Răspuns :
[tex]A=\left(\begin{array}{lll}0 & 1 & 0 \\ 0 & 1 & 0 \\ 1 & 1 & 1\end{array}\right)[/tex]
a)
det(A+I₃)=4
Calculam det(A+I₃)
[tex]det(A+I_3)=A=\left|\begin{array}{lll}1 & 1 & 0 \\ 0 &2 & 0 \\ 1 & 1 & 2\end{array}\right|[/tex]
1 1 0
0 2 0
det(A+I₃)=(4+0+0)-(0+0+0)=4
b)
[tex]A\times A=\left(\begin{array}{lll}0 & 1 & 0 \\ 0 & 1 & 0 \\ 1 & 1 & 1\end{array}\right)\times \left(\begin{array}{lll}0 & 1 & 0 \\ 0 & 1 & 0 \\ 1 & 1 & 1\end{array}\right)=\left(\begin{array}{lll}0 & 1 & 0 \\ 0 & 1 & 0 \\ 1 & 3& 1\end{array}\right)[/tex]
[tex]A\times A\times A=\left(\begin{array}{lll}0 & 1 & 0 \\ 0 & 1 & 0 \\ 1 & 3& 1\end{array}\right) \times \left(\begin{array}{lll}0 & 1 & 0 \\ 0 & 1 & 0 \\ 1 & 1& 1\end{array}\right) =\left(\begin{array}{lll}0 & 1 & 0 \\ 0 & 1 & 0 \\ 1 & 5& 1\end{array}\right)[/tex]
[tex]A\times A\times A+A=\left(\begin{array}{lll}0 & 1 & 0 \\ 0 & 1 & 0 \\ 1 & 5& 1\end{array}\right) +\left(\begin{array}{lll}0 & 1 & 0 \\ 0 & 1 & 0 \\ 1 & 1& 1\end{array}\right) =\left(\begin{array}{lll}0 & 2 & 0 \\ 0 & 2 & 0 \\ 2 & 6& 2\end{array}\right) =2\left(\begin{array}{lll}0 & 1 & 0 \\ 0 & 1 & 0 \\ 1 & 3& 1\end{array}\right) =2A\times A[/tex]
c)
Daca B(x) este inversabila, atunci det(B(x))≠0
[tex]det(B(x))=\left|\begin{array}{ccc}x&1&0\\0&x+1&0\\1&1&x+1\end{array}\right|[/tex]
x 1 0
0 x+1 0
det(B(x))=[x(x+1)²+0+0]-0=x(x+1)²
x(x+1)²≠0
x≠0 si x≠-1
x∈R\{-1,0}
Un alt exercitiu similar de bac il gasesti aici: https://brainly.ro/tema/3845583
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