Given the system defined by:

$A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \\ 5 & 6 \end{bmatrix} \quad \text{and} \quad b = \begin{bmatrix} 7 \\ 8 \\ 9 \end{bmatrix},$

we want to find the least squares solution to $Ax = b$ using two methods:

(a) Without Calculating the Left Inverse

To find the least squares solution without calculating the left inverse, we use the formula:

$x = (A^T A)^{-1} A^T b.$

Let's go through the steps:

1. Calculate $A^T$: $A^T = \begin{bmatrix} 1 & 3 & 5 \\ 2 & 4 & 6 \end{bmatrix}.$

2. Calculate $A^T A$: $A^T A = \begin{bmatrix} 1 & 3 & 5 \\ 2 & 4 & 6 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 3 & 4 \\ 5 & 6 \end{bmatrix} = \begin{bmatrix} 35 & 44 \\ 44 & 56 \end{bmatrix}.$

3. Invert $A^T A$: The inverse of $A^T A$, $(A^T A)^{-1}$, can be computed as:

   $(A^T A)^{-1} = \frac{1}{35 \cdot 56 - 44^2} \begin{bmatrix} 56 & -44 \\ -44 & 35 \end{bmatrix}.$

   Simplifying, we find:

   $(A^T A)^{-1} = \frac{1}{196} \begin{bmatrix} 56 & -44 \\ -44 & 35 \end{bmatrix} = \begin{bmatrix} 0.2857 & -0.2245 \\ -0.2245 & 0.1786 \end{bmatrix}.$

4. Calculate $A^T b$: $A^T b = \begin{bmatrix} 1 & 3 & 5 \\ 2 & 4 & 6 \end{bmatrix} \begin{bmatrix} 7 \\ 8 \\ 9 \end{bmatrix} = \begin{bmatrix} 76 \\ 100 \end{bmatrix}.$

5. Find $x = (A^T A)^{-1} A^T b$: $x = \begin{bmatrix} 0.2857 & -0.2245 \\ -0.2245 & 0.1786 \end{bmatrix} \begin{bmatrix} 76 \\ 100 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \end{bmatrix}.$

So, the least squares solution is:

$x = \begin{bmatrix} 1 \\ 2 \end{bmatrix}.$

(b) By Calculating the Left Inverse

For a matrix $A$ with more rows than columns (tall matrix), there isn’t a true left inverse. Instead, we use the pseudoinverse $A^\dagger = (A^T A)^{-1} A^T$. This is exactly the solution we calculated in part (a).

Thus, we have the same solution:

$x = \begin{bmatrix} 1 \\ 2 \end{bmatrix}.$

Would you like further details or have any questions?

Here are five questions that extend this problem:

1. How does the least squares solution change if we alter the vector $b$?
2. What does it mean geometrically to find a least squares solution for $Ax = b$?
3. How can the least squares solution be interpreted in terms of projections?
4. What conditions are necessary for $A^T A$ to be invertible?
5. How would the least squares solution differ if $A$ were a square matrix?

Tip: When computing least squares solutions, always verify that $A^T A$ is invertible for the approach to work.