91影视

For finding the characteristic equation, we need to find the determinant of (A - 位I). We are given matrix A as: $$ A = \left[\begin{array}{rrr} 1 & -4 & -1 \\ 3 & 2 & 3 \\ 1 & 1 & 3 \\ \end{array}\right] $$ Now let's calculate A - 位I $$ A - 位I = \left[\begin{array}{rrr} 1-位 & -4 & -1 \\ 3 & 2-位 & 3 \\ 1 & 1 & 3-位 \\ \end{array}\right] $$ Now find the determinant: $$ \mbox{det}(A-位 I) = (1-位) \Bigl[(2-位)(3-位) - (3)\cdot(1)\Bigr] + (-4) \Bigl[(3)(1)-(1)(3)\Bigr] + (-1) \Bigl[(3)(1)-(3)\cdot(1)\Bigr] $$

Now let's solve the resulting characteristic equation: $$ \mbox{det}(A-位 I)= - 位^3 + 6 位^2 - 11 位 + 6 = 0 $$ By factoring, we get: $$ (位 - 1)(位 - 2)(位 - 3) = 0 $$ Hence, the eigenvalues are 位鈧� = 1, 位鈧� = 2, and 位鈧� = 3.

Now, let's find the eigenvectors for each eigenvalue. a. For 位鈧� = 1, we need to find the null space of (A - I): $$ A - I = \left[\begin{array}{rrr} 0 & -4 & -1 \\ 3 & 1 & 3 \\ 1 & 1 & 2 \\ \end{array}\right] $$ With row reduction: $$ \left[\begin{array}{rrr} 1 & 1 & 2 \\ 0 & -4 & -1 \\ 0 & 0 & 0 \\ \end{array}\right] $$ From this, we get the eigenvector corresponding to 位鈧� as: $$ v_1 = \left[\begin{array}{c} x_1 \\ x_2 \\ x_3 \end{array}\right]= t\left[\begin{array}{c} 1 \\ 1 \\ -2 \end{array}\right],\ t\neq0 $$ b. For 位鈧� = 2, we need to find the null space of (A - 2I): $$ A - 2I = \left[\begin{array}{rrr} -1 & -4 & -1 \\ 3 & 0 & 3 \\ 1 & 1 & 1 \\ \end{array}\right] $$ With row reduction: $$ \left[\begin{array}{rrr} 1 & 1 & 1 \\ 0 & -1 & 1 \\ 0 & 0 & 0 \\ \end{array}\right] $$ From this, we get the eigenvector corresponding to 位鈧� as: $$ v_2 = \left[\begin{array}{c} x_1 \\ x_2 \\ x_3 \end{array}\right]= t\left[\begin{array}{c} 1 \\ -1 \\ 1 \end{array}\right],\ t\neq0 $$ c. For 位鈧� = 3, we need to find the null space of (A - 3I): $$ A - 3I = \left[\begin{array}{rrr} -2 & -4 & -1 \\ 3 & -1 & 3 \\ 1 & 1 & 0 \\ \end{array}\right] $$ With row reduction: $$ \left[\begin{array}{rrr} 1 & 2 & \frac{1}{2} \\ 0 & 1 & -\frac{1}{2} \\ 0 & 0 & 0 \\ \end{array}\right] $$ From this, we get the eigenvector corresponding to 位鈧� as: $$ v_3 = \left[\begin{array}{c} x_1 \\ x_2 \\ x_3 \end{array}\right]= t\left[\begin{array}{c} -1 \\ 2 \\ 4 \end{array}\right],\ t\neq0 $$ To summarize, the eigenvalues and their corresponding eigenvectors are: 位鈧� = 1 with eigenvector v鈧� = t[1, 1, -2] 位鈧� = 2 with eigenvector v鈧� = t[1, -1, 1] 位鈧� = 3 with eigenvector v鈧� = t[-1, 2, 4]