Solvability Condition Analysis for Matrix Equation

The matrix equation $ \mathbf{Ax} = \mathbf{b} $ is solvable if and only if the vector $ \mathbf{b} $ lies within the column space of matrix $ \mathbf{A} $. This is equivalent to stating that the rank of the augmented matrix $ [\mathbf{A} | \mathbf{b}] $ must be equal to the rank of the matrix $ \mathbf{A} $.

Rank Calculation for Matrix $ \mathbf{A} $

Given matrix $ \mathbf{A} $: $ \mathbf{A} = \begin{bmatrix} 1 & 1 \\ 1 & 3 \\ -2 & -3 \end{bmatrix} $ We perform elementary row operations to determine its rank:

1. Replace Row 2 ($ R_2 $) with $ R_2 - R_1 $: $ \begin{bmatrix} 1 & 1 \\ 0 & 2 \\ -2 & -3 \end{bmatrix} $
2. Replace Row 3 ($ R_3 $) with $ R_3 + 2R_1 $: $ \begin{bmatrix} 1 & 1 \\ 0 & 2 \\ 0 & -1 \end{bmatrix} $
3. Replace Row 3 ($ R_3 $) with $ R_3 + \frac{1}{2}R_2 $: $ \begin{bmatrix} 1 & 1 \\ 0 & 2 \\ 0 & 0 \end{bmatrix} $

The resulting matrix has 2 non-zero rows. Therefore, the rank of $ \mathbf{A} $ is 2.

Rank Condition for Augmented Matrix $ [\mathbf{A} | \mathbf{b}] $

For the system $ \mathbf{Ax} = \mathbf{b} $ to be solvable, the rank of the augmented matrix $ [\mathbf{A} | \mathbf{b}] $ must also be 2. The augmented matrix is:

Applying the same row operations as above:

1. $ R_2 \leftarrow R_2 - R_1 $: $ \begin{bmatrix} 1 & 1 & b_1 \\ 0 & 2 & b_2 - b_1 \\ -2 & -3 & b_3 \end{bmatrix} $
2. $ R_3 \leftarrow R_3 + 2R_1 $: $ \begin{bmatrix} 1 & 1 & b_1 \\ 0 & 2 & b_2 - b_1 \\ 0 & -1 & b_3 + 2b_1 \end{bmatrix} $
3. $ R_3 \leftarrow R_3 + \frac{1}{2}R_2 $: $ \begin{bmatrix} 1 & 1 & b_1 \\ 0 & 2 & b_2 - b_1 \\ 0 & 0 & (b_3 + 2b_1) + \frac{1}{2}(b_2 - b_1) \end{bmatrix} $

Deriving the Condition on $ \mathbf{b} $

For the rank of the augmented matrix to remain 2, the last row must consist entirely of zeros. Specifically, the element in the third column must be zero:

Multiply the equation by 2 to eliminate the fraction:

Combine like terms to get the final condition:

This condition must hold for the system $ \mathbf{Ax} = \mathbf{b} $ to be solvable.