Let A=( 3 -3 1 2 4 1 1 2 1 ), 𝐛=( 6 5 4

Let A=( 3 -3 1 2 4 1 1 2...

Let $A=\left(\begin{array}{rrr}3 & -3 & 1 \\ 2 & 4 & 1 \\ 1 & 2 & 1\end{array}\right), \mathbf{b}=\left(\begin{array}{l}6 \\ 5 \\ 4\end{array}\right)$. Prove, using Gaussian Elimination, that the linear system $A \mathbf{x}=\mathbf{b}$ has a unique solution. Show that the least squares solution (5.34) is the same. Explain why this is necessarily the case.

Key Concepts

Full Rank and Invertibility

A matrix is said to have full rank if its rank is equal to the number of its rows or columns. In the context of a square matrix, full rank implies invertibility. This property ensures that the system of linear equations has a unique solution, and therefore the normal equations used in the least squares method lead to this unique solution rather than an approximation.

Least Squares Method

The least squares method is a technique used to find an approximate solution to an over-determined or under-determined system by minimizing the sum of the squares of the residuals. When the system has a unique solution (typically because the matrix is invertible), the least squares solution coincides with the exact solution of the linear system.

Gaussian Elimination

Gaussian elimination is a systematic method for solving systems of linear equations by reducing a matrix to its row echelon form (or even reduced row echelon form) using elementary row operations. This process helps in identifying the pivot positions which indicate the variables that can be solved step by step, ensuring that if the system is consistent and the matrix has full rank, a unique solution exists.

Unique Solution in Linear Systems

The concept of a unique solution in a linear system occurs when the coefficient matrix is invertible, meaning its determinant is non-zero and it has full rank. In such cases, the system of equations has one and only one solution, which can be found by methods such as Gaussian elimination or by directly computing the inverse of the matrix.

Transcript

00:01 In this video, we're going to go through the answer to question number 14 from chapter 9 .2, where we're asked to find the solution to this system of equations for when r is equal to minus 1 and for when r is equal to 2.

00:16 So firstly, we want to write this in a way that we're familiar with.

00:19 So currently we have some x1s, x2s, x3s on the right -hand side as well as the left -hand side.

00:27 So we want to bring these over to the left -hand side to write it in a way.

00:32 That we're familiar with.

00:36 So when we do that, we now have ours on the left -hand side as well.

00:41 So then we substitute in first for r is equal to minus 1, and we reach this set of equations.

00:50 So to solve these, let's add the first equation to the second equation that can get rid of these x -3s.

01:00 We're now going to have three lots of x -2s, and we're going to have three lots of x -1s.

01:06 If we had to six lots of the first equation to the third equation, we get rid of these x -3s.

01:13 We're going to have minus four plus 12 x2.

01:18 So this is going to become plus eight lots of x -2.

01:25 We're going to have four plus 12 lots of x -1.

01:30 So this is going to become 16 lots of x -1.

01:36 So now we have, if we just look at the new equation two and the new equation three, we have 3x1 plus 3x2 is equal to 0 and 16x1 plus 8 x2 is equal to 0.

01:51 So if we were to subtract, for example, 8 over 3 lots of equation 2 from equation 3, then we could get rid of these x2s and we would end up with some number of x1s is equal to 0.

02:16 So from that we can find out that x1 is going to be 0.

02:23 So now we can get rid of all the x1s because they're all zero.

02:27 So now from equation 2 we have three lots of x2 is equal to 0...

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