Question

Solve and Check
For (1)
2a+b-3c-d= -2
a-b-c+3d= 0
3a+2b+c+5d= 6
a-c-d= 2

and (2)
r+s+2t-u= -3
2r+3s+3t+u= 2
4r+2s-t+u= 5
s+2t+2u=7

Answer 1

Answer on Question #64696 – Math – Linear Algebra

Question

1) Solve and Check

$$\begin{array}{l} 2a + b - 3c - d = -2 \\ a - b - c + 3d = 0 \\ 3a + 2b + c + 5d = 6 \\ a - c - d = 2 \\ \end{array}$$

Solution

Let's rewrite the system of equations in matrix form and solve it by Gaussian elimination

$$\begin{array}{l} 2 \quad 1 \quad -3 \quad -1 -2 \\ 1 \quad -1 \quad -1 \quad 3 \quad 0 \\ 3 \quad 2 \quad 1 \quad 5 \quad 6 \\ 1 \quad 0 \quad -1 \quad -1 \quad 2 \\ \end{array}$$

Change the order of rows in the matrix. Set the fourth row to be the first one

$$\begin{array}{l} 1 \quad 0 \quad -1 \quad -1 \quad 2 \\ 2 \quad 1 \quad -3 \quad -1 -2 \\ 1 \quad -1 \quad -1 \quad 3 \quad 0 \\ 3 \quad 2 \quad 1 \quad 5 \quad 6 \\ \end{array}$$

Subtracting row 1 from row 3 and finally multiplying by (-1) in the modified matrix

$$\begin{array}{l} 1 \quad 0 \quad -1 \quad -1 \quad 2 \\ 2 \quad 1 \quad -3 \quad -1 -2 \\ 0 \quad 1 \quad 0 \quad -4 \quad 2 \\ 3 \quad 2 \quad 1 \quad 5 \quad 6 \\ \end{array}$$

Adding row 1, multiplied by -2, to row 2

$$\begin{array}{l} 1 \quad 0 \quad -1 \quad -1 \quad 2 \\ 0 \quad 1 \quad -1 \quad 1 \quad -6 \\ 0 \quad 1 \quad 0 \quad -4 \quad 2 \\ 3 \quad 2 \quad 1 \quad 5 \quad 6 \\ \end{array}$$

Interchanging the second and third rows

$$\begin{array}{l} 1 \quad 0 \quad -1 \quad -1 \quad 2 \\ 0 \quad 1 \quad 0 \quad -4 \quad 2 \\ 0 \quad 1 \quad -1 \quad 1 \quad -6 \\ 3 \quad 2 \quad 1 \quad 5 \quad 6 \\ \end{array}$$

Adding row 1, multiplied by -3, to row 4

Dividing row 4 by 2

Subtracting the second row from the third row

Multiplying the third row by (-1);

Subtracting row 2 from row 4

Adding row 3, multiplied by -2, to row 4

Dividing row 4 by 18

1 0 -1 -1 2

0 1 0 -4 2

0 0 1 -5 8

0 0 0 1 -1

Adding row 4, multiplied by 5, to row 4

1 0 -1 -1 2

0 1 0 -4 2

0 0 1 0 3

0 0 0 1 -1

Adding row 3 to row 1

1 0 0 -1 5

0 1 0 -4 2

0 0 1 0 3

0 0 0 1 -1

Adding row 4, multiplied by 4, to row 2

1 0 0 -1 5

0 1 0 0 -2

0 0 1 0 3

0 0 0 1 -1

Adding row 4 to row 1

1 0 0 0 4

0 1 0 0 -2

0 0 1 0 3

0 0 0 1 -1

Hence

$$a = 4, b = -2, c = 3, d = -1.$$

Let us check. Substituting this solution into the equations of the system and perform the calculation:

$$\begin{array}{l} 2 \cdot 4 + (-2) - 3 \cdot 3 - (-1) = 8 - 2 - 9 + 1 = -2 \\ 4 - (-2) - 3 + 3 \cdot (-1) = 4 + 2 - 3 - 3 = 0 \\ 3 \cdot 4 + 2 \cdot (-2) + 3 + 5 \cdot (-1) = 12 - 4 + 3 - 5 = 6 \\ 4 - 3 - (-1) = 4 - 3 + 1 = 2 \\ \end{array}$$

Verification is successful.

Answer:

a = 4, b = -2, c = 3, d = -1.

Question

2)

```

r+s+2t-u= -3

2r+3s+3t+u= 2

4r+2s-t+u= 5

s+2t+2u=7

```

Solution

Let's rewrite the system of equations in the matrix form and solve it by the Gaussian elimination

Subtracting row 1, multiplied by 2, from row 2;

Subtracting row 1, multiplied by 4, from row 3

Adding row 2, multiplied by 2, to row 3

Subtracting row 2 from row 4

1 1 2 -1-3

0 1 -1 3 8

0 0 -11 11 33

0 0 3 -1-1

Dividing row 3 by -11

1 1 2 -1-3

0 1 -1 3 8

0 0 1 -1-3

0 0 3 -1-1

Subtracting row 3, multiplied by 3, from row 4

1 1 2 -1-3

0 1 -1 3 8

0 0 1 -1-3

0 0 0 2 8

Dividing row 4 by 2

1 1 2 -1-3

0 1 -1 3 8

0 0 1 -1-3

0 0 0 1 4

Adding row 4 to row 3

1 1 2 -1-3

0 1 -1 3 8

0 0 1 0 1

0 0 0 1 4

Subtracting row 4, multiplied by 3, from row 2

1 1 2 -1-3

0 1 -1 0 -4

0 0 1 0 1

0 0 0 1 4

Adding row 4 to row 1

1 1 2 0 1

0 1 -1 0 -4

0 0 1 0 1

0 0 0 1 4

Adding row 3 to row 2

1 1 2 0 1

0 1 0 0 -3

0 0 1 0 1

0 0 0 1 4

Subtract row 3, multiplied by 2, from row 1

1 1 0 0 -1

0 1 0 0 -3

0 0 1 0 1

0 0 0 1 4

Subtracting row 2 from row 1

1 0 0 0 2

0 1 0 0 -3

0 0 1 0 1

0 0 0 1 4

Hence

$$r = 2, s = -3, t = 1, u = 4$$

Let us check. Substituting this solution into the equations of the system and perform the calculation:

$$\begin{array}{l} 2 + (-3) + 2 \cdot 1 - 4 = 2 - 3 + 2 - 4 = -3 \\ 2 \cdot 2 + 3 \cdot (-3) + 3 \cdot 1 + 4 = 4 - 9 + 3 + 4 = 2 \\ 4 \cdot 2 + 2 \cdot (-3) - 1 + 4 = 8 - 6 - 1 + 4 = 5 \\ (-3) + 2 \cdot 1 + 2 \cdot 4 = -3 + 2 + 8 = 7 \\ \end{array}$$

Verification is successful.

Answer:

$$r = 2, s = -3, t = 1, u = 4.$$

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