Question

Subiecul II, problema 1

Answer 1

[tex]\displaystyle A= \left(\begin{array}{ccc}1&0&-1\\0&1&0\\-1&0&1\end{array}\right) \\ \\ a).det~A=? \\ \\ det~A= \left|\begin{array}{ccc}1&0&-1\\0&1&0\\-1&0&1\end{array}\right|= \\ \\ =1 \cdot 1 \cdot 1+(-1) \cdot 0 \cdot 0+0 \cdot 0 \cdot (-1)-(-1) \cdot1 \cdot (-1)-0 \cdot 0 \cdot 1-1 \cdot 0 \cdot 0= \\ \\ =1+0+0-1-0-0=0 \Rightarrow det~A=0[/tex]

[tex]\displaystyle b).A^3-3A^2+2A=O_3 \\ \\ A^2=A \cdot A= \left(\begin{array}{ccc}1&0&-1\\0&1&0\\-1&0&1\end{array}\right) \cdot \left(\begin{array}{ccc}1&0&-1\\0&1&0\\-1&0&1\end{array}\right) = \left(\begin{array}{ccc}2&0&-2\\0&1&0\\-2&0&2\end{array}\right) [/tex]

[tex]3A^2=3 \cdot \left(\begin{array}{ccc}2&0&-2\\0&1&0\\-2&0&2\end{array}\right)= \left(\begin{array}{ccc}3 \cdot 2&3 \cdot 0&3 \cdot (-2)\\3 \cdot 0&3 \cdot 1&3 \cdot 0\\3 \cdot (-2)&3 \cdot 0&3 \cdot 2\end{array}\right)= \\ \\ = \left(\begin{array}{ccc}6&0&-6\\0&3&0\\-6&0&6\end{array}\right)[/tex]

[tex]\displaystyle A^3=A^2 \cdot A= \left(\begin{array}{ccc}2&0&-2\\0&1&0\\-2&0&2\end{array}\right) \cdot \left(\begin{array}{ccc}1&0&-1\\0&1&0\\-1&0&1\end{array}\right)=\left(\begin{array}{ccc}4&0&-4\\0&1&0\\-4&0&4\end{array}\right)[/tex]

[tex]\displaystyle 2A=2 \cdot \left(\begin{array}{ccc}1&0&-1\\0&1&0\\-1&0&1\end{array}\right)= \left(\begin{array}{ccc}2 \cdot 1&2 \cdot 0&2 \cdot (-1)\\2 \cdot 0&2 \cdot 1&2 \cdot 0\\2 \cdot (-1)&2 \cdot 0&2 \cdot 1\end{array}\right)= \\ \\ = \left(\begin{array}{ccc}2&0&-2\\0&2&0\\-2&0&2\end{array}\right)[/tex]

[tex]\displaystyle A^3-3A^2+2A= \\ \\ = \left(\begin{array}{ccc}4&0&-4\\0&1&0\\-4&0&4\end{array}\right) -\left(\begin{array}{ccc}6&0&-6\\0&3&0\\-6&0&6\end{array}\right) +\left(\begin{array}{ccc}2&0&-2\\0&2&0\\-2&0&2\end{array}\right)= [/tex]

[tex]\displaystyle =\left(\begin{array}{ccc}4-6&0-0&(-4)-(-6)\\0-0&1-3&0-0\\(-4)-(-6)&0-0&4-6\end{array}\right)+\left(\begin{array}{ccc}2&0&-2\\0&2&0\\-2&0&2\end{array}\right)= \\ \\ = \left(\begin{array}{ccc}-2&0&2\\0&-2&0\\2&0&-2\end{array}\right)+ \left(\begin{array}{ccc}2&0&-2\\0&2&0\\-2&0&2\end{array}\right)= \\ \\ = \left(\begin{array}{ccc}(-2)+2&0+0&2+(-2)\\0+0&(-2)+2&0+0\\2+(-2)&0+0&(-2)+2\end{array}\right)= \left(\begin{array}{ccc}0&0&0\\0&0&0\\0&0&0\end{array}\right)=O_3 \Rightarrow [/tex]

$\Rightarrow A^3-3A^2+2A=O_3$

[tex]c).det(A-xI_3)=0 \\ \\ xI_3=x \cdot \left(\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right)= \left(\begin{array}{ccc}x&0&0\\0&x&0\\0&0&x\end{array}\right) \\ \\ A-xI_3= \left(\begin{array}{ccc}1&0&-1\\0&1&0\\-1&0&1\end{array}\right) -\left(\begin{array}{ccc}x&0&0\\0&x&0\\0&0&x\end{array}\right) = \\ \\ = \left(\begin{array}{ccc}1-x&0-0&(-1)-0\\0-0&1-x&0-0\\(-1)-0&0-0&1-x\end{array}\right)= \left(\begin{array}{ccc}1-x&0&-1\\0&1-x&0\\-1&0&1-x\end{array}\right)[/tex]

[tex]det(A-xI_3)= \left|\begin{array}{ccc}1-x&0&-1\\0&1-x&0\\-1&0&1-x\end{array}\right|= \\ \\ =(1-x) \cdot(1-x)\cdot(1-x)+(-1) \cdot 0 \cdot 0+0 \cdot 0 \cdot (-1)- \\ \\ -(-1) \cdot (1-x) \cdot (-1)-0 \cdot 0 \cdot (1-x)-(1-x) \cdot 0 \cdot 0= \\ \\ =(1-x)^3+0+0-(1-x)-0-0=(1-x)^3-1+x \\ \\ (1-x)^3-1+x=0 \\ \\ -x(x-1)(x-2)=0 \\ \\ x=0 \\ \\ x-1=0 \Rightarrow x=1 \\ \\ x-2=0 \Rightarrow x=2[/tex]