Chapter 1 - Linear Equations in Linear Algebra - 1.7 Exercises - Page 63: 44

See solution below. The non pivot columns of A are a linear combination of all the pivot columns of A.

Matrix B has all columns of matrix A except columns 3 and 5. Therefore, let us solve the two equations below: $\begin{bmatrix} 12 & 10 & -3 & 10\\ -7 & -6 & 7 & 5\\ 9 & 9 & -5 & -1\\ -4 & -3 & 6 & 9\\ 8 & 7 & -9 & -8\\ \end{bmatrix}\begin{bmatrix} x_1\\ x_2\\ x_3\\ x_4\\ \end{bmatrix}=\begin{bmatrix} -6\\ 4\\ -9\\ 1\\ -5\\ \end{bmatrix}$ By finding the rref of the augmented matrix, we find $x=\begin{bmatrix} 2\\ -3\\ 0\\ 0\\ \end{bmatrix}$ $\begin{bmatrix} 12 & 10 & -3 & 10\\ -7 & -6 & 7 & 5\\ 9 & 9 & -5 & -1\\ -4 & -3 & 6 & 9\\ 8 & 7 & -9 & -8\\ \end{bmatrix}\begin{bmatrix} x_1\\ x_2\\ x_3\\ x_4\\ \end{bmatrix}=\begin{bmatrix} 7\\ -9\\ 5\\ -8\\ 11\\ \end{bmatrix}$ By finding the rref of the augmented matrix, we find $x=\begin{bmatrix} 2\\ -2\\ -1\\ 0\\ \end{bmatrix}$ Therefore, the columns 3 and 5, not included in Matrix B span the columns of B. This shows that the non pivot columns of A are a linear combination of all the pivot columns of A.