Question:

Let $v_1 = \begin{bmatrix} 1 \\ -1 \end{bmatrix}$. We are given the following relationships: $Pv_1 = \begin{bmatrix} -1 \\ 2 \end{bmatrix}$ $P^2v_1 = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$ Step 1: Determine the matrix P. Let $v_2 = Pv_1 = \begin{bmatrix} -1 \\ 2 \end{bmatrix}$. Then $P^2v_1 = P(Pv_1) = Pv_2 = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$. We can express P using the column vectors $v_1$ and $v_2$: $P \begin{bmatrix} v_1 & v_2 \end{bmatrix} = \begin{bmatrix} Pv_1 & Pv_2 \end{bmatrix}$ Substituting the known vectors: $$P \begin{bmatrix} 1 & -1 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} -1 & 1 \\ 2 & 0 \end{bmatrix}$$ To find P, we multiply both sides by the inverse of the matrix $\begin{bmatrix} 1 & -1 \\ -1 & 2 \end{bmatrix}$. The determinant of $\begin{bmatrix} 1 & -1 \\ -1 & 2 \end{bmatrix}$ is $(1)(2) - (-1)(-1) = 2 - 1 = 1$. The inverse matrix is $\frac{1}{1} \begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix}$. Now, we can calculate P: $$P = \begin{bmatrix} -1 & 1 \\ 2 & 0 \end{bmatrix} \begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix}$$ $$P = \begin{bmatrix} (-1)(2) + (1)(1) & (-1)(1) + (1)(1) \\ (2)(2) + (0)(1) & (2)(1) + (0)(1) \end{bmatrix}$$ $$P = \begin{bmatrix} -2+1 & -1+1 \\ 4+0 & 2+0 \end{bmatrix}$$ $$P = \begin{bmatrix} -1 & 0 \\ 4 & 2 \end{bmatrix}$$ Step 2: Find the eigenvalues $p_1$ and $p_2$. The eigenvalues $p$ are the roots of the characteristic equation $\det(P - pI) = 0$, where I is the identity matrix. $$P - pI = \begin{bmatrix} -1-p & 0 \\ 4 & 2-p \end{bmatrix}$$ The determinant is: $$\det(P - pI) = (-1-p)(2-p) - (0)(4)$$ $$= -(p+1)(2-p)$$ $$= (p+1)(p-2)$$ Setting the determinant to zero to find the eigenvalues: $$(p+1)(p-2) = 0$$ The eigenvalues are $p = -1$ and $p = 2$. Given that $p_1 > p_2$, we assign $p_1 = 2$ and $p_2 = -1$. Step 3: Calculate the value of $5p_1 + 2p_2$. Substitute the values of $p_1$ and $p_2$: $$5p_1 + 2p_2 = 5(2) + 2(-1)$$ $$= 10 - 2$$ $$= 8$$