+0

# system

0
74
2

Find the ordered quintuplet (a,b,c,d,e)  that satisfies the system of equations

a + 2b + 3c + 4d + 5e = 41

2a + 3b + 4c + 5d + e = 15

3a + 4b + 5c + 1d + 2e = 34

4a + 5b + 1c + 2d + 3e = 68

5a + 1b + 2c + 3d + 4e = 57

Jun 1, 2023

#1
+1982
0

We can solve this system of equations using Gaussian elimination. Writing the augmented matrix, we have:

$$\left[\begin{array}{ccccc|c} 1 & 2 & 3 & 4 & 5 & 41 \\ 2 & 3 & 4 & 5 & 1 & 15 \\ 3 & 4 & 5 & 1 & 2 & 34 \\ 4 & 5 & 1 & 2 & 3 & 68 \\ 5 & 1 & 2 & 3 & 4 & 57 \end{array}\right]$$

Subtracting twice the first row from the second row, three times the first row from the third row, four times the first row from the fourth row, and five times the first row from the fifth row, we get:

$$\left[\begin{array}{ccccc|c} 1 & 2 & 3 & 4 & 5 & 41 \\ 0 & -1 & -2 & -3 & -9 & -67 \\ 0 & -2 & -4 & -11 & -13 & -83 \\ 0 & -3 & -11 & -14 & -13 & -108 \\ 0 & -9 & -13 & -17 & -11 & -148 \end{array}\right]$$

Adding twice the second row to the third row, three times thesecond row to the fourth row, and nine times the second row to the fifth row, we get:

$$\left[\begin{array}{ccccc|c} 1 & 2 & 3 & 4 & 5 & 41 \\ 0 & -1 & -2 & -3 & -9 & -67 \\ 0 & 0 & 0 & -17 & -31 & -213 \\ 0 & 0 & -17 & -5 & 10 & -289 \\ 0 & 0 & -31 & 8 & 67 & -691 \end{array}\right]$$

Finally, adding 17 times the third row to the fourth row, and 31 times the third row to the fifth row, we get:

$$\left[\begin{array}{ccccc|c} 1 & 2 & 3 & 4 & 5 & 41 \\ 0 & -1 & -2 & -3 & -9 & -67 \\ 0 & 0 & 0 & -17 & -31 & -213 \\ 0 & 0 & 0 & -74 & -472 & -3111 \\ 0 & 0 & 0 & 441 & 1044 & -1353 \end{array}\right]$$

We can now solve for the variables by back-substitution. From the last row, we have:

441d + 1044e = -1353

Solving for d and e, we get:

d = (-472(441) - 74(1044))/(-74(-17) - 472(0)) = 14
e = (-1353 - 441(14))/1044 = -2

Substituting d and e back into the fourth row, we get:

-74c - 472 = 3111 - 441(14)
-74c = 3897
c = -51

Substituting c, d, and e back into the third row, we get:

-17b - 31(-2) = 213 + 51 + 14
-17b = 311
b = -\frac{311}{17}

Substituting b, c, d, and e back into the second row, we get:

-a - 2(-\frac{311}{17}) - 3(-51) - 4(14) - 5(-2) = -67
-a = -\frac{1847}{17}

Substituting a, b, c, d, and e back into the first row, we get:

\frac{-1847}{17} + 2(-\frac{311}{17}) + 3(-51) + 4(14) + 5(-2) = 41

Therefore, the ordered quintuplet (a, b, c, d, e) that satisfies the system of equations is:

(a, b, c, d, e) = (-\frac{1847}{17}, -\frac{311}{17}, -51, 14, -2)

Jun 1, 2023
#2
0

a + 2b + 3c + 4d + 5e = 41

2a + 3b + 4c + 5d + e = 15

3a + 4b + 5c + 1d + 2e = 34

4a + 5b + 1c + 2d + 3e = 68

5a + 1b + 2c + 3d + 4e = 57

Use eliminations and substitutions to get:

a=(91 / 15),  b=(76 / 15),  c=(-59 / 15),  d=(-14 / 15),  e=(121 / 15)

Jun 1, 2023