The number of 3 2 matrices A , which can be formed using the elements of the set -2,-1,0,1…

Question

The number of $3 	imes 2$ 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

Accepted Answer

Let matrix $A$ of order $3	imes 2$ is

$A=[[|\begin{array}{cc} a & b \ c & d \ e & f \end{array}]]|,a,b,c,d,e,f\in {-2,-1,0,1,2}$

${A}^{T}=[\begin{gathered} a \quad  c \quad  e \ b \quad  d \quad  f \end{gathered}]$

${A}^{T}A=[\begin{gathered} a \quad  c \quad  e \ b \quad  d \quad  f \end{gathered}][[|\begin{array}{cc} a & b \ c & d \ e & f \end{array}]]|=[\begin{gathered} {a}^{2}+{c}^{2}+{e}^{2} ab+cd+ef \ ab+cd+ef {b}^{2}+{d}^{2}+{f}^{2} \end{gathered}]$

sum of diagonal elements of ${A}^{T}A=5$

${a}^{2}+{c}^{2}+{e}^{2}+{b}^{2}+{d}^{2}+{f}^{2}=5$

possible values of squares is ${0,1,4}$

Case 1 : one element is 4,1 element is 1 and four elements are 0 .

one element is $4=2$ choices ${-2,2}$

one element is $1=2$ choices ${-1,1}$

4 element is zero $=1$ choice ${0}$

Selecting one element 4 out of 6 choice ${{a}^{2},{b}^{2},{c}^{2},{d}^{2},{e}^{2},{f}^{2}}$, and have 2 choice ${-2,2}$, total $={}^{6}{C}_{1}	imes 2=12$

Selecting one element which is 1 from remaining 5 element $={}^{5}{C}_{1}$, and have two choice ${-1,1}$, total $={}^{5}{C}_{1}	imes 2=10$

Selecting 4 element which is zero from remaining 4 elements $={}^{4}{C}_{4}=1$ total matrices for Case $1=12	imes 10	imes 1=120$

Case 2 : five elements are 1 and one element is 0 .

squares ${a}^{2},{b}^{2},{c}^{2},{d}^{2},{e}^{2},{f}^{2}\in {1,1,1,1,1,0}$

selection of 5 one from ${a}^{2},{b}^{2},{c}^{2},{d}^{2},{e}^{2},{f}^{2}={}^{6}{C}_{5}$

and for 1 there are two choice for each elements $a,b,c,d,e,f={-1,1}$, total $={2}^{5}$

for 0 only 1 choices

total matrices of case $2={}^{6}{C}_{5}	imes {2}^{5}=192$

total number of such matrices = Case $1+$ Case 2

$=120+192=312$

Solutions · 1