Considere a matriz A = [4 0] parte de baixo [-5 7]
apresente a matriz A - K . I , onde k e [] e I é a matriz identidade de 2a ordem.
quais valores de k tornam nulo o determinante da matriz A -K . I ?
Marque a alternativa que responde aos itens anteriores, respectivamente.
Segue anexo da pergunta para auxiliar. Com as letras de Múltipla Escolha
Anexos:
![](https://pt-static.z-dn.net/files/d74/eef776a3be98f4767e51de2f241831e0.png)
Soluções para a tarefa
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Olá
Alternativa correta, letra C)
A - K.I = 0
sendo "A" a matriz dada no enunciado, "K" uma constante pertencente ao reais, e "I" a matriz identidade
A matriz identidade tem esse formato
![I= \left[\begin{array}{ccc}1&0\\0&1\\\end{array}\right] I= \left[\begin{array}{ccc}1&0\\0&1\\\end{array}\right]](https://tex.z-dn.net/?f=I%3D++%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%26amp%3B0%5C%5C0%26amp%3B1%5C%5C%5Cend%7Barray%7D%5Cright%5D+)
Vamos começar
![\left[\begin{array}{ccc}4&0\\-5&7\\\end{array}\right] ~-~K\cdot\left[\begin{array}{ccc}1&0\\0&1\\\end{array}\right] \\ \\ \\ \text{Primeiro, multiplica o K pela matriz identidade} \\ \\ \\ \left[\begin{array}{ccc}4&0\\-5&7\\\end{array}\right] ~-~\left[\begin{array}{ccc}K\cdot1~~&K\cdot0\\K\cdot0~~&K\cdot1\\\end{array}\right] \\ \\ \\ \left[\begin{array}{ccc}4&0\\-5&7\\\end{array}\right] ~-~\left[\begin{array}{ccc}K~~&0\\0~~&K\\\end{array}\right] \\ \\ \\ \text{Agora subtrai da matriz A} \left[\begin{array}{ccc}4&0\\-5&7\\\end{array}\right] ~-~K\cdot\left[\begin{array}{ccc}1&0\\0&1\\\end{array}\right] \\ \\ \\ \text{Primeiro, multiplica o K pela matriz identidade} \\ \\ \\ \left[\begin{array}{ccc}4&0\\-5&7\\\end{array}\right] ~-~\left[\begin{array}{ccc}K\cdot1~~&K\cdot0\\K\cdot0~~&K\cdot1\\\end{array}\right] \\ \\ \\ \left[\begin{array}{ccc}4&0\\-5&7\\\end{array}\right] ~-~\left[\begin{array}{ccc}K~~&0\\0~~&K\\\end{array}\right] \\ \\ \\ \text{Agora subtrai da matriz A}](https://tex.z-dn.net/?f=++%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D4%26amp%3B0%5C%5C-5%26amp%3B7%5C%5C%5Cend%7Barray%7D%5Cright%5D+%7E-%7EK%5Ccdot%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%26amp%3B0%5C%5C0%26amp%3B1%5C%5C%5Cend%7Barray%7D%5Cright%5D+%5C%5C++%5C%5C++%5C%5C+%5Ctext%7BPrimeiro%2C+multiplica+o+K+pela+matriz+identidade%7D+%5C%5C++%5C%5C++%5C%5C+++%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D4%26amp%3B0%5C%5C-5%26amp%3B7%5C%5C%5Cend%7Barray%7D%5Cright%5D+%7E-%7E%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7DK%5Ccdot1%7E%7E%26amp%3BK%5Ccdot0%5C%5CK%5Ccdot0%7E%7E%26amp%3BK%5Ccdot1%5C%5C%5Cend%7Barray%7D%5Cright%5D+%5C%5C++%5C%5C++%5C%5C+%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D4%26amp%3B0%5C%5C-5%26amp%3B7%5C%5C%5Cend%7Barray%7D%5Cright%5D+%7E-%7E%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7DK%7E%7E%26amp%3B0%5C%5C0%7E%7E%26amp%3BK%5C%5C%5Cend%7Barray%7D%5Cright%5D+%5C%5C++%5C%5C++%5C%5C+%5Ctext%7BAgora+subtrai+da+matriz+A%7D)
![\left[\begin{array}{ccc}4-K~~&0-0\\-5-0~~&7-k\\\end{array}\right] \\ \\ \\ \boxed{\left[\begin{array}{ccc}4-K~~&0\\-5~~&7-k\\\end{array}\right]}~~~~\longleftarrow~ ~~ \text{Primeira resposta} \\ \\ \\ \text{Agora, temos que calcular o determinante da matriz que encontramos} \\ \\ \\ \left[\begin{array}{ccc}4-K~~&0\\-5~~&7-k\\\end{array}\right] \\ \\ \\ ((4-k)\cdot(7-k))~- ~(-5\cdot 0) \\ \\ \text{Aplica a distributiva} \\ \\ (28-4k-7k+k^2)~-~0 \left[\begin{array}{ccc}4-K~~&0-0\\-5-0~~&7-k\\\end{array}\right] \\ \\ \\ \boxed{\left[\begin{array}{ccc}4-K~~&0\\-5~~&7-k\\\end{array}\right]}~~~~\longleftarrow~ ~~ \text{Primeira resposta} \\ \\ \\ \text{Agora, temos que calcular o determinante da matriz que encontramos} \\ \\ \\ \left[\begin{array}{ccc}4-K~~&0\\-5~~&7-k\\\end{array}\right] \\ \\ \\ ((4-k)\cdot(7-k))~- ~(-5\cdot 0) \\ \\ \text{Aplica a distributiva} \\ \\ (28-4k-7k+k^2)~-~0](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D4-K%7E%7E%26amp%3B0-0%5C%5C-5-0%7E%7E%26amp%3B7-k%5C%5C%5Cend%7Barray%7D%5Cright%5D+%5C%5C++%5C%5C++%5C%5C+%5Cboxed%7B%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D4-K%7E%7E%26amp%3B0%5C%5C-5%7E%7E%26amp%3B7-k%5C%5C%5Cend%7Barray%7D%5Cright%5D%7D%7E%7E%7E%7E%5Clongleftarrow%7E+%7E%7E+%5Ctext%7BPrimeira+resposta%7D+%5C%5C++%5C%5C++%5C%5C+%5Ctext%7BAgora%2C+temos+que+calcular+o+determinante+da+matriz+que+encontramos%7D+%5C%5C++%5C%5C++%5C%5C+%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D4-K%7E%7E%26amp%3B0%5C%5C-5%7E%7E%26amp%3B7-k%5C%5C%5Cend%7Barray%7D%5Cright%5D+%5C%5C++%5C%5C++%5C%5C+%28%284-k%29%5Ccdot%287-k%29%29%7E-+%7E%28-5%5Ccdot+0%29+%5C%5C++%5C%5C+%5Ctext%7BAplica+a+distributiva%7D+%5C%5C++%5C%5C+%2828-4k-7k%2Bk%5E2%29%7E-%7E0+)
![\text{Agrupa os termos em comum e iguala a zero}\\ \\ k^2-11k+28=0 \\ \\ \text{Caimos em uma funcao do segundo grau} \\ \\ \text{Resolve por bhaskara} \\ \\ \triangle=(-11)^2-4\cdot(1)\cdot(28)\\\triangle=121-112 \\ \triangle=9 \\ \\ \\ \displaystyle X= \frac{-b\pm \sqrt{\triangle} }{2\cdot a} \text{Agrupa os termos em comum e iguala a zero}\\ \\ k^2-11k+28=0 \\ \\ \text{Caimos em uma funcao do segundo grau} \\ \\ \text{Resolve por bhaskara} \\ \\ \triangle=(-11)^2-4\cdot(1)\cdot(28)\\\triangle=121-112 \\ \triangle=9 \\ \\ \\ \displaystyle X= \frac{-b\pm \sqrt{\triangle} }{2\cdot a}](https://tex.z-dn.net/?f=%5Ctext%7BAgrupa+os+termos+em+comum+e+iguala+a+zero%7D%5C%5C+%5C%5C+k%5E2-11k%2B28%3D0+%5C%5C++%5C%5C+%5Ctext%7BCaimos+em+uma+funcao+do+segundo+grau%7D+%5C%5C++%5C%5C+%5Ctext%7BResolve+por+bhaskara%7D+%5C%5C++%5C%5C+%5Ctriangle%3D%28-11%29%5E2-4%5Ccdot%281%29%5Ccdot%2828%29%5C%5C%5Ctriangle%3D121-112+%5C%5C+%5Ctriangle%3D9+%5C%5C++%5C%5C++%5C%5C+%5Cdisplaystyle+X%3D+%5Cfrac%7B-b%5Cpm+%5Csqrt%7B%5Ctriangle%7D+%7D%7B2%5Ccdot+a%7D+)
![\displaystyle X_1= \frac{11+ \sqrt{9} }{2\cdot(1)} ~\longrightarrow X_1= \frac{11+3}{2} ~~\longrightarrow~X_1= \frac{14}{2} ~~\longrightarrow~~\boxed{X_1=7} \\ \\ \\ X_2=\frac{11- \sqrt{9} }{2\cdot(1)} ~\longrightarrow X_2= \frac{11-3}{2} ~~\longrightarrow~X_2= \frac{8}{2} ~~\longrightarrow~~\boxed{X_2=4} \displaystyle X_1= \frac{11+ \sqrt{9} }{2\cdot(1)} ~\longrightarrow X_1= \frac{11+3}{2} ~~\longrightarrow~X_1= \frac{14}{2} ~~\longrightarrow~~\boxed{X_1=7} \\ \\ \\ X_2=\frac{11- \sqrt{9} }{2\cdot(1)} ~\longrightarrow X_2= \frac{11-3}{2} ~~\longrightarrow~X_2= \frac{8}{2} ~~\longrightarrow~~\boxed{X_2=4}](https://tex.z-dn.net/?f=%5Cdisplaystyle+X_1%3D+%5Cfrac%7B11%2B+%5Csqrt%7B9%7D+%7D%7B2%5Ccdot%281%29%7D+%7E%5Clongrightarrow+X_1%3D+%5Cfrac%7B11%2B3%7D%7B2%7D+%7E%7E%5Clongrightarrow%7EX_1%3D+%5Cfrac%7B14%7D%7B2%7D+%7E%7E%5Clongrightarrow%7E%7E%5Cboxed%7BX_1%3D7%7D+%5C%5C++%5C%5C++%5C%5C+X_2%3D%5Cfrac%7B11-+%5Csqrt%7B9%7D+%7D%7B2%5Ccdot%281%29%7D+%7E%5Clongrightarrow+X_2%3D+%5Cfrac%7B11-3%7D%7B2%7D+%7E%7E%5Clongrightarrow%7EX_2%3D+%5Cfrac%7B8%7D%7B2%7D+%7E%7E%5Clongrightarrow%7E%7E%5Cboxed%7BX_2%3D4%7D+)
Então, para que o determinante da matriz A-K.I seja nulo (zero), K tem que ser
K=4 ou K=7
Alternativa correta, letra C)
A - K.I = 0
sendo "A" a matriz dada no enunciado, "K" uma constante pertencente ao reais, e "I" a matriz identidade
A matriz identidade tem esse formato
Vamos começar
Então, para que o determinante da matriz A-K.I seja nulo (zero), K tem que ser
K=4 ou K=7
mykemeris:
Correto
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