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Mathematics · Determinants

JEE Advanced 2020 — Paper 2 — Question 37

The trace of a square matrix is defined to be the sum of its diagonal entries. If A is a 2×22 \times 2 matrix such that the trace of AA is 3 and the trace of A3A^{3} is -18 , then the value of the determinant of AA is ____\_\_\_\_

Answer: 5

Numerical answer — enter this value.

Step-by-step solution

 Let A=[abcd],a+d=3\text { Let } A=\left[\begin{array}{ll} a & b\\ c & d \end{array}\right], a+d=3 ⇒A2=[abcd][abcd]=[a2+bcab+bdac+cdbc+d2]\Rightarrow \quad A^{2}=\left[\begin{array}{ll} a & b\\ c & d \end{array}\right]\left[\begin{array}{ll} a & b\\ c & d \end{array}\right]=\left[\begin{array}{ll} a^{2}+b c & a b+b d\\ a c+c d & b c+d^{2} \end{array}\right] ⇒A3=[a3+2abc+bcda2b+abd+b2c+bd2a2c+acd+bc2+cd2abc+2bcd+d3]\Rightarrow \quad A^{3}=\left[\begin{array}{cc} a^{3}+2 a b c+b c d & a^{2} b+a b d+b^{2} c+b d^{2}\\ a^{2} c+a c d+b c^{2}+c d^{2} & a b c+2 b c d+d^{3} \end{array}\right] ⇒a3+d3+3abc+3bcd=−18\Rightarrow \quad \mathrm{a}^{3}+\mathrm{d}^{3}+3 \mathrm{abc}+3 \mathrm{bcd}=-18 ⇒(a+d)(a2+d2−ad)+3bc(a+d)=−18\Rightarrow \quad(\mathrm{a}+\mathrm{d})\left(\mathrm{a}^{2}+\mathrm{d}^{2}-\mathrm{ad}\right)+3 \mathrm{bc}(\mathrm{a}+\mathrm{d})=-18 ⇒(a+d)((a+d)2−3ad)+3bc(a+d)=−18\Rightarrow \quad(a+d)\left((a+d)^{2}-3 a d\right)+3 b c(a+d)=-18 ⇒3(9−3ad)+9bc=−18\Rightarrow \quad 3(9-3 \mathrm{ad})+9 \mathrm{bc}=-18 ⇒ad−bc=5\Rightarrow \quad \mathrm{ad}-\mathrm{bc}=5 ⇒∣A∣=5\Rightarrow \quad|\mathrm{A}|=5

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Exam
JEE Advanced 2020
Paper
Paper 2
Subject
Mathematics
Chapter
Determinants
Topic
Determinants