Explicație pas cu pas:
a) det(A) =
[tex]\left|\begin{array}{ccc}3&6\\1&2\end{array}\right| = 3 \times 2 - 6 \times 1 = 6 - 6 = 0[/tex]
→ det(A) = 0
b)
[tex]A \times A = \left(\begin{array}{ccc}3&6\\1&2\end{array}\right) \times \left(\begin{array}{ccc}3&6\\1&2\end{array}\right) =[/tex]
[tex]= \left(\begin{array}{ccc}3 \times 3 + 6 \times 1&3 \times 6 + 6 \times 2\\1 \times 3 + 2 \times 1&1 \times 6 + 2 \times 2\end{array}\right)[/tex]
[tex]= \left(\begin{array}{ccc}9 + 6&18 + 12\\3 + 2&6 + 4\end{array}\right)[/tex]
[tex]= \left(\begin{array}{ccc}15&30\\5&10\end{array}\right)[/tex]
[tex]= \left(\begin{array}{ccc}5 \times 3&5 \times 6\\5 \times 1&5 \times 2\end{array}\right)[/tex]
[tex]= 5 \times \left(\begin{array}{ccc}3&6\\1&2\end{array}\right) = 5 \times A[/tex]
→ A•A = 5•A
c)
[tex]x \times A + (1 - x)I = [/tex]
[tex]= x \times \left(\begin{array}{ccc}3&6\\1&2\end{array}\right) +(1 - x) \left(\begin{array}{ccc}1&0\\0&1\end{array}\right)[/tex]
[tex] = \left(\begin{array}{ccc}3x&6x\\x&2x\end{array}\right) + \left(\begin{array}{ccc}1 - x&0\\0&1 - x\end{array}\right)[/tex]
[tex]= \left(\begin{array}{ccc}3x + 1 - x&6x\\x&2x + 1 - x\end{array}\right) [/tex]
[tex]= \left(\begin{array}{ccc}2x + 1&6x\\x&x + 1\end{array}\right)[/tex]
[tex]det\left( x \times A + (1 - x)I\right) =[/tex]
[tex] = (2x + 1)(x + 1) - 6x \times x [/tex]
[tex]= 2 {x}^{2} + 2x + x + 1 - 6 {x}^{2} [/tex]
[tex]= - 4 {x}^{2} + 3x + 1[/tex]
[tex] = - (4x + 1)(x - 1)[/tex]
[tex]det\left( x \times A + (1 - x)I\right) = 0[/tex]
[tex]= > - (4x + 1)(x - 1) = 0[/tex]
[tex]4x + 1 = 0 = > x = - \frac{1}{4} \\ [/tex]
[tex]x - 1 = 0 = > x = 1[/tex]
