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Solve the homogeneous linear system corresponding to the given coefficient matrix. (If there is no solution, enter NO SOLUTION. If the system has an infinite number of solutions, set x4 = t and x2 = s and solve for x1 and x3 in terms of t and s.)

[1 0 0 1]
[0 0 1 0]
[0 0 0 0]

(x1, x2, x3, x4) = _________

Respuesta :

MrRoyal

Answer:

[tex]x_1 = -t[/tex]

[tex]x_2 = s[/tex]

[tex]x_3 = 0[/tex]

[tex]x_4 = t[/tex]

Step-by-step explanation:

Given

[tex]\left[\begin{array}{cccc}1&0&0&1\\0&0&1&0\\0&0&0&0\end{array}\right][/tex]

Required

Determine x1 to x4

First, we write the augmented matrix

[tex]\left[\begin{array}{cccc}1&0&0&1\\0&0&1&0\\0&0&0&0\end{array}\right] = \left[\begin{array}{c}0&0&0\end{array}\right][/tex]

Taking the position of each column, we have:

[tex]x_1 + 0 + 0 +x_4 = 0[/tex]

[tex]0 + 0 + x_3 +0 = 0[/tex]

[tex]0 + 0 + 0 + 0 = 0[/tex]

There is no explicit equation for [tex]x_2[/tex]

So: [tex]x_2 = s[/tex] ----- arbitrary variable

[tex]x_1 + 0 + 0 +x_4 = 0[/tex] implies that:

[tex]x_1 + x_4 = 0[/tex]

[tex]0 + 0 + x_3 +0 = 0[/tex] implies that

[tex]x_3 = 0[/tex]

[tex]0 + 0 + 0 + 0 = 0[/tex] implies that

[tex]0 = 0[/tex]

So, we have:

[tex]x_1 + x_4 = 0[/tex]

[tex]x_3 = 0[/tex]

[tex]x_1 + x_4 = 0[/tex] becomes

[tex]x_1 = -x_4[/tex]

Let

[tex]x_4 = t[/tex]

So:

[tex]x_1 = -x_4[/tex]

[tex]x_1 = -t[/tex]

Where t , s are real numbers

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