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Softa [21]
3 years ago
8

Which equation best models the fine schedule for overdue books?

Mathematics
2 answers:
Dmitry_Shevchenko [17]3 years ago
7 0

Answer:

B

Step-by-step explanation:

kondor19780726 [428]3 years ago
3 0

Answer:

B

Step-by-step explanation:

I took the test

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For each of the systems of equations that follow, use Gaussian elimination to obtain an equivalent system whose coefficient matr
noname [10]

Answer:

b) inconsistent

d) inconsistent

f) inconsistent

j) X_1 = 2 - 6a , X_2 = 4 + a , X_3 = 3 - a, X_4 = a (free variable)

Step-by-step explanation:

b)

Given: Augmented matrix

\left[\begin{array}{ccc}2&-3&5\\-4&6&8\end{array}\right]

Row operation : R_2 + 2*R_1

\left[\begin{array}{ccc}2&-3&5\\0&0&18\end{array}\right]

Looking at R_2 : 0 not equal to 18 ; Hence, inconsistent

c)

Given: Augmented matrix

\left[\begin{array}{cccc}3&2&-3&4\\1&-2&2&1\\11&2&1&14\end{array}\right]

Row operation : R_2 <---> R_1

\left[\begin{array}{cccc}1&-2&2&1\\3&2&-3&4\\11&2&1&14\end{array}\right]\\

Row operations : R_2 - 3*R_1   &   R_3 - 11*R_1

\left[\begin{array}{cccc}1&-2&2&1\\0&8&-9&1\\0&24&-21&3\end{array}\right]\\

Row operations : R_3 - 3*R_2  

\left[\begin{array}{cccc}1&-2&2&1\\0&8&-9&1\\0&0&6&0\end{array}\right]\\

Looking at R_3 : 0 not equal to 6 ; Hence, inconsistent

f)

Given: Augmented matrix

\left[\begin{array}{cccc}1&-1&2&4\\2&3&-1&1\\7&3&4&7\end{array}\right]\\

Row operations : R_2 - 2*R_1   &   R_3 - 7*R_1

\left[\begin{array}{cccc}1&-1&2&4\\0&5&-5&-7\\0&10&-10&-21\end{array}\right]\\

Row operations : R_2 --> ( 1 / 5 )* R_2  & R_3 --> ( 1 / 10 )* R_3

\left[\begin{array}{cccc}1&-1&2&4\\0&1&-1&-1.4\\0&1&-1&-2.1\end{array}\right]\\

Row operations : R_3 - R_1

\left[\begin{array}{cccc}1&-1&2&4\\0&1&-1&-1.4\\0&0&0&-0.7\end{array}\right]\\

Looking at R_3 : 0 not equal to -0.7 ; Hence, inconsistent

j)

Given: Augmented matrix

\left[\begin{array}{ccccc}1&2&-3&1&1\\-1&-1&4&-1&6\\-2&-4&7&-1&1\end{array}\right]\\

Row operations : R_2 + R_1   &   R_3 + 2*R_1

\left[\begin{array}{ccccc}1&2&-3&1&1\\0&1&1&0&7\\0&0&1&1&3\end{array}\right]\\

corresponding equations are:

X_1 + 2*X_2 - 3*X_3 + X_4 = 1

X_2 + X_3  = 7

X_3 + X_4 = 3

Free parameter : X_4 = a , then:

X_3 = 3 - a

X_2 = 4 + a

X_1 = 1 - a + 3*(3 - a) - 2*(4 + a) = 2 - 6 a

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