal solutionSolution
An odd-dimensional orthogonal matrix with det +1 always has eigenvalue 1, so det(A−I) = 0.

Concept and Intuition

AA^T = I means A is orthogonal, and positive integer determinant forces det(A) = +1. In odd dimension a real rotation must fix an axis (eigenvalue 1) because complex eigenvalues occur in conjugate pairs and the leftover real eigenvalue, together with det = +1, must be +1.

Step-by-Step Solution

AA^T = I ⇒ A orthogonal ⇒ eigenvalues have modulus 1 and det = ±1.

Integer positive det ⇒ det(A) = +1. …
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KEAM 2024Set eng-2024-06074 marksMCQ
$A=\begin{pmatrix}-7&3\\3&-1\end{pmatrix}$
›Reveal solutionSolution

$\det(A^5)=(\det A)^5$ .

$\det A=(-7)(-1)-(3)(3)=7-9=-2$ . Then $\det(A^5)=(\det A)^5=(-2)^5=-32$ …
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KEAM 2021Set eng-2021-P2-B14 marksMCQ
$\begin{vmatrix} x-1 & 2 & 1 \\ 2 & x-1 & 2 \\ 1 & x+2 & x-1 \end{vmatrix} = ax^3 + bx^2 + cx + d$
›Reveal solutionSolution
Setting $x=0$ gives the constant term $d=16$ .

Concept and Intuition

Writing the determinant as $ax^3+bx^2+cx+d$ , the constant $d$ equals the value at $x=0$ , since all $x$ -bearing terms vanish there.

Step-by-Step Solution

At $x=0$ the matrix is $\begin{bmatrix}-1 & 2 & 1\\ 2 & -1 & 2\\ 1 & 2 & -1\end{bmatrix}$ .

Expand: $-1((-1)(-1)-2\cdot2)-2(2\cdot(-1)-2\cdot1)+1(2\cdot2-(-1)\cdot1)$ . …
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KEAM 2025Set eng-2025-04294 marksMCQ
$\alpha + \beta + \gamma = 0$
›Reveal solutionSolution

Let $a=e^{\alpha},b=e^{\beta},c=e^{\gamma}$ , so $abc=e^{\alpha+\beta+\gamma}=1$ . Split the third column $a^3-1=a^3+(-1)$ ; the two resulting Vandermonde determinants are equal and cancel, giving $0$ .

Write the rows as $(a,a^2,a^3-1)$ etc. By column-linearity the determinant splits as $D_1-D_2$ where

$D_1=\begin{vmatrix}a&a^2&a^3\\\\ b&b^2&b^3\\\\ c&c^2&c^3\end{vmatrix}=abc\begin{vmatrix}1&a&a^2\\\\1&b&b^2\\\\1&c&c^2\end{vmatrix}=abc\\,V,$ …
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KEAM 2025Set eng-2025-04254 marksMCQ
$a_1, a_2, a_3, a_4, a_5$
›Reveal solutionSolution

Two columns are proportional, forcing the determinant to zero.

With $a_1 = 2, r = \tfrac12$ : the terms are $2, 1, \tfrac12, \tfrac14, \tfrac18, \tfrac{1}{16}$ .

In the determinant

$\begin{vmatrix} a_1 & a_2 & 1 \\ a_3 & a_4 & 1 \\ a_5 & a_6 & 1 \end{vmatrix},$ …
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KEAM 2026Set eng-2026-04224 marksMCQ
$\begin{vmatrix} \sin 30^\circ & \cos 30^\circ & \sin(30^\circ + 75^\circ) \\ \sin 45^\circ & \cos 45^\circ & \sin(45^\circ + 75^\circ) \\ \sin 60^\circ & \cos 60^\circ & \sin(60^\circ + 75^\circ) \end{vmatrix}$
›Reveal solutionSolution

The third column is a fixed linear combination of the first two, forcing a zero determinant.

Each entry of column 3 is $\sin(\theta+75^\circ)=\sin\theta\cos75^\circ+\cos\theta\sin75^\circ$ . …
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KEAM 2022Set eng-2022-P2-B14 marksMCQ
$A = \begin{bmatrix} 3 & 4 \\ 1 & -2 \end{bmatrix}$
›Reveal solutionSolution
$|B^T|=14$ .

Concept and Intuition

Determinants multiply: $|AB|=|A||B|$ , and a transpose has the same determinant, $|B^T|=|B|$ .

Step-by-Step Solution

$|A|=3(-2)-4(1)=-6-4=-10$ .

$|AB|=(-5)(-13)-(41)(5)=65-205=-140$ .

$|B|=\frac{|AB|}{|A|}=\frac{-140}{-10}=14$ .

$|B^T|=|B|=14$ .

Common Mistakes …
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←Previous questionQ32 — If f(x) = 0 x - a x - b x + a 0 x - c x + b x + c 0 , then (A) f(a) = …→Next questionQ34 — If A and B are invertible matrices, then which of the following is not…

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

Concept understanding — Determinant Evaluation Using Identities

Determinant Evaluation Using Identities

The geometric intuition

The three row (or column) operations

The strategy

Worked example

Method: Testing Invertibility via the Determinant

Steps

Common Mistakes

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Common Mistakes …

Q.If A=(2λ−3025113)A = \begin{pmatrix} 2 & \lambda & -3 \\ 0 & 2 & 5 \\ 1 & 1 & 3 \end{pmatrix}A=​201​λ21​−353​​, then A−1A^{-1}A−1 exists if (A) λ=2\lambda = 2λ=2 (B) λ≠2\lambda \neq 2λ=2 (C) λ≠−2\lambda \neq -2λ=−2 (D) None of these

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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

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