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.

COMEDK 2025Set 2025-A1 markMCQ
$a_{21}$
›Reveal solutionSolution
The cofactor of $a_{21}$ is found by taking $(-1)^{2+1}$ times the determinant of the submatrix obtained by deleting row 2 and column 1. The result is $-4$ , so the correct option is (C).

The cofactor of an element in a matrix is not just the minor (the determinant of the submatrix left after removing that element’s row and column). It also includes a sign factor $(-1)^{i+j}$ , where $i$ and $j$ are the row and column indices. This sign alternates like a chessboard pattern. For $a_{21}$ (row 2, column 1), the sign is negative because $2+1=3$ is odd. So we compute the minor and then flip its sign.

Identify the element and its position. The element $a_{21}$ is in row 2, column 1. In the given matrix

$\Delta = \begin{vmatrix} 1 & 4 & 4 \\ -3 & 5 & 9 \\ 2 & 1 & 2 \end{vmatrix},$
$a_{21} = -3$ . But the cofactor depends only on position, not on the value of the element itself.

Delete row 2 and column 1. Removing row 2 and column 1 leaves the submatrix:

$\begin{pmatrix} 4 & 4 \\ 1 & 2 \end{pmatrix}.$

Compute the minor $M_{21}$ . The minor is the determinant of that $2\times2$ submatrix:

$M_{21} = \begin{vmatrix} 4 & 4 \\ 1 & 2 \end{vmatrix} = (4)(2) - (4)(1) = 8 - 4 = 4.$

Apply the sign factor. …
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COMEDK 2024Set 2024-A1 markMCQ
$\left|\begin{array}{ccc} \cos (\alpha+\beta) & -\sin (\alpha+\beta) & \cos 2 \beta \\ \sin \alpha & \cos \alpha & \sin \beta \\ -\cos \alpha & \sin \alpha & \cos \beta \end{array}\right|$
›Reveal solutionSolution

Expanding along the first row collapses the determinant to $1+\cos 2\beta$ , which contains no $\alpha$ — so it is independent of $\alpha$ : option (D).

Cofactor expansion along row 1

The three minors are
$M_{11}=\begin{vmatrix}\cos\alpha&\sin\beta\\\sin\alpha&\cos\beta\end{vmatrix}=\cos\alpha\cos\beta-\sin\alpha\sin\beta=\cos(\alpha+\beta),$ $M_{12}=\begin{vmatrix}\sin\alpha&\sin\beta\\-\cos\alpha&\cos\beta\end{vmatrix}=\sin\alpha\cos\beta+\cos\alpha\sin\beta=\sin(\alpha+\beta),$ $M_{13}=\begin{vmatrix}\sin\alpha&\cos\alpha\\-\cos\alpha&\sin\alpha\end{vmatrix}=\sin^2\alpha+\cos^2\alpha=1.$
With the cofactor sign pattern $(+,-,+)$ and the row-1 entries $\cos(\alpha+\beta),\ -\sin(\alpha+\beta),\ \cos2\beta$ : …
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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

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

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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