Question

Ex S3 subpunctul a)
……..

Answer 1

$\displaystyle A=\left(\begin{array}{ccc}1&x&2\\2&-1&x\\m&x&3\end{array}\right)\\\\ Matricea~A~este~inversabila~daca~detA\not =0.\\\\ detA=\left|\begin{array}{ccc}1&x&2\\2&-1&x\\m&x&3\end{array}\right|=1 \cdot (-1) \cdot 3+2 \cdot 2 \cdot x+x \cdot x \cdot m-2 \cdot (-1) \cdot m-\\~~~~~~~~~~~~~~~~~~~~~~~~~~~-x \cdot 2 \cdot 3-1\cdot x \cdot x=-3+4x+mx^2+2m-6x-x^2=\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=mx^2-x^2-2x+2m-3$

$mx^2-x^2-2x+2m-3\not=0 \Rightarrow x^2\big(m-1\big)-2x+2m-3\not=0\\\\ \Delta=b^2-4ac \\\\ a=m-1,~b=-2,~c=2m-3\\\\ \Delta=(-2)^2-4 \cdot (m-1) \cdot (2m-3)=4-4\Big(2m^2-3m-2m+3\Big)=\\\\=4-4\Big(2m^2-5m+3\Big)=4-8m^2+20m-12=-8m^2+20m-8$

$\displaystyle 8m^2+20m-8 \not =0 \Rightarrow -4(2m-1)(m-2)\not = 0 \Rightarrow \\\\ \Rightarrow m\not = \frac{1}{2} ;~m \not = 2\\\\ m\in \mathbb{R}-\Bigg\{\frac{1}{2} ;2\Bigg\}$