Mathematics · Matrices

JEE Main 2024 — 8 April, Shift 1 — Question 8

Let A=\left[ \begin{array}{*{35}{l}}2 & a & 0 \\1 & 3 & 1 \\0 & 5 & b \\\end{array} \right].If A3=4A2−A−21IA^{3}=4 A^{2}-A-21 I, where I is the identity matrix of order 3×33 \times 3, then 2a+3b2 a+3 b is equal to :

  1. Option A:

    −10-10

  2. Option B:

    −13-13

    Correct
  3. Option C:

    −9-9

  4. Option D:

    −12-12

Answer: B

Step-by-step solution

A3−4A2+A+21I=0A^{3}-4 A^{2}+A+21 I=0

tr⁡(A)=4\operatorname{tr}(A)=4 (from equation)

2+3+b=4⇒b=−12+3+b =4\Rightarrow b=-1

∣A∣=−21|\mathrm{A}|=-21 (from equation)

−16+a=−21⇒a=−5-16+a=-21 \Rightarrow a=-5

2a+3b=−13\boxed{2 a+3 b=-13}

Answer key and solution verified before publishing.

Practise Matrices

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Exam
JEE Main 2024
Subject
Mathematics
Chapter
Matrices
Topic
Characteristic Equation & roots,application of cayley - hamilton theorem
Let A= [ begin array 35 l 2 & a & 0 \\1 & 3 & 1 \\0 & 5 & b \\end… | JEE Main 2024 PYQ with Solution · DhiX AI