The number of 3 2 matrices A=pmatrixa&b&d&fpmatrix with a,b,c,d,e,f-2,-1,0,1,2, for which …

Question

The number of $3	imes 2$ matrices $A=\begin{pmatrix}a&b\c&d\e&f\end{pmatrix}$ with $a,b,c,d,e,f\in\{-2,-1,0,1,2\}$, for which the sum of all the diagonal elements of $A^TA$ is 5, is \_\_\_\_.

Answer: 312

Accepted Answer

$A^TA=\begin{pmatrix}a^2+c^2+e^2 & ab+cd+ef\ ab+cd+ef & b^2+d^2+f^2\end{pmatrix}$, so the trace condition is
$a^2+b^2+c^2+d^2+e^2+f^2=5.$

Each square lies in $\{0,1,4\}$ (values $\pm2	o4$, $\pm1	o1$, $0	o0$).

Case 1: one square is 4, one square is 1, and the remaining four squares are 0.
Choose which variable is $\pm2$: $6$ ways, with 2 sign choices $=12$; choose which of the remaining 5 variables is $\pm1$: $5$ ways, with 2 sign choices $=10$; the rest are forced to 0. Total $=12	imes10=120$.

Case 2: five squares equal 1 and one square equals 0.
Choose which variable is 0: $\binom{6}{5}=6$ ways to choose the five that are $\pm1$ (equivalently $\binom65$ for the zero), and each of the five nonzero variables has 2 sign choices: $6	imes2^5=192$.

Total number of matrices $=120+192=312$.

Solutions · 1