Let $H (λ) = λ 222 2 λ 22 22 λ 2 222 λ$ .

hard 5 marks Class 12 · Mathematics · Determinants

Answer

Determinant

= (λ + 6) (λ - 2)^{3}

; singular

λ = - 6, 2

;

det (adj H (2)) = 0

The correct answer is Determinant $= (λ + 6) (λ - 2)^{3}$ ; singular $λ = - 6, 2$ ; $det (adj H (2)) = 0$ — The determinant of the zero matrix is $0$ , so $det (adj H (2)) = 0$ .

Step-by-step solution

Observe that $H (λ)$ has the form $a I + b J$ , where $I$ is the $4 \times 4$ identity, $J$ is the matrix of all ones, $a = λ - 2$ , $b = 2$ .
Matrix $J$ has eigenvalue $4$ (with eigenvector $[1, 1, 1, 1]^{T}$ ) and eigenvalue $0$ (multiplicity $3$ ).
Therefore eigenvalues of $H (λ) = a I + b J$ are $a + 4 b = (λ - 2) + 8 = λ + 6$ (once) and $a$ repeated three times, i.e., $λ - 2$ (thrice).
The determinant equals the product of eigenvalues: $det H (λ) = (λ + 6) (λ - 2)^{3}$ .
The matrix is singular when its determinant is zero, so set each factor to zero: $λ + 6 = 0$ gives $λ = - 6$ ; $λ - 2 = 0$ gives $λ = 2$ . Hence $λ = - 6$ or $2$ .
For $λ = 2$ , $H (2)$ becomes a matrix with every entry equal to $2$ , i.e., rank $1$ . When a $4 \times 4$ matrix has rank $\leq 2$ , its adjoint is the zero matrix. Thus $adj H (2) = 0_{4 \times 4}$ .
The determinant of the zero matrix is $0$ , so $det (adj H (2)) = 0$ .

Common mistake: Treating the off‑diagonal constant as a separate variable instead of using eigenvalue insight, leading to an incorrect polynomial.

Concept tested: Determinant of matrix $aI+bJ$ and adjoint of singular matrix

Published 19 September 2026 · How this solution was prepared and checked

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Let H(λ)=​λ222​2λ22​22λ2​222λ​​.

Step-by-step solution

Let $H (λ) = λ 222 2 λ 22 22 λ 2 222 λ$ .