Question

Se considera matricea A=(2 -2 1 -1).... a)calculati det A
b) determinati numerele reale p pentru care A•A=pA
c) determinati matricele B=(0 b b o) , stiind ca det de (A+B)=0 unde b este un numar real

Answer 1

[tex]\displaystyle A= \left(\begin{array}{ccc}2&-2\\1&-1\\\end{array}\right) \\ \\ a).det~A=? \\ \\ det~A= \left|\begin{array}{ccc}2&-2\\1&-1\\\end{array}\right|=2 \cdot (-1)-(-2) \cdot 1=-2+2=0 \Rightarrow det~A=0[/tex]

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

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

$\displaystyle c).B= \left(\begin{array}{ccc}0&b\\b&0\\\end{array}\right) ;~det(A+B)=0 \\ \\ A+B= \left(\begin{array}{ccc}2&-2\\1&-1\\\end{array}\right) + \left(\begin{array}{ccc}0&b\\b&0\\\end{array}\right) = \left(\begin{array}{ccc}2&-2+b\\1+b&-1\\\end{array}\right) \\ \\ det(A+B)= \left|\begin{array}{ccc}2&-2+b\\1+b&-1\\\end{array}\right|=2 \cdot (-1)-(-2+b)\cdot (1+b)= -b^2+b$

$\displaystyle det(A+B)=0 \Rightarrow -b^2+b=0 \Rightarrow b=0~sau~b=1 \Rightarrow \\ \\ \Rightarrow B= \left(\begin{array}{ccc}0&0\\0&0\\\end{array}\right)~sau~B= \left(\begin{array}{ccc}0&1\\1&0\\\end{array}\right)$