The total number of 3 × 3 matrices A having enteries from the set (0, 1, 2, 3) such that the sum of all the diagonal entries of A A ^ T is 9 , is equal to

Correct answer: 0 Let A = [ cases a & b & c d & e & f g & h & i cases ], A ^ T = [ cases a & d & g b & e & h c & f & i cases ] Diagonal elements of A A ^ T is a ^ 2 + b ^ 2 + c ^ 2 , d ^ 2 + e ^ 2 + f ^ 2 , g ^ 2 + b ^ 2 + c ^ 2 . Sum = a ^ 2 + b ^ 2 + c ^ 2 + d ^ 2 + e ^ 2 + f ^ 2 + g ^ 2 + h ^ 2 + i ^ 2 = 9 a, b, c, d, e, f, g, h, i ∈ \ 0, 1, 2, 3\ Case No. of Matrices (1) All – 1 ^ ' s (9 !)/(9 !) = 1 (2) One → 3 ^ ' s , remaining- 0 ^ ' s (9 !)/(1 ! × 8 !) = 9 (3) One- 2 ^ ' s Five- 1 ^ ' s Three- 0 ^ ' s (9 !)/(1 ! × 5 ! × 3 !) = 8 × 63 (4) Two - 2 ' s One- 1 ^ ' s Six- 0 ' s (9 !)/(2 ! × 6 !) = 63 × 4 Then, the total no. of ways = 1 + 9 + 8 × 63 + 63 × 4 = 766

JEE Main 202116 MarchMorning Shift

Exam: JEE Main
Year: 2021
Date: 16 March
Shift: Morning Shift
Subject: Mathematics
Chapter: Matrices

JEE Main 2021 · 16 March · Morning Shift · Mathematics · Matrices

The total number of 3 × 3 matrices A having enteries from the set...

The total number of

3 \times 3

matrices

A

having enteries from the set

\left(0, 1, 2, 3\right)

such that the sum of all the diagonal entries of

A{A}^{T}

9

, is equal to

Correct answer

Step-by-step solution

Let $A = \left[$ \begin{cases}a & b & c \\ d & e & f \\ g & h & i\end{cases} $\right], {A}^{T} = \left[$ \begin{cases}a & d & g \\ b & e & h \\ c & f & i\end{cases} $\right]$

Diagonal elements of $A{A}^{T}\text{is}{a}^{2} + {b}^{2} + {c}^{2}, {d}^{2} + {e}^{2} + {f}^{2}, {g}^{2} + {b}^{2} + {c}^{2} .$

Sum $= {a}^{2} + {b}^{2} + {c}^{2} + {d}^{2} + {e}^{2} + {f}^{2} + {g}^{2} + {h}^{2} + {i}^{2} = 9$

$a, b, c, d, e, f, g, h, i \in \left\{0, 1, 2, 3\right\}$

	Case	No. of Matrices
(1)	All – ${1}^{'}s$	$\dfrac{9 !}{9 !} = 1$
(2)	One $\to {3}^{'}s$ , remaining- ${0}^{'}s$	$\dfrac{9 !}{1 ! \times 8 !} = 9$
(3)	One- ${2}^{'}s$ Five- ${1}^{'}s$ Three- ${0}^{'}s$	$\dfrac{9 !}{1 ! \times 5 ! \times 3 !} = 8 \times 63$
(4)	Two - $2 ' s$ One- ${1}^{'}s$ Six- $0 ' s$	$\dfrac{9 !}{2 ! \times 6 !} = 63 \times 4$

Then, the total no. of ways $= 1 + 9 + 8 \times 63 + 63 \times 4$

$= 766$

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