Matrices practice questions — Honors Math

$I_3$ has $1$s on the main diagonal and $0$s elsewhere.

$\mathrm{tr}(A)=2+3+5=10$.

$A^2=\left(\begin{array}{cc}1+0& 1+1\\ 0+0& 0+1\end{array}\right)=\left(\begin{array}{cc}1& 2\\ 0& 1\end{array}\right)$.

$A$ is $2\times 3$, so $A^T$ is $3\times 2$.

Matrix multiplication is not commutative in general. (Concrete example: $A=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$, $B=\left(\begin{array}{cc}0& 0\\ 1& 0\end{array}\right)$ — check $AB\mathrm{\ne }BA$.)

$(AB{)}_{11}=1\cdot 3+2\cdot 1=5$; $(AB{)}_{12}=1\cdot 0+2\cdot 4=8$; $(AB{)}_{21}=0\cdot 3+1\cdot 1=1$; $(AB{)}_{22}=0\cdot 0+1\cdot 4=4$. So $AB=\left(\begin{array}{cc}5& 8\\ 1& 4\end{array}\right)$.

Multiply each entry by $3$: $\left(\begin{array}{cc}6& -3\\ 0& 9\end{array}\right)$.

Add componentwise: $\left(\begin{array}{cc}1& 1\\ 8& 6\end{array}\right)$.

By induction, $A^n=\left(\begin{array}{cc}1& n\\ 0& 1\end{array}\right)$, so $(A^n{)}_{12}=n$.

We need $a_{12}=a_{21}$, i.e. $a=3$.