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The number of $3 \times 3$ matrices $A,$ with entries from the set $\{-1,0,1\}$ such that the sum of the diagonal elements of $AA^{T}$ is $3$, is __ .

What I tried:

Let $\displaystyle A=\begin{pmatrix} a& b & c\\ d& e & f \\ g& h & i \end{pmatrix}$ and $\displaystyle A^{T}=\begin{pmatrix} a& d & g \\ b& e & h \\ c & f & i \end{pmatrix}$

$$AA^{T}=\begin{pmatrix} \sum a^2 & .. & ..\\ .. & \sum d^2 &.. \\ ..& ...& \sum g^2 \end{pmatrix}$$

Given $a^2+b^2+c^2+d^2+e^2+f^2+g^2+h^2+i^2=3$

given $a,b,c,d,e,f,g,h,i\in \{-1,0,1\}$

Out of $ 9 ,$ any three is $1$ and rest all are zero(like $1,1,1,0,0,0,0,0,0$)

$$\binom{9}{3}\cdot \frac{9!}{6!\cdot 3!}=$$

but answer is different. How do I solve it? Help me, please.

Rodrigo de Azevedo
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jacky
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2 Answers2

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You are very close.

First choose three of the slots out of the $9$ to set the corresponding square, $x^2$ to $1$. Each each of the location, we have two choices.

Hence $$2^3 \cdot \binom{9}{3}$$

Siong Thye Goh
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For the first part you could have relied on the fact that $tr(AA^T)=\|A\|_F^2=\sum_i\sum_ja_{ij}^2.$

Out of the 9 elements of $A$, exactly 3 must be nonzero, and their sign can be positive or negative, so $$ 2^3 {9\choose 3}. $$

cangrejo
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