Răspuns :
[tex]A(a)=\left(\begin{array}{ccc}1 & 2 & -1 \\ -2 & -3 & 0 \\ 2 & 4 & a\end{array}\right)[/tex]
a)
Calculam det(A(a)), adaugam primele doua linii ale determinantului si obtinem:
[tex]det(A(a))=\left|\begin{array}{ccc}1 & 2 & -1 \\ -2 & -3 & 0 \\ 2 & 4 & a\end{array}\right|[/tex]
1 2 -1
-2 -3 0
det(A(a))=(-3a+8+0)-(6+0-4a)=-3a+8-6+4a=a+2
b)
a=0
Inversa matricei A(a)
[tex]A(a)^{-1}=\frac{1}{detA}\cdot A^*[/tex]
Transpusa matricei
[tex]A(0)^t=\left(\begin{array}{ccc}1 & -2 & 2 \\ 2 & -3 &4 \\ -1 & 0 & 0\end{array}\right)[/tex]
[tex]A(0)^*=\left(\begin{array}{ccc}0& -4 & -3 \\ 0 & 2 & 2 \\ -2 & 0 & 1\end{array}\right)[/tex]
det(A(0))=0+2=2
[tex]A(0)^{-1}=\left(\begin{array}{ccc}0& -2 & -\frac{3}{2} \\ 0 & 1 & 1 \\ -1 & 0 & \frac{1}{2} \end{array}\right)[/tex]
c)
a≠-2
det(A(a))≠0⇒ metoda lui Cramer
Determinantului sistemului
Δ=a+2
Inlocuim coloana coeficientilor lui x cu coloana termenilor liberi
[tex]\Delta_x=\left|\begin{array}{ccc}-1 & 2 & -1 \\ 1 & -3 & 0 \\ -2 & 4 & a\end{array}\right|[/tex]
-1 2 -1
1 -3 0
[tex]\Delta_x=3a-4+0-(-6+0+2a)=a+2\\\\x=\frac{\Delta_x}{\Delta} =1[/tex]
Inlocuim coloana coeficientilor lui y cu coloana termenilor liberi
[tex]\Delta_y=\left|\begin{array}{ccc}1 & -1 & -1 \\ -2 & 1 & 0 \\ 2 &-2 & a\end{array}\right|[/tex]
1 -1 -1
-2 1 0
[tex]\Delta_y=a-4+0-(-2+0+2a)=-a-2=-(a+2)\\\\y=\frac{\Dellta_y}{\Delta}=-1[/tex]
Inlocuim coloana coeficientilor lui z cu coloana termenilor liberi
[tex]\Delta_z=\left|\begin{array}{ccc}1 & 2 & -1 \\ -2 & -3 & 1 \\ 2 & 4 & -2\end{array}\right|[/tex]
1 2 -1
-2 -3 1
[tex]\Delta_z=6+8+4-(6+4+8)=0\\\\z=\frac{\Delta_z}{\Delta}=0[/tex]
Un alt exercitiu cu matrice gasesti aici: https://brainly.ro/tema/9918940
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