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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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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