Question

Exercitiul 16 C 17,18

Answer 1

$16.~A=\left(\begin{array}{ccc}3&-4\\-2&3\end{array}\right);~~B=\left(\begin{array}{ccc}-1&x\\1&-1\end{array}\right);~~x\in\mathbb{R}\\\\c)~det(B+(detB)\cdot A)=0;~~~~~~~x=?\\\\detB=\left|\begin{array}{ccc}-1&x\\1&-1\end{array}\right|=(-1)\cdot (-1)-x\cdot 1=1-x\\\\(detB)\cdot A=(1-x) \cdot \left(\begin{array}{ccc}3&-4\\-2&3\end{array}\right)=\left(\begin{array}{ccc}(1-x)\cdot3&(1-x)\cdot (-4)\\(1-x)\cdot(-2)&(1-x)\cdot3\end{array}\right)=\\\\=\left(\begin{array}{ccc}3-3x&-4+4x\\-2+2x&3-3x\end{array}\right)$

$B+(detB)\cdot A=\left(\begin{array}{ccc}-1&x\\1&-1\end{array}\right)+\left(\begin{array}{ccc}3-3x&-4+4x\\-2+2x&3-3x\end{array}\right)=\\\\=\left(\begin{array}{ccc}-1+3-3x&x-4+4x\\1-2+2x&-1+3-3x\end{array}\right)=\left(\begin{array}{ccc}2-3x&5x-4\\-1+2x&2-3x\end{array}\right)$

$det(B+(detB)\cdot A)=\left|\begin{array}{ccc}2-3x&5x-4\\-1+2x&2-3x\end{array}\right|=\\\\=(2-3x)(2-3x)-(5x-4)(-1+2x)=\\\\=4-6x-6x+9x^2+5x-10x^2-4+8x=-x^2+x\\\\det(B+(detB)\cdot A)=0\Rightarrow -x^2+x=0\Rightarrow x(-x+1)=0\Rightarrow \\\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow x=0\\\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow -x+1=0\Rightarrow -x=-1\Rightarrow x=1$

$17.~~A=\left(\begin{array}{ccc}4&3\\1&3\end{array}\right);~~B=\left(\begin{array}{ccc}2x+1&x\\1&x+1\end{array}\right);~~x\in\mathbb{R}\\\\a)~detA=9\\\\detA=\left|\begin{array}{ccc}4&3\\1&3\end{array}\right|=4\cdot 3-3\cdot 1=12-3=9$

$b)~A+B(1) \cdot B(-1)=2B(0)\\\\B(1)=\left(\begin{array}{ccc}2 \cdot 1+1&1\\1&1+1\end{array}\right)=\left(\begin{array}{ccc}3&1\\1&2\end{array}\right)\\\\B(-1)=\left(\begin{array}{ccc}2 \cdot (-1)+1&-1\\1&-1+1\end{array}\right)=\left(\begin{array}{ccc}-1&-1\\1&0\end{array}\right)\\\\B(0)=\left(\begin{array}{ccc}2 \cdot 0+1&0\\1&0+1\end{array}\right)=\left(\begin{array}{ccc}1&0\\1&1\end{array}\right)$

$2B(0)=2 \cdot \left(\begin{array}{ccc}1&0\\1&1\end{array}\right)=\left(\begin{array}{ccc}2\cdot1&2\cdot0\\2\cdot1&2\cdot1\end{array}\right)=\left(\begin{array}{ccc}2&0\\2&2\end{array}\right)$

$B(1) \cdot B(-1)=\left(\begin{array}{ccc}3&1\\1&2\end{array}\right) \cdot \left(\begin{array}{ccc}-1&-1\\1&0\end{array}\right)=\\\\=\left(\begin{array}{ccc}3\cdot(-1)+1\cdot1&3\cdot(-1)+1\cdot0\\1\cdot(-1)+2\cdot1&1\cdot(-1)+2\cdot0\end{array}\right)=\left(\begin{array}{ccc}-2&-3\\1&-1\end{array}\right)$

$A+B(1)\cdot B(-1)=\left(\begin{array}{ccc}4&3\\1&3\end{array}\right)+\left(\begin{array}{ccc}-2&-3\\1&-1\end{array}\right)=\\\\=\left(\begin{array}{ccc}4+(-2)&3+(-3)\\1+1&3+(-1)\end{array}\right)=\left(\begin{array}{ccc}2&0\\2&2\end{array}\right)=2B(0)$

$c)~B(1)+B(2)+...+B(9)=9B(x);~~~~x=?\\\\ 9B(x)=9 \cdot \left(\begin{array}{ccc}2x+1&x\\1&x+1\end{array}\right)=\left(\begin{array}{ccc}9(2x+1)&9x\\9\cdot1&9(x+1)\end{array}\right)=\\\\=\left(\begin{array}{ccc}18x+9&9x\\9&9x+9\end{array}\right)$

$B(1)+B(2)+...+B(9)=\\\\=\left(\begin{array}{ccc}2 \cdot 1+1&1\\1&1+1\end{array}\right)+\left(\begin{array}{ccc}2\cdot2+1&2\\1&2+1\end{array}\right)+...\left(\begin{array}{ccc}2\cdot9+1&9\\1&9+1\end{array}\right)=\\\\=\left(\begin{array}{ccc}(2\cdot1+1)+(2\cdot2+1)+...(2\cdot9+1)&1+2+...+9\\1+1+1+1+1+1+1+1+1&(1+1)+(2+1)+...(9+1)\end{array}\right)=\\\\=\left(\begin{array}{ccc}2(1+2+...+9)+9&1+2+...+9\\9&(1+2+...+9)+9\end{array}\right)=$

$=\left(\begin{array}{ccc}\not2 \cdot 9 \cdot \cfrac{9+1}{\not2}+9 &9\cdot\cfrac{9+1}{2} \\9&9\cdot\cfrac{9+1}{2}+9 \end{array}\right)=\left(\begin{array}{ccc}9 \cdot 10+9&\cfrac{9 \cdot 10}{2} \\9&\cfrac{9 \cdot 10}{2}+9 \end{array}\right)=\\\\=\left(\begin{array}{ccc}90+9&\cfrac{90}{2} \\9&\cfrac{90}{2}+9 \end{array}\right)=\left(\begin{array}{ccc}99&45\\9&54\end{array}\right)$

$B(1)+B(2)+...+B(9)=9B(x)\Rightarrow \\\\\Rightarrow \left(\begin{array}{ccc}99&45\\9&54\end{array}\right)=\left(\begin{array}{ccc}18x+9&9x\\9&9x+9\end{array}\right)\Rightarrow \\\\ \Rightarrow 99=18x+9 \Rightarrow 18x=90 \Rightarrow x=\cfrac{90}{18} \Rightarrow x=5$

$18.~~A=\left(\begin{array}{ccc}3&-6\\2&-3\end{array}\right);~~I_2=\left(\begin{array}{ccc}1&0\\0&1\end{array}\right);~~B(a)=\left(\begin{array}{ccc}0&a-2\\1&3a\end{array}\right);~~x\in\mathbb{R}\\\\a)~detA=3\\\\detA=\left|\begin{array}{ccc}3&-6\\2&-3\end{array}\right|=3 \cdot (-3)-(-6)\cdot 2=-9+12=3$