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

2×2 Matrices Satisfying A²-4A+3I=0, Trace 4: 19 | JEE 2026

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

Question
Let S={A=[abcd]:a,b,c,d∈{0,1,2,3,4} and A2−4A+3I=O}S=\left\{A=\begin{bmatrix}a&b\\c&d\end{bmatrix}:a,b,c,d\in\{0,1,2,3,4\}\ \text{and}\ A^2-4A+3I=O\right\} be a set of 2×22\times2 matrices. Then the number of matrices in SS, for which the sum of the diagonal elements is 44, is
A2020
B1717
C2121
D1919correct
Solution
Step 1: A2−4A+3I=OA^2-4A+3I=O means the characteristic equation is λ2−4λ+3=0\lambda^2-4\lambda+3=0, so tr(A)=a+d=4\mathrm{tr}(A)=a+d=4 and det⁡(A)=ad−bc=3\det(A)=ad-bc=3. Step 2: Go through (a,d)(a,d) with a+d=4a+d=4 (and ad−bc=3ad-bc=3): - a=0,d=4a=0,d=4: ad=0⇒bc=−3ad=0\Rightarrow bc=-3 — impossible (b,c≥0b,c\ge0). No solution. - a=1,d=3a=1,d=3: ad=3⇒bc=0ad=3\Rightarrow bc=0. Number of (b,c)(b,c) with bc=0bc=0 in {0,…,4}\{0,\ldots,4\}: 99 ways. - a=2,d=2a=2,d=2: ad=4⇒bc=1⇒(b,c)=(1,1)ad=4\Rightarrow bc=1\Rightarrow(b,c)=(1,1): 11 way. - a=3,d=1a=3,d=1: ad=3⇒bc=0ad=3\Rightarrow bc=0: 99 ways. - a=4,d=0a=4,d=0: bc=−3bc=-3 — impossible. No solution. Step 3: Total =9+1+9=19=9+1+9=19. Correct answer: (4)
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