Determinantes
Determine x tal que:
a) |3x x+2|=11
|5 4 |
b) |2x-1 x+3|=0
| 4 x |
c) |x 3 1|
|1 4 5|=55
|x - 2 3|
d) | 3 1 4|
|x+2 -1 5| =35
| 2 3 -2|
Se possivel deixar os calculos eu agradeço.
Soluções para a tarefa
Respondido por
13
E aí Ganguiamente,
para equação matricial 2x2, use o mesmo procedimento, usado para de terminante de 2ª ordem:
![\left|\begin{array}{ccc}3x&x+2\\5&4\\\end{array}\right|=11\\\\\\
4*3x-5(x+2)=11\\
12x-5x-10=11\\
7x=11+10\\
7x=21\\
x=21/7\\
x=3\\\\
\boxed{S=\{7\}} \left|\begin{array}{ccc}3x&x+2\\5&4\\\end{array}\right|=11\\\\\\
4*3x-5(x+2)=11\\
12x-5x-10=11\\
7x=11+10\\
7x=21\\
x=21/7\\
x=3\\\\
\boxed{S=\{7\}}](https://tex.z-dn.net/?f=++%5Cleft%7C%5Cbegin%7Barray%7D%7Bccc%7D3x%26amp%3Bx%2B2%5C%5C5%26amp%3B4%5C%5C%5Cend%7Barray%7D%5Cright%7C%3D11%5C%5C%5C%5C%5C%5C%0A4%2A3x-5%28x%2B2%29%3D11%5C%5C%0A12x-5x-10%3D11%5C%5C%0A7x%3D11%2B10%5C%5C%0A7x%3D21%5C%5C%0Ax%3D21%2F7%5C%5C%0Ax%3D3%5C%5C%5C%5C%0A%5Cboxed%7BS%3D%5C%7B7%5C%7D%7D+)
____________________
![\left|\begin{array}{ccc}2x-1&x+3\\4&x\\\end{array}\right|=0\\\\\\
(2x-1)*x-4(x+3)=0\\
2 x^{2} -x-4x-12=0\\
2 x^{2} -5x-12=0 \left|\begin{array}{ccc}2x-1&x+3\\4&x\\\end{array}\right|=0\\\\\\
(2x-1)*x-4(x+3)=0\\
2 x^{2} -x-4x-12=0\\
2 x^{2} -5x-12=0](https://tex.z-dn.net/?f=++%5Cleft%7C%5Cbegin%7Barray%7D%7Bccc%7D2x-1%26amp%3Bx%2B3%5C%5C4%26amp%3Bx%5C%5C%5Cend%7Barray%7D%5Cright%7C%3D0%5C%5C%5C%5C%5C%5C%0A%282x-1%29%2Ax-4%28x%2B3%29%3D0%5C%5C%0A2+x%5E%7B2%7D+-x-4x-12%3D0%5C%5C%0A2+x%5E%7B2%7D+-5x-12%3D0+)
![\boxed{\Delta=b ^{2}-4ac}\\\\
\Delta=(-5)^{2}-4*2*(-12)\\
\Delta=25+96\\
\Delta=121 \boxed{\Delta=b ^{2}-4ac}\\\\
\Delta=(-5)^{2}-4*2*(-12)\\
\Delta=25+96\\
\Delta=121](https://tex.z-dn.net/?f=%5Cboxed%7B%5CDelta%3Db+%5E%7B2%7D-4ac%7D%5C%5C%5C%5C%0A%5CDelta%3D%28-5%29%5E%7B2%7D-4%2A2%2A%28-12%29%5C%5C%0A%5CDelta%3D25%2B96%5C%5C%0A%5CDelta%3D121++)
![\boxed{x= \frac{-b\pm \sqrt{\Delta} }{2a}}\\\\\\
x= \frac{-(-5)\pm \sqrt{121} }{2*2}\to~x= \frac{5\pm11}{4}\to\begin{cases}x'= \frac{5-11}{4}\to~x'= \frac{-6}{4}\to~x'=- \frac{3}{2}\\\\
x''= \frac{5+11}{4}\to~x''= \frac{16}{4}\to~x''=4 \end{cases} \boxed{x= \frac{-b\pm \sqrt{\Delta} }{2a}}\\\\\\
x= \frac{-(-5)\pm \sqrt{121} }{2*2}\to~x= \frac{5\pm11}{4}\to\begin{cases}x'= \frac{5-11}{4}\to~x'= \frac{-6}{4}\to~x'=- \frac{3}{2}\\\\
x''= \frac{5+11}{4}\to~x''= \frac{16}{4}\to~x''=4 \end{cases}](https://tex.z-dn.net/?f=%5Cboxed%7Bx%3D+%5Cfrac%7B-b%5Cpm+%5Csqrt%7B%5CDelta%7D+%7D%7B2a%7D%7D%5C%5C%5C%5C%5C%5C%0Ax%3D+%5Cfrac%7B-%28-5%29%5Cpm+%5Csqrt%7B121%7D+%7D%7B2%2A2%7D%5Cto%7Ex%3D+%5Cfrac%7B5%5Cpm11%7D%7B4%7D%5Cto%5Cbegin%7Bcases%7Dx%27%3D+%5Cfrac%7B5-11%7D%7B4%7D%5Cto%7Ex%27%3D+%5Cfrac%7B-6%7D%7B4%7D%5Cto%7Ex%27%3D-+%5Cfrac%7B3%7D%7B2%7D%5C%5C%5C%5C%0Ax%27%27%3D+%5Cfrac%7B5%2B11%7D%7B4%7D%5Cto%7Ex%27%27%3D+%5Cfrac%7B16%7D%7B4%7D%5Cto%7Ex%27%27%3D4+++%5Cend%7Bcases%7D+++)
Portanto, a solução da equação matricial será:
![\boxed{S=\{- \frac{3}{2} ,~4\}} \boxed{S=\{- \frac{3}{2} ,~4\}}](https://tex.z-dn.net/?f=%5Cboxed%7BS%3D%5C%7B-+%5Cfrac%7B3%7D%7B2%7D+%2C%7E4%5C%7D%7D)
____________________
E para matricial 3x3, utilize o mesmo procedimento usado para matriz de 3ª ordem, (regra de Sarruz):
![\left|\begin{array}{ccc}x&3&1\\1&4&5\\x&-2&3\end{array}\right| \left\begin{array}{ccc}x&3\\1&4\\x&-2\end{array}\right=55\\\\\\
d.p.~\to~12x+15x-2\\
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=55\\
d.s.~\to~-4x+10x-9\\\\
(d.p.)+(d.s.)~\to~33x-11=55\\\\
33x=55+11\\
33x=66\\
x=66/33\\
x=2\\\\
\boxed{S=\{2\}}
\left|\begin{array}{ccc}x&3&1\\1&4&5\\x&-2&3\end{array}\right| \left\begin{array}{ccc}x&3\\1&4\\x&-2\end{array}\right=55\\\\\\
d.p.~\to~12x+15x-2\\
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=55\\
d.s.~\to~-4x+10x-9\\\\
(d.p.)+(d.s.)~\to~33x-11=55\\\\
33x=55+11\\
33x=66\\
x=66/33\\
x=2\\\\
\boxed{S=\{2\}}](https://tex.z-dn.net/?f=++%5Cleft%7C%5Cbegin%7Barray%7D%7Bccc%7Dx%26amp%3B3%26amp%3B1%5C%5C1%26amp%3B4%26amp%3B5%5C%5Cx%26amp%3B-2%26amp%3B3%5Cend%7Barray%7D%5Cright%7C++%5Cleft%5Cbegin%7Barray%7D%7Bccc%7Dx%26amp%3B3%5C%5C1%26amp%3B4%5C%5Cx%26amp%3B-2%5Cend%7Barray%7D%5Cright%3D55%5C%5C%5C%5C%5C%5C%0Ad.p.%7E%5Cto%7E12x%2B15x-2%5C%5C%0A%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%3D55%5C%5C%0Ad.s.%7E%5Cto%7E-4x%2B10x-9%5C%5C%5C%5C%0A%28d.p.%29%2B%28d.s.%29%7E%5Cto%7E33x-11%3D55%5C%5C%5C%5C%0A33x%3D55%2B11%5C%5C%0A33x%3D66%5C%5C%0Ax%3D66%2F33%5C%5C%0Ax%3D2%5C%5C%5C%5C%0A%5Cboxed%7BS%3D%5C%7B2%5C%7D%7D%0A)
____________________
![\left|\begin{array}{ccc}3&1&4\\x+2&-1&5\\2&3&-2\end{array}\right| \left\begin{array}{ccc}3&1\\x+2&-1\\2&3\end{array}\right=35\\\\\\
d.p.\to~6+10+12x+24\\
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=35\\
d.s.\to~8-45+2x+4\\\\\\
d.p.\to~40+12x\\
~~~~~~~~~~~~~~~~~~~~~~~~~~~=35\\
d.s.\to~-33+2x\\\\\\
(d.p.)+(d.s.)~\to~7+14x=35\\
14x=35-7\\
14x=28\\
x=28/14\\
x=2\\\\\\
\boxed{S=\{2\}} \left|\begin{array}{ccc}3&1&4\\x+2&-1&5\\2&3&-2\end{array}\right| \left\begin{array}{ccc}3&1\\x+2&-1\\2&3\end{array}\right=35\\\\\\
d.p.\to~6+10+12x+24\\
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=35\\
d.s.\to~8-45+2x+4\\\\\\
d.p.\to~40+12x\\
~~~~~~~~~~~~~~~~~~~~~~~~~~~=35\\
d.s.\to~-33+2x\\\\\\
(d.p.)+(d.s.)~\to~7+14x=35\\
14x=35-7\\
14x=28\\
x=28/14\\
x=2\\\\\\
\boxed{S=\{2\}}](https://tex.z-dn.net/?f=++%5Cleft%7C%5Cbegin%7Barray%7D%7Bccc%7D3%26amp%3B1%26amp%3B4%5C%5Cx%2B2%26amp%3B-1%26amp%3B5%5C%5C2%26amp%3B3%26amp%3B-2%5Cend%7Barray%7D%5Cright%7C++%5Cleft%5Cbegin%7Barray%7D%7Bccc%7D3%26amp%3B1%5C%5Cx%2B2%26amp%3B-1%5C%5C2%26amp%3B3%5Cend%7Barray%7D%5Cright%3D35%5C%5C%5C%5C%5C%5C%0Ad.p.%5Cto%7E6%2B10%2B12x%2B24%5C%5C%0A%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%3D35%5C%5C%0Ad.s.%5Cto%7E8-45%2B2x%2B4%5C%5C%5C%5C%5C%5C%0Ad.p.%5Cto%7E40%2B12x%5C%5C%0A%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%7E%3D35%5C%5C%0Ad.s.%5Cto%7E-33%2B2x%5C%5C%5C%5C%5C%5C%0A%28d.p.%29%2B%28d.s.%29%7E%5Cto%7E7%2B14x%3D35%5C%5C%0A14x%3D35-7%5C%5C%0A14x%3D28%5C%5C%0Ax%3D28%2F14%5C%5C%0Ax%3D2%5C%5C%5C%5C%5C%5C%0A%5Cboxed%7BS%3D%5C%7B2%5C%7D%7D+)
Espero ter ajudado e tenha ótimos estudos =))
para equação matricial 2x2, use o mesmo procedimento, usado para de terminante de 2ª ordem:
____________________
Portanto, a solução da equação matricial será:
____________________
E para matricial 3x3, utilize o mesmo procedimento usado para matriz de 3ª ordem, (regra de Sarruz):
____________________
Espero ter ajudado e tenha ótimos estudos =))
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