To solve the given problem, we need to determine the value of $m$ such that the vectors $\mathbf{a} = (1, 1, -2)$, $\mathbf{b} = (m, 1, -1)$, and $\mathbf{c} = (m + 1, -4, 3)$ are linearly dependent.

Three vectors $\mathbf{a}$, $\mathbf{b}$, and $\mathbf{c}$ are linearly dependent if the determinant of the matrix formed by these vectors as columns is zero. We form the matrix:

1 & m & m+1 \\ 1 & 1 & -4 \\ -2 & -1 & 3 \end{pmatrix}$$ Now, we calculate the determinant of this matrix: $$\text{det}(A) = \begin{vmatrix} 1 & m & m+1 \\ 1 & 1 & -4 \\ -2 & -1 & 3 \end{vmatrix}$$ Using the cofactor expansion along the first row: $$\text{det}(A) = 1 \cdot \begin{vmatrix} 1 & -4 \\ -1 & 3 \end{vmatrix} - m \cdot \begin{vmatrix} 1 & -4 \\ -2 & 3 \end{vmatrix} + (m+1) \cdot \begin{vmatrix} 1 & 1 \\ -2 & -1 \end{vmatrix}$$ Calculating each 2x2 determinant: $$\begin{vmatrix} 1 & -4 \\ -1 & 3 \end{vmatrix} = (1 \cdot 3) - (-4 \cdot -1) = 3 - 4 = -1$$ $$\begin{vmatrix} 1 & -4 \\ -2 & 3 \end{vmatrix} = (1 \cdot 3) - (-4 \cdot -2) = 3 - 8 = -5$$ $$\begin{vmatrix} 1 & 1 \\ -2 & -1 \end{vmatrix} = (1 \cdot -1) - (1 \cdot -2) = -1 + 2 = 1$$ Substituting these values back into the determinant calculation: $$\text{det}(A) = 1 \cdot (-1) - m \cdot (-5) + (m+1) \cdot 1$$ $$\text{det}(A) = -1 + 5m + m + 1 = 6m$$ For the vectors to be linearly dependent, $$\text{det}(A) = 0$$: $$6m = 0 \implies m = 0$$ Therefore, the value of $$m$$ that makes the vectors linearly dependent is $$m = 0$$. Would you like more details or have any questions? --- Here are 8 related questions to expand your understanding: 1. What does it mean for vectors to be linearly dependent? 2. How do you form a matrix from a set of vectors? 3. What is the significance of the determinant of a matrix? 4. How do you calculate the determinant of a 3x3 matrix? 5. What are some properties of determinants? 6. How can you determine if more than three vectors are linearly dependent? 7. How does the concept of linear dependence relate to vector spaces? 8. Can the method used in this problem be applied to higher-dimensional spaces? **Tip:** Always double-check your determinant calculations, as a small error can lead to an incorrect conclusion about linear dependence.