Răspuns:
Explicație pas cu pas:
[tex]A=\left(\begin{array}{ccc}2&1\\1&2\\\end{array}\right)[/tex]
[tex]A^{2} =\left(\begin{array}{ccc}2&1\\1&2\\\end{array}\right)\cdot \left(\begin{array}{ccc}2&1\\1&2\\\end{array}\right)=\left(\begin{array}{ccc}5&4\\4&5\\\end{array}\right)[/tex]
[tex]A^{2} -4A=\left(\begin{array}{ccc}5&4\\4&5\\\end{array}\right)-\left(\begin{array}{ccc}8&4\\4&8\\\end{array}\right)=\left(\begin{array}{ccc}-3&0\\0&-3\\\end{array}\right)=-3\left(\begin{array}{ccc}1&0\\0&1\\\end{array}\right)=-3I_{2}[/tex]
[tex]A+x\cdot I_{2}=\left(\begin{array}{ccc}2&1\\1&2\\\end{array}\right)+\left(\begin{array}{ccc}x&0\\0&x\\\end{array}\right)=\left(\begin{array}{ccc}2+x&1\\1&2+x\\\end{array}\right)[/tex]
[tex]det(A+x\cdot I_{2})=\left|\begin{array}{ccc}2+x&1\\1&2+x\\\end{array}\right|=(2+x)^{2} -1[/tex]
(2+x)²-1=0
4+4x+x²-1=0
x²+4x+3=0
Δ=16-12=4
[tex]x_{1}=\frac{-4+2}{2} =-1\\\\x_{2}=\frac{-4-2}{2} =-3[/tex]
[tex]A^{-1} =\frac{1}{detA} \cdot A^{*}[/tex]
detA=4-1=3
[tex]A^{t} = \left(\begin{array}{ccc}2&1\\1&2\\\end{array}\right)[/tex]
[tex]A^{*} =\left(\begin{array}{ccc}2&-1\\-1&2\\\end{array}\right)[/tex]
[tex]A^{-1} =\frac{1}{3} \cdot \left(\begin{array}{ccc}2&-1\\-1&2\\\end{array}\right)[/tex]
