Question

Exercitiile 13 si 14 va rog muult!

Answer 1

$13.~~A(x)=\left(\begin{array}{ccc}x-1&x\\1&-1\end{array}\right);~~x\in \mathbb{R}\\\\a)~A(4)=\left(\begin{array}{ccc}4-1&4\\1&-1\end{array}\right)=\left(\begin{array}{ccc}3&4\\1&-1\end{array}\right)\\\\detA(4)=\left|\begin{array}{ccc}3&4\\1&-1\end{array}\right|=3 \cdot (-1)-4 \cdot 1=-3-4=-7$

$b)~det(A(1)\cdot A(1)+2A(x))=11;~~~x=?\\\\A(1)=\left(\begin{array}{ccc}1-1&1\\1&-1\end{array}\right)=\left(\begin{array}{ccc}0&1\\1&-1\end{array}\right)\\\\A(1)\cdot A(1)=\left(\begin{array}{ccc}0&1\\1&-1\end{array}\right)\cdot \left(\begin{array}{ccc}0&1\\1&-1\end{array}\right)=\\\\=\left(\begin{array}{ccc}0\cdot0+1\cdot1&0\cdot1+1\cdot(-1)\\1\cdot0+(-1)\cdot1&1\cdot1+(-1)\cdot(-1)\end{array}\right)=\left(\begin{array}{ccc}0+1&0-1\\0-1&1+1\end{array}\right)=$

$=\left(\begin{array}{ccc}1&-1\\-1&2\end{array}\right)$

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

$det(A(1)\cdot A(1)+2A(x))=\left|\begin{array}{ccc}2x-1&2x-1\\1&0\end{array}\right|=\\\\=(2x-1)\cdot 0 -(2x-1) \cdot 1=0-2x+1=1-2x\\\\ det(A(1)\cdot A(1)+2A(x))=11\Rightarrow 1-2x=11\Rightarrow -2x=11-1\Rightarrow \\\\\Rightarrow -2x=10\Rightarrow x=-\cfrac{10}{2} \Rightarrow x=-5$

$c)~A(0)\cdot A(x)\cdot A(1)=3A(y);~~x,y=?\\\\A(0)=\left(\begin{array}{ccc}0-1&0\\1&-1\end{array}\right)=\left(\begin{array}{ccc}-1&0\\1&-1\end{array}\right)\\\\A(0)\cdot A(x)=\left(\begin{array}{ccc}-1&0\\1&-1\end{array}\right) \cdot \left(\begin{array}{ccc}x-1&x\\1&-1\end{array}\right)=\\\\=\left(\begin{array}{ccc}(-1)\cdot (x-1)+0\cdot1&(-1)\cdot x+0\cdot (-1)\\1 \cdot (x-1)+(-1)\cdot1&1\cdot x+(-1)\cdot (-1)\end{array}\right)=$

$=\left(\begin{array}{ccc}-x+1+0&-x+0\\x-1-1&x+1\end{array}\right)=\left(\begin{array}{ccc}-x+1&-x\\x-2&x+1\end{array}\right)$

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

$3A(y)=3 \cdot \left(\begin{array}{ccc}y-1&y\\1&-1\end{array}\right)=\left(\begin{array}{ccc}3(y-1)&3\cdot y\\3 \cdot 1&3 \cdot (-1)\end{array}\right)=\left(\begin{array}{ccc}3y-3&3y\\3&-3\end{array}\right)$

$A(0)\cdot A(x) \cdot A(1)=3A(y)\Rightarrow \left(\begin{array}{ccc}-x&1\\x+1&-3\end{array}\right)=\left(\begin{array}{ccc}3y-3&3y\\3&-3\end{array}\right)\\\\ 1=3y\Rightarrow 3y=1 \Rightarrow y=\cfrac{1}{3} \\\\x+1=3 \Rightarrow x=3-1\Rightarrow x=2$

$14.~~A=\left(\begin{array}{ccc}-3&2\\-6&4\end{array}\right),~~B(x)=\left(\begin{array}{ccc}x&2\\-7&x-4\end{array}\right),~~x\in\mathbb{R}\\\\a)~detA=0\\\\detA=\left|\begin{array}{ccc}-3&2\\-6&4\end{array}\right|=(-3)\cdot 4-2 \cdot (-6)=-12+12=0$

$b)~detB(x)+det(B(7)-A)=0,~~~~~x=?\\\\detB(x)=\left|\begin{array}{ccc}x&2\\-7&x-4\end{array}\right|=x \cdot (x-4)-2 \cdot (-7)=x^2-4x+14\\\\B(7)=\left(\begin{array}{ccc}7&2\\-7&7-4\end{array}\right)=\left(\begin{array}{ccc}7&2\\-7&3\end{array}\right)\\\\B(7)-A=\left(\begin{array}{ccc}7&2\\-7&3\end{array}\right)-\left(\begin{array}{ccc}-3&2\\-6&4\end{array}\right)=\left(\begin{array}{ccc}7-(-3)&2-2\\-7-(-6)&3-4\end{array}\right)=\\\\=\left(\begin{array}{ccc}10&0\\-1&-1\end{array}\right)$

$det(B(7)-A)=\left|\begin{array}{ccc}10&0\\-1&-1\end{array}\right|=10\cdot (-1)-0\cdot(-1)=-10$

$detB(x)+det(B(7)-A)=0\Rightarrow x^2-4x+14-10=0\Rightarrow \\\\\Rightarrow x^2-4x+4=0\\\\a=1,~b=-4,~c=4\\\\\Delta=b^2-4ac=(-4)^2-4 \cdot 1 \cdot 4=16-16=0\\\\\Delta=0\Rightarrow x_1=x_2=-\cfrac{b}{2a} =-\cfrac{-4}{2 \cdot 1} =2\\\\x=2$