91Ó°ÊÓ

Let's form a matrix \(A\) using the given vectors as columns: \( A = \begin{pmatrix} 1 & 1 & 2 & 2 \\ 1 & 2 & 5 & 6 \\ 1 & 3 & 6 & 8 \\ 1 & 2 & 4 & 5 \\ \end{pmatrix} \)

Now, let's row reduce the matrix (also called Gaussian elimination) to get it into row echelon form: 1. Subtract row 1 from other rows: \( \begin{pmatrix} 1 & 1 & 2 & 2 \\ 0 & 1 & 3 & 4 \\ 0 & 2 & 4 & 6 \\ 0 & 1 & 2 & 3 \\ \end{pmatrix} \) 2. Subtract multiples of row 2 from other rows to get zeros below the second pivot (the \(1\) in second row and second column): \( \begin{pmatrix} 1 & 1 & 2 & 2 \\ 0 & 1 & 3 & 4 \\ 0 & 0 & -2 & -2 \\ 0 & 0 & -1 & -1 \\ \end{pmatrix} \) 3. Multiply row 3 by \(-\frac{1}{2}\) and row 4 by \(-1\) to get simpler matrix: \( \begin{pmatrix} 1 & 1 & 2 & 2 \\ 0 & 1 & 3 & 4 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 1 & 1 \\ \end{pmatrix} \) 4. Subtract multiples of row 3 from other rows to get zeros below the third pivot (the \(1\) in third row and third column): \( \begin{pmatrix} 1 & 1 & 0 & 0 \\ 0 & 1 & 2 & 2 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 \\ \end{pmatrix} \)

Now, the matrix is in row echelon form. We can count the number of non-zero rows: 1. First row: \(1\) 2. Second row: \(2\) 3. Third row: \(3\) 4. Fourth row: \(0\) (zero row) Thus, the number of non-zero rows is 3.

Since there are only 3 non-zero rows in the row reduced matrix, it means that the given set of vectors is not linearly independent and hence they do not form a basis of \(\mathbf{R}^4\). The number of non-zero rows corresponds to the dimension of the subspace spanned by these vectors which is 3. So, the dimension of the subspace spanned by the given vectors is 3.