Question

Find the least squares solutions to the following linear systems: (a) x+2y=1,3x−y=0,−x+2y=3, (b) 4x−2y=1,2x+3y=−4,x−2y=−1,2x+2y=2, (c) 2u+v−2w=1,3u−2w=0,u−v+3w=2, (d) x−z=−1,2x−y+3z=1,y−3z=0,−5x+2y+z=3, (e) x 1

​
+x 2
​
=2,x 2
​
+x 4
​
=1,x 1
​
+x 3
​
=0,x 3
​
−x 4
​
=1,x 1
​
−x 4
​
=2.

Answer 1

To find the least squares solutions to a linear system, we can use matrix algebra. We can write the system of equations as a matrix equation: Ax = b, where A is the coefficient matrix, x is the variable matrix, and b is the constant matrix.

Step-by-step explanation:

To find the least squares solutions to a linear system, we can use matrix algebra. We can write the system of equations as a matrix equation: Ax = b, where A is the coefficient matrix, x is the variable matrix, and b is the constant matrix.

In order to find x, we can use the formula: x = (A^T A)^-1 A^T b, where ^T denotes the transpose. This formula calculates the least squares solutions.

Let's apply this formula to each of the given linear systems to find the least squares solutions.

Answer 2

(a) No solution.

(b) Infinite solutions: (x = t), (y = -1), where (t) is a free parameter.

(c) Unique solution:
`\(u = (2)/(5)\)`,
`\(v = (1)/(5)\), \(w = 1\).`

(d) Unique solution: (x = -2), (y = -1), (z = 1).

(e) Unique solution: (x_1 = 2), (x_2 = 1), (x_3 = 1), (x_4) is a free parameter.

Step-by-step explanation:

To find the least squares solutions for the given linear systems, we need to set up the augmented matrix [A|B] and then perform Gaussian elimination to obtain the row-echelon form. Afterward, we can use back substitution to find the solutions. I'll go through each system:

(a) (x + 2y = 1), (3x - y = 0), (-x + 2y = 3)

Let's form the augmented matrix:

`\[\begin{bmatrix}1 & 2 & | & 1 \\3 & -1 & | & 0 \\-1 & 2 & | & 3\end{bmatrix}\]`

Performing Gaussian elimination:

`\[\begin{bmatrix}1 & 2 & | & 1 \\0 & -7 & | & -3 \\0 & 0 & | & 2\end{bmatrix}\]`

The system is inconsistent (the third row corresponds to (0x + 0y = 2), so there is no solution.

(b) (4x - 2y = 1), (2x + 3y = -4), (x - 2y = -1), (2x + 2y = 2)

`\[\begin{bmatrix}4 & -2 & | & 1 \\2 & 3 & | & -4 \\1 & -2 & | & -1 \\2 & 2 & | & 2\end{bmatrix}\]`

Performing Gaussian elimination:

`\[\begin{bmatrix}1 & 0 & | & 0 \\0 & 1 & | & -1 \\0 & 0 & | & 0 \\0 & 0 & | & 0\end{bmatrix}\]`

This system is dependent (the third and fourth rows are the same). It has infinitely many solutions. The solution can be parametrized as (x = t), (y = -1), where (t) is a free parameter.

(c) (2u + v - 2w = 1), (3u - 2w = 0), (u - v + 3w = 2)

`\[\begin{bmatrix}2 & 1 & -2 & | & 1 \\3 & 0 & -2 & | & 0 \\1 & -1 & 3 & | & 2\end{bmatrix}\]`

Performing Gaussian elimination:

`\[\begin{bmatrix}1 & 0 & 4/5 & | & 2/5 \\0 & 1 & 2/5 & | & 1/5 \\0 & 0 & 1 & | & 1\end{bmatrix}\]`

The solution is (w = 1), (v = 1/5), (u = 2/5).

(d) (x - z = -1), (2x - y + 3z = 1), (y - 3z = 0), (-5x + 2y + z = 3)

`\[\begin{bmatrix}1 & 0 & -1 & | & -1 \\2 & -1 & 3 & | & 1 \\0 & 1 & -3 & | & 0 \\-5 & 2 & 1 & | & 3\end{bmatrix}\]`

Performing Gaussian elimination:

`\[\begin{bmatrix}1 & 0 & 0 & | & -2 \\0 & 1 & 0 & | & -1 \\0 & 0 & 1 & | & 1\end{bmatrix}\]`

The solution is (x = -2), (y = -1), (z = 1).

(e) (x_1 + x_2 = 2), (x_2 + x_4 = 1), (x_1 + x_3 = 0), (x_3 - x_4 = 1), (x_1 - x_4 = 2)

`\[\begin{bmatrix}1 & 1 & 0 & 0 & | & 2 \\0 & 1 & 0 & 1 & | & 1 \\1 & 0 & 1 & 0 & | & 0 \\0 & 0 & 1 & -1 & | & 1 \\1 & 0 & 0 & -1 & | & 2\end{bmatrix}\]`

Performing Gaussian elimination:

`\[\begin{bmatrix}1 & 0 & 0 & -1 & | & 2 \\0 & 1 & 0 & 1 & | & 1 \\0 & 0 & 1 & -1 & | & 1 \\0 & 0 & 0 & 0 & | & 0 \\0 & 0 & 0 & 0 & | & 0\end{bmatrix}\]`

(a) (x + 2y = 1), (3x - y = 0), (-x + 2y = 3)

Let's form the augmented matrix:

`\[\begin{bmatrix}1 & 2 & | & 1 \\3 & -1 & | & 0 \\-1 & 2 & | & 3\end{bmatrix}\]`

Performing Gaussian elimination:

`\[\begin{bmatrix}1 & 2 & | & 1 \\0 & -7 & | & -3 \\0 & 0 & | & 2\end{bmatrix}\]`

The system is inconsistent (the third row corresponds to \(0x + 0y = 2\)), so there is no solution.

(b) (4x - 2y = 1), (2x + 3y = -4), (x - 2y = -1), (2x + 2y = 2)

`\[\begin{bmatrix}4 & -2 & | & 1 \\2 & 3 & | & -4 \\1 & -2 & | & -1 \\2 & 2 & | & 2\end{bmatrix}\]`

Performing Gaussian elimination:

`\[\begin{bmatrix}1 & 0 & | & 0 \\0 & 1 & | & -1 \\0 & 0 & | & 0 \\0 & 0 & | & 0\end{bmatrix}\]`

(c) (2u + v - 2w = 1), (3u - 2w = 0), (u - v + 3w = 2)

`\[\begin{bmatrix}2 & 1 & -2 & | & 1 \\3 & 0 & -2 & | & 0 \\1 & -1 & 3 & | & 2\end{bmatrix}\]`

Performing Gaussian elimination:

`\[\begin{bmatrix}1 & 0 & 4/5 & | & 2/5 \\0 & 1 & 2/5 & | & 1/5 \\0 & 0 & 1 & | & 1\end{bmatrix}\]`

The solution is (w = 1), (v = 1/5), (u = 2/5).

(d) (x - z = -1), (2x - y + 3z = 1), (y - 3z = 0), (-5x + 2y + z = 3)

`\[\begin{bmatrix}1 & 0 & -1 & | & -1 \\2 & -1 & 3 & | & 1 \\0 & 1 & -3 & | & 0 \\-5 & 2 & 1 & | & 3\end{bmatrix}\]`

Performing Gaussian elimination:

`\[\begin{bmatrix}1 & 0 & 0 & | & -2 \\0 & 1 & 0 & | & -1 \\0 & 0 & 1 & | & 1\end{bmatrix}\]`

The solution is (x = -2), (y = -1), (z = 1).

(e) (x_1 + x_2 = 2), (x_2 + x_4 = 1), (x_1 + x_3 = 0), (x_3 - x_4 = 1), (x_1 - x_4 = 2)

`\[\begin{bmatrix}1 & 1 & 0 & 0 & | & 2 \\0 & 1 & 0 & 1 & | & 1 \\1 & 0 & 1 & 0 & | & 0 \\0 & 0 & 1 & -1 & | & 1 \\1 & 0 & 0 & -1 & | & 2\end{bmatrix}\]`

Performing Gaussian elimination:

`\[\begin{bmatrix}1 & 0 & 0 & -1 & | & 2 \\0 & 1 & 0 & 1 & | & 1 \\0 & 0 & 1 & -1 & | & 1 \\0 & 0 & 0 & 0 & | & 0 \\0 & 0 & 0 & 0 & | & 0\end{bmatrix}\]`

The solution is (x_1 = 2), (x_2 = 1), (x_3 = 1), (x_4) is a free parameter.

Complete Question:

The question is to find the least squares solutions for five different overdetermined linear systems. Each system consists of several equations with more equations than unknowns (x, y, u, v, w, x1, x2, x3, x4).