Question

preciso de ajuda com essas questões!!!!
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Answer 1

Resposta passo a passo:

Questão 01:

$\left[\begin{array}{cc}x&1\\-1&2\end{array}\right] + \left[\begin{array}{cc}-5&y\\0&-1\end{array}\right] = \left[\begin{array}{cc}3&-2\\z&t\end{array}\right]$

$\left[\begin{array}{cc}x+(-5)=3&1+y=-2\\-1+0=z&2+(-1)=t\end{array}\right] \\\\\\\left[\begin{array}{cc}x-5=3&y=-2-1\\-1=z&2-1=t\end{array}\right] \\\\\\\left[\begin{array}{cc}x=3+5&y=-3\\-1=z&1=t\end{array}\right]$

$\bold{\left[\begin{array}{cc}x=8&y=-3\\z=-1&t=1\end{array}\right]}$

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Questão 02:

2x - 3B = 2Aˣ - C

2x = 2A - C + 3B

As dimensões de A e C são diferentes, portanto não é possível subtrair as matrizes.

A matriz A não pode ser multiplicada por ela mesma.

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Questão 03:

x + B = A²

x = A² - B

x = A . A - B

$x = \left[\begin{array}{cc}-1&0\\1&2\end{array}\right] .\left[\begin{array}{cc}-1&0\\1&2\end{array}\right]-\left[\begin{array}{cc}3&-1\\1&2\end{array}\right]\\\\\\x=\left[\begin{array}{cc}1&0\\1&4\end{array}\right]-\left[\begin{array}{cc}3&-1\\1&2\end{array}\right]\\\\\\x=\left[\begin{array}{cc}1-3&0+1\\1-1&4-2\end{array}\right]$

$\bold{x=\left[\begin{array}{cc}-2&1\\0&2\end{array}\right]}$

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Questão 04:

A)

$\left[\begin{array}{cc}2x&4\\-1&-5\end{array}\right] =6\\\\\\2x(-5)-4(-1)=6\\\\-10x+4=6\\\\10x=4-6\\\\10x=-2\\\\5x=-1$

$\bold{x=-\frac{1}{5}}$

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B)

Para comparar as matrizes em si, as dimensões deveriam ser iguais, então irei descobrir o possível valor de x, comparando seus determinantes:

$\left[\begin{array}{ccc}1&0&1\\6&x&0\\-x&x&x\end{array}\right] =\left[\begin{array}{cc}5&3\\3&1\end{array}\right] \\\\\\1.x.x+0.0.-x+1.6.x-(-x).x.1-x.0.1-x.6.0=5.1-3.3\\\\x^{2} +0+6x+x^{2} -0-0=5-9\\\\x^{2} +6x+x^{2} =-4\\\\2x^{2} +6x=-4\\\\2x^{2} +6x+4=0\\\\x^{2} +3x+2=0$

$x=\frac{-3\pm\sqrt{3^{2} -4.1.2} }{2.1}\\\\x=\frac{-3\pm\sqrt{9 -8} }{2}\\\\x=\frac{-3\pm\sqrt{1} }{2}\\\\x=\frac{-3\pm1 }{2}\\\\\\x'=\frac{-3-1}{2} \\\\x'=\frac{-4}{2} \\\\\bold{x'=-2}\\\\\\x''=\frac{-3+1}{2} \\\\x''=\frac{-2}{2} \\\\\bold{x''=-1}$

Obs: As matrizes NÃO são iguais, mas podem possuir um mesmo determinante, caso x = -2 ou x = -1.