Matrices & DeterminantsmediumPYQ · JEE Main · 4 Apr 2026 · Shift 1 (Morning)Free

adj(adj(2(adjA)⁻¹)) Element Sum = -3 | JEE Main 2026

JEE Maths question with a full step-by-step solution.

Question
Let A=[112−201135]A=\begin{bmatrix}1&1&2\\-2&0&1\\1&3&5\end{bmatrix}. Then the sum of all elements of the matrix adj(adj(2(adj A)−1))\mathrm{adj}\big(\mathrm{adj}\big(2(\mathrm{adj}\,A)^{-1}\big)\big) is equal to
A33
B44
C−4-4
D−3-3correct
Solution
Step 1: Compute ∣A∣|A|:
∣A∣=1(0−3)−1(−10−1)+2(−6−0)=−3+11−12=−4.|A|=1(0-3)-1(-10-1)+2(-6-0)=-3+11-12=-4.
 Step 2: Since (adj A)−1=A∣A∣(\mathrm{adj}\,A)^{-1}=\dfrac{A}{|A|}, let
B=2(adj A)−1=2A∣A∣=2A−4=−A2.B=2(\mathrm{adj}\,A)^{-1}=\frac{2A}{|A|}=\frac{2A}{-4}=-\frac{A}{2}.
 Step 3: For a 3×33\times3 matrix, ∣B∣=(−12)3∣A∣=−18(−4)=12|B|=\left(-\dfrac12\right)^3|A|=-\dfrac18(-4)=\dfrac12. Step 4: Use adj(adj B)=∣B∣ n−2B=∣B∣ B\mathrm{adj}(\mathrm{adj}\,B)=|B|^{\,n-2}B=|B|\,B (with n=3n=3):
adj(adj B)=12(−A2)=−A4.\mathrm{adj}(\mathrm{adj}\,B)=\frac12\left(-\frac{A}{2}\right)=-\frac{A}{4}.
 Step 5: Sum of elements of A=1+1+2−2+0+1+1+3+5=12A=1+1+2-2+0+1+1+3+5=12, so sum of elements of −A4=−124=−3-\dfrac{A}{4}=-\dfrac{12}{4}=-3. Correct answer: (4)
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