91Ó°ÊÓ

To determine if \(\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\) form a basis for \(\mathbb{R}^3\), we need to check if they are linearly independent. This can be checked by setting up the matrix with these vectors as columns and finding the determinant. If the determinant is non-zero, they form a basis. The matrix is: \[A = \begin{bmatrix} 1 & 2 & 0 \1 & -1 & 1 \1 & -1 & -1\end{bmatrix}.\]The determinant of \(A\) is computed as follows: \[\text{det}(A) = 1\big((-1)(-1) - (1)(1)\big) - 2\big((1)(-1) - (1)(1)\big) + 0\big((1)(1) - (-1)(-1)\big).\]Simplifying this gives:\[\text{det}(A) = 1(1 - 1) - 2(-1 - 1).\]\[\text{det}(A) = 0 + 4 = 4.\]Since the determinant is not zero, \(\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\) form a basis for \(\mathbb{R}^3\).

To confirm if these vectors form an orthogonal basis, calculate the dot product between each pair of vectors. An orthogonal basis requires the dot product of different vectors to be zero.1. \(\mathbf{v}_1 \cdot \mathbf{v}_2 = 1 \times 2 + 1 \times (-1) + 1 \times (-1) = 2 - 1 - 1 = 0\)2. \(\mathbf{v}_1 \cdot \mathbf{v}_3 = 1 \times 0 + 1 \times 1 + 1 \times (-1) = 0 + 1 - 1 = 0\)3. \(\mathbf{v}_2 \cdot \mathbf{v}_3 = 2 \times 0 + (-1) \times 1 + (-1) \times (-1) = 0 - 1 + 1 = 0\)Since all dot products between different vectors are zero, \(\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\) are indeed orthogonal.

Since the vectors form an orthogonal basis, we can express \(\mathbf{w}\) as a linear combination of these basis vectors. For an orthogonal basis, the coefficients can be found using the formula:\[c_i = \frac{\mathbf{w} \cdot \mathbf{v}_i}{\mathbf{v}_i \cdot \mathbf{v}_i},\]where \(c_i\) is the coefficient of the vector \(\mathbf{v}_i\).- Calculating \(c_1\): \(\mathbf{w} \cdot \mathbf{v}_1 = 1 \times 1 + 2 \times 1 + 3 \times 1 = 1 + 2 + 3 = 6\) \(\mathbf{v}_1 \cdot \mathbf{v}_1 = 1^2 + 1^2 + 1^2 = 3\) \(c_1 = \frac{6}{3} = 2\)- Calculating \(c_2\): \(\mathbf{w} \cdot \mathbf{v}_2 = 1 \times 2 + 2 \times (-1) + 3 \times (-1) = 2 - 2 - 3 = -3\) \(\mathbf{v}_2 \cdot \mathbf{v}_2 = 2^2 + (-1)^2 + (-1)^2 = 6\) \(c_2 = \frac{-3}{6} = -\frac{1}{2}\)- Calculating \(c_3\): \(\mathbf{w} \cdot \mathbf{v}_3 = 1 \times 0 + 2 \times 1 + 3 \times (-1) = 0 + 2 - 3 = -1\) \(\mathbf{v}_3 \cdot \mathbf{v}_3 = 0^2 + 1^2 + (-1)^2 = 2\) \(c_3 = \frac{-1}{2}\)Thus, \(\mathbf{w} = 2\mathbf{v}_1 - \frac{1}{2}\mathbf{v}_2 - \frac{1}{2}\mathbf{v}_3\).

The coordinate vector of \(\mathbf{w}\) with respect to the basis \(\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\}\) is given by the coefficients we found:\[[\mathbf{w}]_{\mathcal{B}} = \begin{bmatrix} 2 \-\frac{1}{2} \-\frac{1}{2} \end{bmatrix}.\]This vector contains the weights of \(\mathbf{v}_1, \mathbf{v}_2,\) and \(\mathbf{v}_3\) in expressing \(\mathbf{w}\).