Determinants, Adjoints, Inverses and Linear Systems (JEE Advanced)

Question 1

Let A be a 3×3 real matrix with det A=−2. The value of det(3 adj(2 adj(A⁻¹))) is ______. Exactly one option is correct.

Accepted Answer

C. For an invertible 3×3 matrix B, B adjB=(det B)I, so adjB=(det B)B⁻¹.
With B=A⁻¹: det(A⁻¹)=−1/2, hence adj(A⁻¹)=−1/2 A and 2 adj(A⁻¹)=−A.
Every cofactor of a 3×3 matrix is a 2×2 determinant, so adj(kM)=k²adjM. Thus adj(−A)=adjA.
Finally det(3 adjA)=3³det(adjA)=27(det A)²=27·4=108.
Trap: writing adj(−A)=−adjA gives −108.
Difficulty: Hard

Question 2

Let a,b,c be the three roots (real or complex) of t³−4t²+5t−7=0, taken in any order. The value of beginvmatrix a&b&c; b&c&a; c&a&bendvmatrix is ______. Exactly one option is correct.

Accepted Answer

B. Apply C₁→ C₁+C₂+C₃: every entry of the first column becomes a+b+c, so
D=(a+b+c)beginvmatrix1&b&c; 1&c&a; 1&a&bendvmatrix. Subtracting R₁ from R₂,R₃ and expanding,
D=(a+b+c)[(c−b)(b−c)−(a−c)(a−b)]=−(a+b+c)(a²+b²+c²−ab−bc−ca).
This is symmetric in a,b,c, so the order of the roots does not matter.
By Vieta, a+b+c=4, ab+bc+ca=5, so a²+b²+c²=16−10=6 and a²+b²+c²−(ab+bc+ca)=1.
Hence D=−4·1=−4.
Difficult

Question 3

Let A be a 3×3 real matrix with rows R₁,R₂,R₃ and det A=5. For a real number λ, let B be the 3×3 matrix whose rows are R₁+2R₂−R₃, 2R₁+R₂+R₃ and R₁−R₂+λ R₃ (in this order). The value of λ for which det

Accepted Answer

D. B=TA with T=beginbmatrix1&2&−1; 2&1&1; 1&−1&λendbmatrix, so det B=det T·det A.
det T=1(λ+1)−2(2λ−1)+(−1)(−2−1)=6−3λ.
det(2A)=2³det A=40. So 5(6−3λ)=40, giving 6−3λ=8, λ=−2/3.
Trap: using det(2A)=2det A gives λ=4/3.
Difficulty: Hard

Determinants, Adjoints, Inverses and Linear Systems: questions with solutions

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