Question

Utilize as operações elementares para encontrar a inversa:

b) 3 4 -1
A= 1 0 3
2 5 -4

c) √2 3√2 0
A= -4√2 √2 0
0 0 1

Answer 1

a) Seja $\left(\begin{array}{ccc}a&b&c\\d&e&f\\g&h&i\end{array}\right)$ a matriz inversa de $\left(\begin{array}{ccc}3&4&-1\\1&0&3\\2&5&-4\end{array}\right)$.

Desse modo, devemos ter
 
$\left(\begin{array}{ccc}a&b&c\\d&e&f\\g&h&i\end{array}\right)\cdot\left(\begin{array}{ccc}3&4&-1\\1&0&3\\2&5&-4\end{array}\right)=\left(\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right)$

Com isso, obtemos:

$\left(\begin{array}{ccc}3a+4d-g&3b+4e-h&3c+4f-i\\a+3g&b+3h&c+3i\\2a+5d-4g&2b+5e-4h&2c+5f-4i\end{array}\right)=\left(\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right)$

Temos três sistemas:

$\bullet~~\begin{cases} 3a+4d-g=1 \\ a+3g=0 \\ 2a+5d-4g=0 \end{cases}$

Da segunda equação, $a+3g=0 \iff \boxed{a=-3g}$. Substituindo na primeira e na terceira equações:
$3a+4d-g=1 \iff 3\cdot(-3g)+4d-g=1 \iff -9g+4d-g=1 \iff 4d-10g=1~(i)$

$2a+5d-4g=0 \iff 2\cdot(-3g)+5d-4g=0 \iff -6g+5d-4g=0 \iff 5d-10g=0~(ii)$
Subtraindo $(i)$ de $(ii)$:
$5d-10g-(4d-10g)=0-1 \iff 5d-4d-10g+10g=-1 \iff \boxed{d=-1}$

Substituindo em $(i)$:

$4d-10g=1 \iff 4\cdot(-1)-10g=1 \iff -4-10g=1 \iff 10g=-5 \iff 2g=-1 \iff \boxed{g=-\dfrac{1}{2}}$

Lembrando que $a=-3g$, segue que:

$a=-3g \iff a=-3\cdot\left(-\dfrac{1}{2}\right) \iff \boxed{a=\dfrac{3}{2}}$.

$\bullet~~\begin{cases} 3b+4e-h=0 \\ b+3h=1 \\ 2b+5e-4h=0 \end{cases}$

Da segunda equação, $b+3h=1 \iff \boxed{b=1-3h}$. 

Substituindo na primeira e na terceira equações:

$3b+4e-h=0 \iff 3\cdot(1-3h)+4e-h=0 \iff 3-9h+4e-h=0 \iff 4e-10h=-3~(i)$

$2b+5e-4h=0 \iff 2\cdot(1-3h)+5e-4h=0 \iff 2-6h+5e-4h=0 \iff 5e-10h=-2~(ii)$

Subtraindo $(i)$ de $(ii)$:

$5e-10h-(4e-10h)=-2-(-3) \iff 5e-4e-10h+10h=1 \iff \boxed{e=1}$

Substituindo em $(i)$:

$4e-10h=-3 \iff 4\cdot1-10h=-3 \iff 4-10h=-3 \iff 10h=7 \iff \boxed{h=\dfrac{7}{10}}$

Lembrando que $b=1-3h$, segue que:

$b=1-3h \iff b=1-3\cdot\left(\dfrac{7}{10}\right) \iff b=1-\dfrac{21}{10} \iff \boxed{b=\dfrac{11}{10}}$.

$\bullet~~\begin{cases} 3c+4f-i=0 \\ c+3i=0 \\ 2c+5f-4i=1 \end{cases}$

Da segunda equação, $c+3i=0 \iff \boxed{c=-3i}$. 

Substituindo na primeira e na terceira equações:

$3c+4f-i=0 \iff 3\cdot(-3i)+4f-i=0 \iff -9i+4f-i=0 \iff 4f-10i=0~(i)$

$2c+5f-4i=1 \iff 2\cdot(-3i)+5f-4i=1 \iff -6i+5f-4i=1 \iff 5f-10i=1~(ii)$

Subtraindo $(i)$ de $(ii)$:

$5f-10i-(4f-10i)=1-0 \iff 5f-4f-10i+10i=1 \iff \boxed{f=1}$

Substituindo em $(i)$:

$4f-10i=0 \iff 4\cdot1-10i=0 \iff 4-10i=0 \iff 10i=4 \iff \boxed{i=\dfrac{2}{5}}$

Lembrando que $c=-3i$, segue que:

$c=1-3i \iff c=-3\cdot\left(\dfrac{2}{5}\right) \iff \boxed{c=-\dfrac{6}{5}}$.

$\boxed{\left(\begin{array}{ccc}3&4&-1\\1&0&3\\2&5&-4\end{array}\right)^{-1}=\left(\begin{array}{ccc}\frac{3}{2}&-\frac{11}{10}&-\frac{6}{5}\\-1&1&1\\-\frac{1}{2}&\frac{7}{10}&\frac{2}{5}\end{array}\right)}$


b) $\left(\begin{array}{ccc}\sqrt{2}a+3\sqrt{2}d&\sqrt{2}b+3\sqrt{2}e&\sqrt{2}c+3\sqrt{2}f\\-4\sqrt{2}a+\sqrt{2}d&-4\sqrt{2}b+\sqrt{2}e&-4\sqrt{2}c+\sqrt{2}f\\g&h&i\end{array}\right)=\left(\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right)$
$\bullet~~\begin{cases} \sqrt{2}a+3\sqrt{2}d=1 \\ -4\sqrt{2}a+\sqrt{2}d=0 \\ \boxed{g=0} \end{cases}$

$\bullet~~\begin{cases} \sqrt{2}b+3\sqrt{2}e=0 \\ -4\sqrt{2}b+\sqrt{2}e=1 \\ \boxed{h=0} \end{cases}$

$\bullet~~\begin{cases} \sqrt{2}c+3\sqrt{2}f=0 \\ -4\sqrt{2}c+\sqrt{2}f=0 \\ \boxed{i=1} \end{cases}$

Obtemos: $a=e=\dfrac{\sqrt{2}}{26}$, $b=\dfrac{-3\sqrt{2}}{26}$, $c=f=0$, $d=\dfrac{2\sqrt{2}}{13}$ 

$\boxed{\left(\begin{array}{ccc}\sqrt{2}&3\sqrt{2}&0\\-4\sqrt{2}&\sqrt{2}&0\\0&0&1\end{array}\right)^{-1}=\left(\begin{array}{ccc}\frac{\sqrt{2}}{26}&-\frac{3\sqrt{2}}{26}&0\\\\ \frac{2\sqrt{2}}{13}&\frac{\sqrt{2}}{26}&0\\\\0&0&1\end{array}\right)}$