Mathematics · Matrices

JEE Main 2026 — 24 January, Morning Shift — Question 23

The number of $3 \times 2$ matrices $A$ , which can be formed using the elements of the set $\{-2,-1,0,1,2\}$ such that the sum of all the diagonal elements of $\mathrm{A}^{\mathrm{T}} \mathrm{A}$ is 5 , is $\_\_\_\_$

Answer: 312

Numerical answer — enter this value.

Step-by-step solution

A=\left(\begin{array}{cc} a_{1} & b_{1} \\ a_{2} & b_{2} \\ a_{3} & b_{3} \end{array}\right)_{3\times 2}

A^{T}A= \left(\begin{array}{ccc} a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \end{array}\right)_{2\times 3} \left(\begin{array}{cc} a_{1} & b_{1} \\ a_{2} & b_{2} \\ a_{3} & b_{3} \end{array}\right)_{3\times 2}

= \left(\begin{array}{cc} a_{1}^{2}+a_{2}^{2}+a_{3}^{2} & a_{1}b_{1}+a_{2}b_{2}+a_{3}b_{3} \\ a_{1}b_{1}+a_{2}b_{2}+a_{3}b_{3} & b_{1}^{2}+b_{2}^{2}+b_{3}^{2} \end{array}\right)

\operatorname{Tr}(A^{T}A) = a_{1}^{2}+a_{2}^{2}+a_{3}^{2} + b_{1}^{2}+b_{2}^{2}+b_{3}^{2} =5

\{2,1,0,0,0,0\},\quad \{2,-1,0,0,0,0\},\quad \{-2,1,0,0,0,0\},\quad \{-2,-1,0,0,0,0\},\quad \{1,1,1,1,1,0\}

\text{No. of ways} = \frac{6!}{4!}\times 4 + 2\times\frac{6!}{5!} + 2\times\frac{6!}{4!} + 2\times\frac{6!}{3!2!}

= \frac{6!}{3!} + 2\times 6 + 2\times 15 + 2\times\frac{6!}{3!}

=120+120+12+60=312

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 24 January, Morning Shift
Subject: Mathematics
Chapter: Matrices
Topic: Types of matrices & its properties