Question

preciso de ajuda nessas questões !!​

Answer 1

Resposta passo a passo:

Questão 01

a)

$A=\left[\begin{array}{cc}3&-2\\1&4\end{array}\right]$

$detA=4\times3-1\times(-2)\\\\detA=12+2\\\\\bold{detA=14}$

b)

$B=\left[\begin{array}{cc}-1&2\\5&-3\end{array}\right]$

$detB=-3\times(-1)-5\times(-2)\\\\detB=3+10\\\\\bold{detB=13}$

c)

$C=\left[\begin{array}{cc}5&-4\\2&-3\end{array}\right]$

$detC=-3\times5-2\times(-4)\\\\detC=-15+8\\\\\bold{detC=-7}$

d)

$D=\left[\begin{array}{cc}\frac{1}{2} &\frac{1}{6} \\3&2\end{array}\right]$

$detD=2\times\frac{1}{2} -3\times\frac{1}{6} \\\\detD=\frac{2}{2} -\frac{3}{6} \\\\detD=\frac{2}{2}-\frac{1}{2} \\\\detD=\frac{2-1}{2} \\\\\bold{detD=\frac{1}{2}}$

$----------------$

Questão 02

a)

$\left[\begin{array}{cc}2x &2 \\-3&1\end{array}\right]=12$

$2x\times1-(-3)\times2=12\\\\2x-(-6)=12\\\\2x+6=12\\\\\frac{2x+6}{2} =\frac{12}{2} \\\\x+3=6\\\\x=6-3\\\\\bold{x=3}$

b)

$\left[\begin{array}{cc}2x &1 \\3&5\end{array}\right]=\left[\begin{array}{cc}3 &4 \\2&1\end{array}\right]$

$2x\times 5 - 3\times1=3\times1-2\times4\\\\10x-3=3-8\\\\10x-3=-5\\\\10x=-5+3\\\\10x=-2\\\\x=\frac{-2}{10} \\\\\bold{x=-\frac{1}{5}}$

c)

$\left[\begin{array}{cc}x+2 &2x-1 \\3&4\end{array}\right]=\left[\begin{array}{cc}x &2 \\8&3\end{array}\right]$

$(x+2)\times4-(2x-1)\times3=3x-8\times2\\\\4x+8-6x+3=3x-16\\\\-2x+11=3x-16\\\\11+16=3x+2x\\\\5x=27\\\\\bold{x=\frac{27}{5}}$

d)

$\left[\begin{array}{cc}7 &x \\x&6\end{array}\right]=\left[\begin{array}{cc}1 &2 \\3&3\end{array}\right]+\left[\begin{array}{cc}3 &1 \\7&1\end{array}\right]$

$\left[\begin{array}{cc}7 &x \\x&6\end{array}\right]=\left[\begin{array}{cc}1+3 &2+1 \\3+7&3+1\end{array}\right]\\\\\\\left[\begin{array}{cc}7 &x \\x&6\end{array}\right]=\left[\begin{array}{cc}4 &3 \\10&4\end{array}\right]$

$7\times6-x^{2} =4\times4-10\times3\\\\42-x^{2} =16-30\\\\42-x^{2} =-14\\\\42+14-x^{2} =0\\\\56-x^{2} =0\\\\56=x^{2} \\\\x=\pm\sqrt{56} \\\\x=\pm\sqrt{4\times14} \\\\x=\pm\sqrt{4} \sqrt{14} \\\\\bold{x=\pm2\sqrt{14} }$