Question

Exercitiul
21 va rooog

Answer 1

$21.~~A=\left(\begin{array}{ccc}3&-2\\5&-3\end{array}\right),~I_2=\left(\begin{array}{ccc}1&0\\0&1\end{array}\right),~B(x)=A-xI_2,~x\in\mathbb{R}\\\\ a)~det(B(0))=1\\\\ B(0)=\left(\begin{array}{ccc}3&-2\\5&-3\end{array}\right)-0 \cdot \left(\begin{array}{ccc}1&0\\0&1\end{array}\right)=\left(\begin{array}{ccc}3&-2\\5&-3\end{array}\right)\\\\ det(B(0))=\left|\begin{array}{ccc}3&-2\\5&-3\end{array}\right|=3 \cdot (-3)-(-2)\cdot 5=-9+10=1$

$b)~A \cdot A+I_2=O_2,~O_2=\left(\begin{array}{ccc}0&0\\0&0\end{array}\right)\\\\ A \cdot A=\left(\begin{array}{ccc}3&-2\\5&-3\end{array}\right)\cdot \left(\begin{array}{ccc}3&-2\\5&-3\end{array}\right)=\\\\=\left(\begin{array}{ccc}3\cdot3+(-2)\cdot 5&3\cdot(-2)+(-2)\cdot(-3)\\5\cdot3+(-3)\cdot5&5\cdot(-2)+(-3)\cdot(-3)\end{array}\right)=\left(\begin{array}{ccc}9-10&-6+6\\15-15&-10+9\end{array}\right)=\\\\=\left(\begin{array}{ccc}-1&0\\0&-1\end{array}\right)$

$A \cdot A+I_2=\left(\begin{array}{ccc}-1&0\\0&-1\end{array}\right)+\left(\begin{array}{ccc}1&0\\0&1\end{array}\right)=\left(\begin{array}{ccc}-1+1&0+0\\0+0&-1+1\end{array}\right)=\left(\begin{array}{ccc}0&0\\0&0\end{array}\right)$

$c)~B(x)\geq 1~(\forall)~x\in\mathbb{R}\\\\ B(x)=A-xI_2\\\\ xI_2=x\left(\begin{array}{ccc}1&0\\0&1\end{array}\right)=\left(\begin{array}{ccc}x\cdot1&x\cdot0\\x\cdot0&x\cdot 1\end{array}\right)=\left(\begin{array}{ccc}x&0\\0&x\end{array}\right)$

$A-xI_2=\left(\begin{array}{ccc}3&-2\\5&-3\end{array}\right)-\left(\begin{array}{ccc}x&0\\0&x\end{array}\right)=\left(\begin{array}{ccc}3-x&-2\\5&-3-x\end{array}\right)=B(x)$

$detB(x)=\left|\begin{array}{ccc}3-x&-2\\5&-3-x\end{array}\right|=(3-x)(-3-x)-(-2)\cdot 5=\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=x^2-9+10=x^2+1\geq 1 ~(\forall)~x\in\mathbb{R}$