If A=≤ft[{array}{ccc}4 & λ & -3 \\ 0 & 2 & 5 \\ 1 & 1 & 3{array}] then A^{-1} exists if :

Correct Answer is : λ≠−2λ ≠-2λ=−2

Correct Answer is : λ≠−2λ ≠-2λ=−2

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COMEDK 2025 Evening Solved Paper

Question : 170 of 180

Marks: +1, -0

A=\left[\begin{array}{ccc}4 & \lambda & -3 \\ 0 & 2 & 5 \\ 1 & 1 & 3\end{array}\right]

A^{-1}

exists if :

Solution:

Concept:

The inverse of a square matrix exists if and only if its determinant is non‑zero.

Explanation:

Compute

\det(A)

by expanding along the first row.

\det(A) = 4\begin{vmatrix}2&5\\1&3\end{vmatrix} - \lambda\begin{vmatrix}0&5\\1&3\end{vmatrix} + (-3)\begin{vmatrix}0&2\\1&1\end{vmatrix}

Evaluate each

2\times2

determinant:

\begin{vmatrix}2&5\\1&3\end{vmatrix} = (2)(3)-(5)(1)=6-5=1

\begin{vmatrix}0&5\\1&3\end{vmatrix} = (0)(3)-(5)(1)=0-5=-5

\begin{vmatrix}0&2\\1&1\end{vmatrix} = (0)(1)-(2)(1)=0-2=-2

Substitute:

4(1) - \lambda(-5) + (-3)(-2) = 4 + 5\lambda + 6 = 5\lambda + 10 = 5(\lambda+2)

For

A^{-1}

to exist,

\det(A) \neq 0

, i.e.,

5(\lambda+2) \neq 0

→

\lambda+2 \neq 0

→

\lambda \neq -2

Answer:

Option D:

\lambda \neq -2

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