Question

Am pus 30 puncte (15 pt fiecare ), va rog sa rezolvati cu explicatii pentru ca sa inteleg metoda de rezolvare sa o pot aplica la alte exercitii asemanatoare si eu. Va mulumesc.

Answer 1

$a)det A= \left[\begin{array}{ccc}1&-5\\2&6 \end{array}\right] =1\cdot6-[(2)\cdot(-5)]=6+10=16$

$b) A=\left(\begin{array}{ccc}1&-5\\2&6 \end{array}\right) ;B=\left(\begin{array}{ccc}6&5\\-2&1 \end{array}\right) \\ \\ \left(\begin{array}{ccc}1&-5\\2&6 \end{array}\right)\cdotB\cdot\left(\begin{array}{ccc}6&5\\-2&1 \end{array}\right)= \\ \\= \left(\begin{array}{ccc} 1\cdot6-10 & 1\cdot5+1\cdot(-5) \\\\2\cdot6+6\cdot(-2)&2\cdot5+6\cdot1 \end{array}\right) =\left(\begin{array}{ccc}-4&0\\0&16 \end{array}\right) \\ \\ \left(\begin{array}{ccc}-4&0\\0&16 \end{array}\right) =a \cdot I_{2}$

$\left(\begin{array}{ccc}16&0\\0&16 \end{array}\right) =a\cdot \left(\begin{array}{ccc}1&0\\0&1 \end{array}\right)$

[tex]det A =\left[\begin{array}{ccc}16&0\\0&16 \end{array}\right] ; \left[\begin{array}{ccc}16&0\\0&16 \end{array}\right]= \left[\begin{array}{ccc}a&0\\0&a \end{array}\right] \\ \\[/tex]

$16\cdot16=a ^{2} \\ \\ a=16$

Answer 2

a)

[tex]\it det(A) =\ \ \begin{vmatrix} 1 \ \ \ -5 \\ \\ 2 \ \ \ \ \ \ 6\end{vmatrix} =1\cdot6 - 2\cdot(-5) = 6+10=16[/tex]

b)

$\it A\cdot B = aI_2 \ \ \ \ \ (*)$

[tex]\it A\cdot B = \ \ \ \begin{pmatrix} 1\ \ \ -5 \\ \\ 2 \ \ \ \ \ \ 6 \end{pmatrix} \cdot \begin{pmatrix} 6 \ \ \ \ \ \ 5 \\ \\ -2 \ \ \ \ \ 1 \end{pmatrix} = \begin{pmatrix} 16 \ \ \ \ \ \ 0 \\ \\ 0 \ \ \ \ \ 16 \end{pmatrix} = 16\cdot \begin{pmatrix} 1 \ \ \ \ \ \ 0 \\ \\ 0 \ \ \ \ \ 1 \end{pmatrix} = \\ \\ \\ = 16I_2\stackrel{(*)}{\Longrightarrow}a=16[/tex]