Answer on Question #55838 – Math – Linear Algebra
1. Solve for x x x y y y 3 x + 4 y = 9 3x + 4y = 9 3 x + 4 y = 9 2 x + 3 y = 8 2x + 3y = 8 2 x + 3 y = 8
x = 7.5 , y = − 4.5 x = 7.0 , y = − 4.5 x = 4.5 , y = − 7.5 x = 7.5 , y = − 4.1 \begin{array}{l}
x = 7.5, y = -4.5 \\
x = 7.0, y = -4.5 \\
x = 4.5, y = -7.5 \\
x = 7.5, y = -4.1 \\
\end{array} x = 7.5 , y = − 4.5 x = 7.0 , y = − 4.5 x = 4.5 , y = − 7.5 x = 7.5 , y = − 4.1 Solution
{ 3 x + 4 y = 9 2 x + 3 y = 8 \left\{
\begin{array}{l}
3x + 4y = 9 \\
2x + 3y = 8
\end{array}
\right. { 3 x + 4 y = 9 2 x + 3 y = 8 ∣ 3 4 ∣ 9 ∣ 2 3 ∣ 8 ∣ \left| \begin{array}{ccc}
3 & 4 & |9| \\
2 & 3 & |8|
\end{array} \right. ∣ ∣ 3 2 4 3 ∣9∣ ∣8∣
Using Cramer’s rule
Δ = ∣ 3 4 2 3 ∣ = 3 ⋅ 3 − 2 ⋅ 4 = 9 − 8 = 1 \Delta = \left| \begin{array}{cc}
3 & 4 \\
2 & 3
\end{array} \right| = 3 \cdot 3 - 2 \cdot 4 = 9 - 8 = 1 Δ = ∣ ∣ 3 2 4 3 ∣ ∣ = 3 ⋅ 3 − 2 ⋅ 4 = 9 − 8 = 1 Δ x = ∣ 9 4 8 3 ∣ = 9 ⋅ 3 − 8 ⋅ 4 = 27 − 32 = − 5 \Delta_x = \left| \begin{array}{cc}
9 & 4 \\
8 & 3
\end{array} \right| = 9 \cdot 3 - 8 \cdot 4 = 27 - 32 = -5 Δ x = ∣ ∣ 9 8 4 3 ∣ ∣ = 9 ⋅ 3 − 8 ⋅ 4 = 27 − 32 = − 5 Δ y = ∣ 3 9 2 8 ∣ = 3 ⋅ 8 − 2 ⋅ 9 = 24 − 18 = 6 \Delta_y = \left| \begin{array}{cc}
3 & 9 \\
2 & 8
\end{array} \right| = 3 \cdot 8 - 2 \cdot 9 = 24 - 18 = 6 Δ y = ∣ ∣ 3 2 9 8 ∣ ∣ = 3 ⋅ 8 − 2 ⋅ 9 = 24 − 18 = 6 x = Δ x Δ = − 5 1 = − 5 x = \frac{\Delta_x}{\Delta} = \frac{-5}{1} = -5 x = Δ Δ x = 1 − 5 = − 5 y = Δ y Δ = 6 1 = 6 y = \frac{\Delta_y}{\Delta} = \frac{6}{1} = 6 y = Δ Δ y = 1 6 = 6
**Answer:** x = − 5 , y = 6 x = -5, y = 6 x = − 5 , y = 6
2. If A . x = λ x A.x = \lambda x A . x = λ x A = ∣ 1 2 1 2 1 3 − 2 1 2 ∣ A = \left| \begin{array}{ccc} 1 & 2 & 1 \\ 2 & 1 & 3 \\ -2 & 1 & 2 \end{array} \right| A = ∣ ∣ 1 2 − 2 2 1 1 1 3 2 ∣ ∣ A A A
If
λ = 2 \lambda = 2 λ = 2
, what is b b b
Solution
A ^ x = λ x \hat{A}x = \lambda x A ^ x = λ x A ^ = [ 2 1 1 2 3 2 − 2 1 2 ] \hat{A} = \left[ \begin{array}{ccc}
2 & 1 & 1 \\
2 & 3 & 2 \\
-2 & 1 & 2
\end{array} \right] A ^ = ⎣ ⎡ 2 2 − 2 1 3 1 1 2 2 ⎦ ⎤
Let’s determine the eigenvalues of matrix A A A
A ^ x = λ E ^ x \hat{A}x = \lambda \hat{E}x A ^ x = λ E ^ x A ^ x − λ E ^ x = 0 \hat{A}x - \lambda \hat{E}x = 0 A ^ x − λ E ^ x = 0 ( A ^ − λ E ^ ) x = 0 (\hat{A} - \lambda \hat{E})x = 0 ( A ^ − λ E ^ ) x = 0 det [ 2 − λ 1 1 2 3 − λ 2 − 2 1 2 − λ ] = 0 \det \begin{bmatrix}
2 - \lambda & 1 & 1 \\
2 & 3 - \lambda & 2 \\
-2 & 1 & 2 - \lambda
\end{bmatrix} = 0 det ⎣ ⎡ 2 − λ 2 − 2 1 3 − λ 1 1 2 2 − λ ⎦ ⎤ = 0 ∣ 2 − λ 1 1 2 3 − λ 2 − 2 1 2 − λ ∣ = 0 \begin{vmatrix}
2 - \lambda & 1 & 1 \\
2 & 3 - \lambda & 2 \\
-2 & 1 & 2 - \lambda
\end{vmatrix} = 0 ∣ ∣ 2 − λ 2 − 2 1 3 − λ 1 1 2 2 − λ ∣ ∣ = 0 ( 2 − λ ) ⋅ ( 3 − λ ) ⋅ ( 2 − λ ) + 1 ⋅ 2 ⋅ ( − 2 ) + 2 ⋅ 1 ⋅ 1 − ( − 2 ) ⋅ ( 3 − λ ) ⋅ 1 − 1 ⋅ 2 ⋅ ( 2 − λ ) − 1 ⋅ 2 ⋅ ( 2 − λ ) = 0 (2 - \lambda) \cdot (3 - \lambda) \cdot (2 - \lambda) + 1 \cdot 2 \cdot (-2) + 2 \cdot 1 \cdot 1 - (-2) \cdot (3 - \lambda) \cdot 1 - 1 \cdot 2 \cdot (2 - \lambda) - 1 \cdot 2 \cdot (2 - \lambda) = 0 ( 2 − λ ) ⋅ ( 3 − λ ) ⋅ ( 2 − λ ) + 1 ⋅ 2 ⋅ ( − 2 ) + 2 ⋅ 1 ⋅ 1 − ( − 2 ) ⋅ ( 3 − λ ) ⋅ 1 − 1 ⋅ 2 ⋅ ( 2 − λ ) − 1 ⋅ 2 ⋅ ( 2 − λ ) = 0 ( 2 − λ ) 2 ⋅ ( 3 − λ ) − 4 + 2 + 6 − 2 λ + 4 λ − 8 = 0 (2 - \lambda)^2 \cdot (3 - \lambda) - 4 + 2 + 6 - 2\lambda + 4\lambda - 8 = 0 ( 2 − λ ) 2 ⋅ ( 3 − λ ) − 4 + 2 + 6 − 2 λ + 4 λ − 8 = 0 ( 2 − λ ) 2 ⋅ ( 3 − λ ) + 2 λ − 4 = 0 (2 - \lambda)^2 \cdot (3 - \lambda) + 2\lambda - 4 = 0 ( 2 − λ ) 2 ⋅ ( 3 − λ ) + 2 λ − 4 = 0 ( 2 − λ ) 2 ⋅ ( 3 − λ ) − 2 ( 2 − λ ) = 0 (2 - \lambda)^2 \cdot (3 - \lambda) - 2(2 - \lambda) = 0 ( 2 − λ ) 2 ⋅ ( 3 − λ ) − 2 ( 2 − λ ) = 0 ( 2 − λ ) ⋅ ( ( 2 − λ ) ⋅ ( 3 − λ ) − 2 ) = 0 (2 - \lambda) \cdot ((2 - \lambda) \cdot (3 - \lambda) - 2) = 0 ( 2 − λ ) ⋅ (( 2 − λ ) ⋅ ( 3 − λ ) − 2 ) = 0 ( 2 − λ ) ⋅ ( λ 2 − 5 λ + 6 − 2 ) = 0 (2 - \lambda) \cdot (\lambda^2 - 5\lambda + 6 - 2) = 0 ( 2 − λ ) ⋅ ( λ 2 − 5 λ + 6 − 2 ) = 0 ( 2 − λ ) ⋅ ( λ 2 − 5 λ + 4 ) = 0 (2 - \lambda) \cdot (\lambda^2 - 5\lambda + 4) = 0 ( 2 − λ ) ⋅ ( λ 2 − 5 λ + 4 ) = 0 λ 1 = 2 \lambda_1 = 2 λ 1 = 2 λ 2 − 5 λ + 4 = 0 \lambda^2 - 5\lambda + 4 = 0 λ 2 − 5 λ + 4 = 0 D = 25 − 16 = 9 D = 25 - 16 = 9 D = 25 − 16 = 9 λ 2 = 4 λ 3 = 1 \lambda_2 = 4 \quad \lambda_3 = 1 λ 2 = 4 λ 3 = 1 λ = 1 ; λ = 2 , λ = 4 \lambda = 1; \lambda = 2, \lambda = 4 λ = 1 ; λ = 2 , λ = 4
Determine the eigen vector corresponding to each eigen value:
λ = 1 : \lambda = 1: λ = 1 : ∣ 2 − 1 1 1 0 2 3 − 1 2 0 − 2 1 2 − 1 0 ∣ = 0 \begin{vmatrix}
2 - 1 & 1 & 1 & 0 \\
2 & 3 - 1 & 2 & 0 \\
-2 & 1 & 2 - 1 & 0
\end{vmatrix} = 0 ∣ ∣ 2 − 1 2 − 2 1 3 − 1 1 1 2 2 − 1 0 0 0 ∣ ∣ = 0 ∣ 1 1 1 0 2 2 2 0 − 2 1 1 0 ∣ = ∣ 1 1 1 0 0 0 0 0 0 3 3 0 ∣ = ∣ 1 1 1 0 0 3 3 0 ∣ \begin{vmatrix}
1 & 1 & 1 & 0 \\
2 & 2 & 2 & 0 \\
-2 & 1 & 1 & 0
\end{vmatrix} = \begin{vmatrix}
1 & 1 & 1 & 0 \\
0 & 0 & 0 & 0 \\
0 & 3 & 3 & 0
\end{vmatrix} = \begin{vmatrix}
1 & 1 & 1 & 0 \\
0 & 3 & 3 & 0
\end{vmatrix} ∣ ∣ 1 2 − 2 1 2 1 1 2 1 0 0 0 ∣ ∣ = ∣ ∣ 1 0 0 1 0 3 1 0 3 0 0 0 ∣ ∣ = ∣ ∣ 1 0 1 3 1 3 0 0 ∣ ∣ { x 1 + x 2 + x 3 = 0 0 + x 2 + x 3 = 0 = = > { x 1 = 0 x 2 + x 3 = 0 \begin{cases}
x_1 + x_2 + x_3 = 0 \\
0 + x_2 + x_3 = 0
\end{cases}
= = > \begin{cases}
x_1 & = 0 \\
x_2 + x_3 = 0
\end{cases} { x 1 + x 2 + x 3 = 0 0 + x 2 + x 3 = 0 ==> { x 1 x 2 + x 3 = 0 = 0 x 1 = 0 x_1 = 0 x 1 = 0 x 2 = − x 3 x_2 = -x_3 x 2 = − x 3 X = [ x 1 x 2 x 3 ] = [ 0 − x 3 x 3 ] = x 3 [ 0 − 1 1 ] , x 3 ≠ 0. X = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 0 \\ -x_3 \\ x_3 \end{bmatrix} = x_3 \begin{bmatrix} 0 \\ -1 \\ 1 \end{bmatrix}, x_3 \neq 0. X = ⎣ ⎡ x 1 x 2 x 3 ⎦ ⎤ = ⎣ ⎡ 0 − x 3 x 3 ⎦ ⎤ = x 3 ⎣ ⎡ 0 − 1 1 ⎦ ⎤ , x 3 = 0. λ = 2 : \lambda = 2: λ = 2 : ∣ 2 − 2 1 1 0 2 3 − 2 2 0 − 2 1 2 − 2 0 ∣ = 0 \begin{vmatrix}
2 - 2 & 1 & 1 & 0 \\
2 & 3 - 2 & 2 & 0 \\
-2 & 1 & 2 - 2 & 0
\end{vmatrix}
= 0 ∣ ∣ 2 − 2 2 − 2 1 3 − 2 1 1 2 2 − 2 0 0 0 ∣ ∣ = 0 ∣ 0 1 1 0 2 1 2 0 − 2 1 0 0 ∣ = ∣ 0 1 1 0 1 1 2 1 0 0 2 2 0 ∣ = ∣ 0 1 1 0 1 1 2 1 0 0 1 1 0 ∣ \begin{vmatrix}
0 & 1 & 1 & 0 \\
2 & 1 & 2 & 0 \\
-2 & 1 & 0 & 0
\end{vmatrix}
= \begin{vmatrix}
0 & 1 & 1 & 0 \\
1 & \frac{1}{2} & 1 & 0 \\
0 & 2 & 2 & 0
\end{vmatrix}
= \begin{vmatrix}
0 & 1 & 1 & 0 \\
1 & \frac{1}{2} & 1 & 0 \\
0 & 1 & 1 & 0
\end{vmatrix} ∣ ∣ 0 2 − 2 1 1 1 1 2 0 0 0 0 ∣ ∣ = ∣ ∣ 0 1 0 1 2 1 2 1 1 2 0 0 0 ∣ ∣ = ∣ ∣ 0 1 0 1 2 1 1 1 1 1 0 0 0 ∣ ∣ { 0 + x 2 + x 3 = 0 x 1 + 1 2 x 2 + x 3 = 0 = ⇒ { x 1 = − 1 2 x 3 x 2 = − x 3 \begin{cases}
0 + x_2 + x_3 = 0 \\
x_1 + \frac{1}{2}x_2 + x_3 = 0
\end{cases}
= \Rightarrow \begin{cases}
x_1 = -\frac{1}{2}x_3 \\
x_2 = -x_3
\end{cases} { 0 + x 2 + x 3 = 0 x 1 + 2 1 x 2 + x 3 = 0 =⇒ { x 1 = − 2 1 x 3 x 2 = − x 3 x 1 = − 1 2 x 3 x_1 = -\frac{1}{2}x_3 x 1 = − 2 1 x 3 x 2 = − x 3 x_2 = -x_3 x 2 = − x 3 X = [ x 1 x 2 x 3 ] = [ − 1 2 x 3 − x 3 x 3 ] = x 3 [ − 1 2 − 1 1 ] , x 3 ≠ 0. X = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} -\frac{1}{2}x_3 \\ -x_3 \\ x_3 \end{bmatrix} = x_3 \begin{bmatrix} -\frac{1}{2} \\ -1 \\ 1 \end{bmatrix}, x_3 \neq 0. X = ⎣ ⎡ x 1 x 2 x 3 ⎦ ⎤ = ⎣ ⎡ − 2 1 x 3 − x 3 x 3 ⎦ ⎤ = x 3 ⎣ ⎡ − 2 1 − 1 1 ⎦ ⎤ , x 3 = 0. λ = 4 : \lambda = 4: λ = 4 : ∣ 2 − 4 1 1 0 2 3 − 4 2 0 − 2 1 2 − 4 0 ∣ = 0 \begin{vmatrix}
2 - 4 & 1 & 1 & 0 \\
2 & 3 - 4 & 2 & 0 \\
-2 & 1 & 2 - 4 & 0
\end{vmatrix}
= 0 ∣ ∣ 2 − 4 2 − 2 1 3 − 4 1 1 2 2 − 4 0 0 0 ∣ ∣ = 0 ∣ − 2 1 1 0 2 − 1 2 0 − 2 1 − 2 0 ∣ = ∣ 1 − 1 2 − 1 2 1 − 1 2 1 1 − 1 2 1 ∣ = ∣ 1 − 1 2 − 1 2 1 − 1 2 1 ∣ \begin{vmatrix}
-2 & 1 & 1 & 0 \\
2 & -1 & 2 & 0 \\
-2 & 1 & -2 & 0
\end{vmatrix}
= \begin{vmatrix}
1 & -\frac{1}{2} & -\frac{1}{2} \\
1 & -\frac{1}{2} & 1 \\
1 & -\frac{1}{2} & 1
\end{vmatrix}
= \begin{vmatrix}
1 & -\frac{1}{2} & -\frac{1}{2} \\
1 & -\frac{1}{2} & 1
\end{vmatrix} ∣ ∣ − 2 2 − 2 1 − 1 1 1 2 − 2 0 0 0 ∣ ∣ = ∣ ∣ 1 1 1 − 2 1 − 2 1 − 2 1 − 2 1 1 1 ∣ ∣ = ∣ ∣ 1 1 − 2 1 − 2 1 − 2 1 1 ∣ ∣ { x 1 − 1 2 x 2 − 1 2 x 3 = 0 x 1 − 1 2 x 2 + x 3 = 0 = ⇒ { x 1 = 1 2 x 2 x 3 = 0 \begin{cases}
x_1 - \frac{1}{2}x_2 - \frac{1}{2}x_3 = 0 \\
x_1 - \frac{1}{2}x_2 + x_3 = 0
\end{cases}
= \Rightarrow \begin{cases}
x_1 = \frac{1}{2}x_2 \\
x_3 = 0
\end{cases} { x 1 − 2 1 x 2 − 2 1 x 3 = 0 x 1 − 2 1 x 2 + x 3 = 0 =⇒ { x 1 = 2 1 x 2 x 3 = 0 x 1 = 1 2 x 2 x_1 = \frac{1}{2}x_2 x 1 = 2 1 x 2 x 3 = 0 x_3 = 0 x 3 = 0 X = [ x 1 x 2 x 3 ] = [ 1 2 x 2 x 2 0 ] = x 2 [ 1 2 1 0 ] , x 2 ≠ 0. X = \left[ \begin{array}{c} x _ {1} \\ x _ {2} \\ x _ {3} \end{array} \right] = \left[ \begin{array}{c} \frac {1}{2} x _ {2} \\ x _ {2} \\ 0 \end{array} \right] = x _ {2} \left[ \begin{array}{c} \frac {1}{2} \\ 1 \\ 0 \end{array} \right], x _ {2} \neq 0. X = ⎣ ⎡ x 1 x 2 x 3 ⎦ ⎤ = ⎣ ⎡ 2 1 x 2 x 2 0 ⎦ ⎤ = x 2 ⎣ ⎡ 2 1 1 0 ⎦ ⎤ , x 2 = 0.
If λ = 2 \lambda = 2 λ = 2
b = [ − 1 2 x 3 − x 3 x 3 ] = x 3 [ − 1 2 − 1 1 ] , x 3 ≠ 0 b = \left[ \begin{array}{c} - \frac {1}{2} x _ {3} \\ - x _ {3} \\ x _ {3} \end{array} \right] = x _ {3} \left[ \begin{array}{c} - \frac {1}{2} \\ - 1 \\ 1 \end{array} \right], x _ {3} \neq 0 b = ⎣ ⎡ − 2 1 x 3 − x 3 x 3 ⎦ ⎤ = x 3 ⎣ ⎡ − 2 1 − 1 1 ⎦ ⎤ , x 3 = 0
Answer:
λ = 1 : a = [ 0 − x 3 x 3 ] \lambda = 1: a = \left[ \begin{array}{c} 0 \\ - x _ {3} \\ x _ {3} \end{array} \right] λ = 1 : a = ⎣ ⎡ 0 − x 3 x 3 ⎦ ⎤ λ = 2 b = [ − 1 2 x 3 − x 3 x 3 ] \lambda = 2 \quad b = \left[ \begin{array}{c} - \frac {1}{2} x _ {3} \\ - x _ {3} \\ x _ {3} \end{array} \right] λ = 2 b = ⎣ ⎡ − 2 1 x 3 − x 3 x 3 ⎦ ⎤ λ = 4 c = [ 1 2 x 2 x 2 0 ] \lambda = 4 \quad c = \left[ \begin{array}{c} \frac {1}{2} x _ {2} \\ x _ {2} \\ 0 \end{array} \right] λ = 4 c = ⎣ ⎡ 2 1 x 2 x 2 0 ⎦ ⎤
If λ = 2 \lambda = 2 λ = 2 X = [ − 1 2 x 3 − x 3 x 3 ] X = \left[ \begin{array}{c} - \frac{1}{2} x_{3} \\ -x_{3} \\ x_{3} \end{array} \right] X = ⎣ ⎡ − 2 1 x 3 − x 3 x 3 ⎦ ⎤
3. If A . x = λ x A.x = \lambda x A . x = λ x A = ∣ ∣ 211232 − 212 ∣ ∣ ∣ A = ||211232 - 212||| A = ∣∣211232 − 212∣∣∣ A A A
If
λ = 4 \lambda = 4 λ = 4
, what is c
Solution
A ^ x = λ x \hat {A} x = \lambda x A ^ x = λ x A ^ x = [ 2 1 1 2 3 2 − 2 1 2 ] \hat {A} x = \left[ \begin{array}{c c c} 2 & 1 & 1 \\ 2 & 3 & 2 \\ - 2 & 1 & 2 \end{array} \right] A ^ x = ⎣ ⎡ 2 2 − 2 1 3 1 1 2 2 ⎦ ⎤
Let's determine the eigen values of matrix A
A ^ x = λ E ^ x \hat{A}x = \lambda \hat{E}x A ^ x = λ E ^ x A ^ x − λ E ^ x = 0 \hat{A}x - \lambda \hat{E}x = 0 A ^ x − λ E ^ x = 0 ( A ^ − λ E ^ ) x = 0 (\hat{A} - \lambda \hat{E})x = 0 ( A ^ − λ E ^ ) x = 0 det [ 2 − λ 1 1 2 3 − λ 2 − 2 1 2 − λ ] = 0 \det \begin{bmatrix} 2 - \lambda & 1 & 1 \\ 2 & 3 - \lambda & 2 \\ -2 & 1 & 2 - \lambda \end{bmatrix} = 0 det ⎣ ⎡ 2 − λ 2 − 2 1 3 − λ 1 1 2 2 − λ ⎦ ⎤ = 0 ∣ 2 − λ 1 1 2 3 − λ 2 − 2 1 2 − λ ∣ = 0 \begin{vmatrix} 2 - \lambda & 1 & 1 \\ 2 & 3 - \lambda & 2 \\ -2 & 1 & 2 - \lambda \end{vmatrix} = 0 ∣ ∣ 2 − λ 2 − 2 1 3 − λ 1 1 2 2 − λ ∣ ∣ = 0 ( 2 − λ ) ⋅ ( 3 − λ ) ⋅ ( 2 − λ ) + 1 ⋅ 2 ⋅ ( − 2 ) + 2 ⋅ 1 ⋅ 1 − ( − 2 ) ⋅ ( 3 − λ ) ⋅ 1 − 1 ⋅ 2 ⋅ ( 2 − λ ) − 1 ⋅ 2 ⋅ ( 2 − λ ) = 0 (2 - \lambda) \cdot (3 - \lambda) \cdot (2 - \lambda) + 1 \cdot 2 \cdot (-2) + 2 \cdot 1 \cdot 1 - (-2) \cdot (3 - \lambda) \cdot 1 - 1 \cdot 2 \cdot (2 - \lambda) - 1 \cdot 2 \cdot (2 - \lambda) = 0 ( 2 − λ ) ⋅ ( 3 − λ ) ⋅ ( 2 − λ ) + 1 ⋅ 2 ⋅ ( − 2 ) + 2 ⋅ 1 ⋅ 1 − ( − 2 ) ⋅ ( 3 − λ ) ⋅ 1 − 1 ⋅ 2 ⋅ ( 2 − λ ) − 1 ⋅ 2 ⋅ ( 2 − λ ) = 0 ( 2 − λ ) 2 ⋅ ( 3 − λ ) − 4 + 2 + 6 − 2 λ + 4 λ − 8 = 0 (2 - \lambda)^2 \cdot (3 - \lambda) - 4 + 2 + 6 - 2\lambda + 4\lambda - 8 = 0 ( 2 − λ ) 2 ⋅ ( 3 − λ ) − 4 + 2 + 6 − 2 λ + 4 λ − 8 = 0 ( 2 − λ ) 2 ⋅ ( 3 − λ ) + 2 λ − 4 = 0 (2 - \lambda)^2 \cdot (3 - \lambda) + 2\lambda - 4 = 0 ( 2 − λ ) 2 ⋅ ( 3 − λ ) + 2 λ − 4 = 0 ( 2 − λ ) 2 ⋅ ( 3 − λ ) − 2 ( 2 − λ ) = 0 (2 - \lambda)^2 \cdot (3 - \lambda) - 2(2 - \lambda) = 0 ( 2 − λ ) 2 ⋅ ( 3 − λ ) − 2 ( 2 − λ ) = 0 ( 2 − λ ) ⋅ ( ( 2 − λ ) ⋅ ( 3 − λ ) − 2 ) = 0 (2 - \lambda) \cdot ((2 - \lambda) \cdot (3 - \lambda) - 2) = 0 ( 2 − λ ) ⋅ (( 2 − λ ) ⋅ ( 3 − λ ) − 2 ) = 0 ( 2 − λ ) ⋅ ( λ 2 − 5 λ + 6 − 2 ) = 0 (2 - \lambda) \cdot (\lambda^2 - 5\lambda + 6 - 2) = 0 ( 2 − λ ) ⋅ ( λ 2 − 5 λ + 6 − 2 ) = 0 ( 2 − λ ) ⋅ ( λ 2 − 5 λ + 4 ) = 0 (2 - \lambda) \cdot (\lambda^2 - 5\lambda + 4) = 0 ( 2 − λ ) ⋅ ( λ 2 − 5 λ + 4 ) = 0 λ 1 = 2 \lambda_1 = 2 λ 1 = 2 λ 2 − 5 λ + 4 = 0 \lambda^2 - 5\lambda + 4 = 0 λ 2 − 5 λ + 4 = 0 D = 25 − 16 = 9 D = 25 - 16 = 9 D = 25 − 16 = 9 λ 2 = 4 λ 3 = 1 \lambda_2 = 4 \quad \lambda_3 = 1 λ 2 = 4 λ 3 = 1 λ = 1 , λ = 2 , λ = 4. \lambda = 1, \lambda = 2, \lambda = 4. λ = 1 , λ = 2 , λ = 4.
Determine the eigen vector corresponding to each eigen value
λ = 1 : \lambda = 1: λ = 1 : ∣ 2 − 1 1 1 0 2 3 − 1 2 0 − 2 1 2 − 1 0 ∣ = 0 \begin{vmatrix} 2 - 1 & 1 & 1 & 0 \\ 2 & 3 - 1 & 2 & 0 \\ -2 & 1 & 2 - 1 & 0 \end{vmatrix} = 0 ∣ ∣ 2 − 1 2 − 2 1 3 − 1 1 1 2 2 − 1 0 0 0 ∣ ∣ = 0 ∣ 1 1 1 0 2 2 2 0 − 2 1 1 0 ∣ = ∣ 1 1 1 0 0 0 0 0 0 3 3 0 ∣ = ∣ 1 1 1 0 0 3 3 0 ∣ \begin{vmatrix} 1 & 1 & 1 & 0 \\ 2 & 2 & 2 & 0 \\ -2 & 1 & 1 & 0 \end{vmatrix} = \begin{vmatrix} 1 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 3 & 3 & 0 \end{vmatrix} = \begin{vmatrix} 1 & 1 & 1 & 0 \\ 0 & 3 & 3 & 0 \end{vmatrix} ∣ ∣ 1 2 − 2 1 2 1 1 2 1 0 0 0 ∣ ∣ = ∣ ∣ 1 0 0 1 0 3 1 0 3 0 0 0 ∣ ∣ = ∣ ∣ 1 0 1 3 1 3 0 0 ∣ ∣ { x 1 + x 2 + x 3 = 0 0 + x 2 + x 3 = 0 ⟺ { x 1 = 0 x 2 + x 3 = 0 x 1 = 0 x 2 = − x 3 \begin{array}{l}
\left\{
\begin{array}{l}
x_1 + x_2 + x_3 = 0 \\
0 + x_2 + x_3 = 0
\end{array}
\right.
\Longleftrightarrow
\left\{
\begin{array}{l}
x_1 = 0 \\
x_2 + x_3 = 0
\end{array}
\right. \\
x_1 = 0 \\
x_2 = -x_3 \\
\end{array} { x 1 + x 2 + x 3 = 0 0 + x 2 + x 3 = 0 ⟺ { x 1 = 0 x 2 + x 3 = 0 x 1 = 0 x 2 = − x 3 X = [ x 1 x 2 x 3 ] = [ 0 − x 3 x 3 ] = x 3 [ 0 − 1 1 ] , x 3 ≠ 0. X = \left[ \begin{array}{l}
x_1 \\
x_2 \\
x_3
\end{array} \right]
= \left[ \begin{array}{l}
0 \\
- x_3 \\
x_3
\end{array} \right]
= x_3 \left[ \begin{array}{l}
0 \\
-1 \\
1
\end{array} \right],
x_3 \neq 0. X = ⎣ ⎡ x 1 x 2 x 3 ⎦ ⎤ = ⎣ ⎡ 0 − x 3 x 3 ⎦ ⎤ = x 3 ⎣ ⎡ 0 − 1 1 ⎦ ⎤ , x 3 = 0. λ = 2 : \lambda = 2: λ = 2 : ∣ 2 − 2 1 1 2 3 − 2 2 − 2 1 2 − 2 ∣ ∣ 0 0 0 ∣ = 0 \left| \begin{array}{ccc}
2 - 2 & 1 & 1 \\
2 & 3 - 2 & 2 \\
-2 & 1 & 2 - 2
\end{array} \right|
\left| \begin{array}{l}
0 \\
0 \\
0
\end{array} \right|
= 0 ∣ ∣ 2 − 2 2 − 2 1 3 − 2 1 1 2 2 − 2 ∣ ∣ ∣ ∣ 0 0 0 ∣ ∣ = 0 \left| \begin{array}{ccc}
0 & 1 & 1 \\
2 & 1 & 2 \\
-2 & 1 & 0
\end{array} \right|
= \left| \begin{array}{ccc}
0 & 1 & 1 \\
1 & \frac{1}{2} & 1 \\
0 & 2 & 2
\end{array} \right|
\left| \begin{array}{l}
0 \\
0 \\
0
\end{array} \right|
= \left| \begin{array}{ccc}
0 & 1 & 1 \\
1 & \frac{1}{2} & 1
\end{array} \right|
\left| \begin{array}{l}
0 \\
0 \\
0
\end{array} \right|
\right| \left\{
\begin{array}{l}
0 + x_2 + x_3 = 0 \\
x_1 + \frac{1}{2} x_2 + x_3 = 0
\end{array}
\right.
\Longleftrightarrow
\left\{
\begin{array}{l}
x_1 = -\frac{1}{2} x_3 \\
x_2 = -x_3
\end{array}
\right. \\
x_1 = -\frac{1}{2} x_3 \\
x_2 = -x_3 \\
\end{array} X = [ x 1 x 2 x 3 ] = [ − 1 2 x 3 − x 3 x 3 ] = x 3 [ − 1 2 − 1 1 ] X = \left[ \begin{array}{l}
x_1 \\
x_2 \\
x_3
\end{array} \right]
= \left[ \begin{array}{l}
-\frac{1}{2} x_3 \\
-x_3 \\
x_3
\end{array} \right]
= x_3 \left[ \begin{array}{l}
-\frac{1}{2} \\
-1 \\
1
\end{array} \right] X = ⎣ ⎡ x 1 x 2 x 3 ⎦ ⎤ = ⎣ ⎡ − 2 1 x 3 − x 3 x 3 ⎦ ⎤ = x 3 ⎣ ⎡ − 2 1 − 1 1 ⎦ ⎤ λ = 4 : \lambda = 4: λ = 4 : ∣ 2 − 4 1 1 2 3 − 4 2 − 2 1 2 − 4 ∣ ∣ 0 0 0 ∣ = 0 \left| \begin{array}{ccc}
2 - 4 & 1 & 1 \\
2 & 3 - 4 & 2 \\
-2 & 1 & 2 - 4
\end{array} \right|
\left| \begin{array}{l}
0 \\
0 \\
0
\end{array} \right|
= 0 ∣ ∣ 2 − 4 2 − 2 1 3 − 4 1 1 2 2 − 4 ∣ ∣ ∣ ∣ 0 0 0 ∣ ∣ = 0 \left| \begin{array}{ccc}
-2 & 1 & 1 \\
2 & -1 & 2 \\
-2 & 1 & -2
\end{array} \right|
\left| \begin{array}{l}
0 \\
0 \\
0
\end{array} \right|
= \left| \begin{array}{ccc}
1 & -\frac{1}{2} & -\frac{1}{2} \\
1 & -\frac{1}{2} & 1 \\
1 & -\frac{1}{2} & 1
\end{array} \right|
\left| \begin{array}{l}
0 \\
0 \\
0
\end{array} \right|
= \left| \begin{array}{ccc}
1 & -\frac{1}{2} & -\frac{1}{2} \\
1 & -\frac{1}{2} & 1
\end{array} \right|
\left| \begin{array}{l}
0 \\
0 \\
0
\end{array} \right|
\right| { x 1 − 1 2 x 2 − 1 2 x 3 = 0 x 1 − 1 2 x 2 + x 3 = 0 ⟺ { x 1 = 1 2 x 2 x 3 = 0 \left\{
\begin{array}{l}
x_1 - \frac{1}{2} x_2 - \frac{1}{2} x_3 = 0 \\
x_1 - \frac{1}{2} x_2 + x_3 = 0
\end{array}
\right.
\Longleftrightarrow
\left\{
\begin{array}{l}
x_1 = \frac{1}{2} x_2 \\
x_3 = 0
\end{array}
\right. { x 1 − 2 1 x 2 − 2 1 x 3 = 0 x 1 − 2 1 x 2 + x 3 = 0 ⟺ { x 1 = 2 1 x 2 x 3 = 0 x 1 = 1 2 x 2 x _ {1} = \frac {1}{2} x _ {2} x 1 = 2 1 x 2 x 3 = 0 x _ {3} = 0 x 3 = 0 X = [ x 1 x 2 x 3 ] = [ 1 2 x 2 x 2 0 ] = x 2 [ 1 2 1 0 ] , x 2 ≠ 0. X = \left[ \begin{array}{c} x _ {1} \\ x _ {2} \\ x _ {3} \end{array} \right] = \left[ \begin{array}{c} \frac {1}{2} x _ {2} \\ x _ {2} \\ 0 \end{array} \right] = x _ {2} \left[ \begin{array}{c} \frac {1}{2} \\ 1 \\ 0 \end{array} \right], x _ {2} \neq 0. X = ⎣ ⎡ x 1 x 2 x 3 ⎦ ⎤ = ⎣ ⎡ 2 1 x 2 x 2 0 ⎦ ⎤ = x 2 ⎣ ⎡ 2 1 1 0 ⎦ ⎤ , x 2 = 0.
**Answer:**
λ = 1 : a = [ 0 − x 3 x 3 ] \lambda = 1 \colon a = \left[ \begin{array}{c} 0 \\ - x _ {3} \\ x _ {3} \end{array} \right] λ = 1 : a = ⎣ ⎡ 0 − x 3 x 3 ⎦ ⎤ λ = 2 b = [ − 1 2 x 3 − x 3 x 3 ] \lambda = 2 \quad b = \left[ \begin{array}{c} - \frac {1}{2} x _ {3} \\ - x _ {3} \\ x _ {3} \end{array} \right] λ = 2 b = ⎣ ⎡ − 2 1 x 3 − x 3 x 3 ⎦ ⎤ λ = 4 c = [ 1 2 x 2 x 2 0 ] \lambda = 4 \quad c = \left[ \begin{array}{c} \frac {1}{2} x _ {2} \\ x _ {2} \\ 0 \end{array} \right] λ = 4 c = ⎣ ⎡ 2 1 x 2 x 2 0 ⎦ ⎤
4. Solve the set of linear equations by the matrix method: a + 3 b + 2 c = 3 a + 3b + 2c = 3 a + 3 b + 2 c = 3 2 a − b − 3 c = − 8 2a - b - 3c = -8 2 a − b − 3 c = − 8 5 a + 2 b + c = 9 5a + 2b + c = 9 5 a + 2 b + c = 9 c c c
**Solution**
{ a + 3 b + c = 3 2 a − b − 3 c = − 8 5 a + 2 b + c = 9 \left\{ \begin{array}{l} a + 3 b + c = 3 \\ 2 a - b - 3 c = - 8 \\ 5 a + 2 b + c = 9 \end{array} \right. ⎩ ⎨ ⎧ a + 3 b + c = 3 2 a − b − 3 c = − 8 5 a + 2 b + c = 9 ∣ 1 3 1 2 − 1 − 3 5 2 1 ∣ 3 − 8 9 \left| \begin{array}{c c c} 1 & 3 & 1 \\ 2 & - 1 & - 3 \\ 5 & 2 & 1 \end{array} \right| \begin{array}{c} 3 \\ - 8 \\ 9 \end{array} ∣ ∣ 1 2 5 3 − 1 2 1 − 3 1 ∣ ∣ 3 − 8 9
Using Cramer's rule
Δ = ∣ 1 3 1 2 − 1 − 3 5 2 1 ∣ = \Delta = \left| \begin{array}{c c c} 1 & 3 & 1 \\ 2 & - 1 & - 3 \\ 5 & 2 & 1 \end{array} \right| = Δ = ∣ ∣ 1 2 5 3 − 1 2 1 − 3 1 ∣ ∣ = = 1 ⋅ ( − 1 ) ⋅ 1 + 3 ⋅ ( − 3 ) ⋅ 5 + 2 ⋅ 2 ⋅ 1 − 5 ⋅ ( − 1 ) ⋅ 1 − 1 ⋅ 2 ⋅ ( − 3 ) − 2 ⋅ 3 ⋅ 1 = = 1 \cdot (- 1) \cdot 1 + 3 \cdot (- 3) \cdot 5 + 2 \cdot 2 \cdot 1 - 5 \cdot (- 1) \cdot 1 - 1 \cdot 2 \cdot (- 3) - 2 \cdot 3 \cdot 1 = = 1 ⋅ ( − 1 ) ⋅ 1 + 3 ⋅ ( − 3 ) ⋅ 5 + 2 ⋅ 2 ⋅ 1 − 5 ⋅ ( − 1 ) ⋅ 1 − 1 ⋅ 2 ⋅ ( − 3 ) − 2 ⋅ 3 ⋅ 1 = = − 1 − 45 + 4 + 5 + 6 − 6 = − 37 = - 1 - 4 5 + 4 + 5 + 6 - 6 = - 3 7 = − 1 − 45 + 4 + 5 + 6 − 6 = − 37 Δ a = ∣ 3 3 1 − 8 − 1 − 3 9 2 1 ∣ = \Delta_ {a} = \left| \begin{array}{c c c} 3 & 3 & 1 \\ - 8 & - 1 & - 3 \\ 9 & 2 & 1 \end{array} \right| = Δ a = ∣ ∣ 3 − 8 9 3 − 1 2 1 − 3 1 ∣ ∣ = = 3 ⋅ ( − 1 ) ⋅ 1 + 3 ⋅ ( − 3 ) ⋅ 9 + ( − 8 ) ⋅ 2 ⋅ 1 − 9 ⋅ ( − 1 ) ⋅ 1 − 3 ⋅ 2 ⋅ ( − 3 ) − ( − 8 ) ⋅ 3 ⋅ 1 = = − 3 − 81 − 16 + 9 + 18 + 24 = − 49 Δ b = ∣ 1 3 1 2 − 8 − 3 5 9 1 ∣ = = 1 ⋅ ( − 8 ) ⋅ 1 + 3 ⋅ ( − 3 ) ⋅ 5 + 2 ⋅ 9 ⋅ 1 − 5 ⋅ ( − 8 ) ⋅ 1 − 1 ⋅ 9 ⋅ ( − 3 ) − 2 ⋅ 3 ⋅ 1 = = − 8 − 45 + 18 + 40 + 27 − 6 = 26 Δ c = ∣ 1 3 3 2 − 1 − 8 5 2 9 ∣ = = 1 ⋅ ( − 1 ) ⋅ 9 + 3 ⋅ ( − 8 ) ⋅ 5 + 2 ⋅ 2 ⋅ 3 − 5 ⋅ ( − 1 ) ⋅ 3 − 1 ⋅ 2 ⋅ ( − 8 ) − 2 ⋅ 3 ⋅ 9 = = − 9 − 120 + 12 + 15 + 16 − 54 = − 140 a = Δ a Δ = − 49 − 37 = 1 12 37 b = Δ b Δ = 26 − 37 c = Δ c Δ = − 140 − 37 = 3 29 37 \begin{array}{l}
= 3 \cdot (-1) \cdot 1 + 3 \cdot (-3) \cdot 9 + (-8) \cdot 2 \cdot 1 - 9 \cdot (-1) \cdot 1 - 3 \cdot 2 \cdot (-3) - (-8) \cdot 3 \cdot 1 = \\
= -3 - 81 - 16 + 9 + 18 + 24 = -49 \\
\Delta_{b} = \left| \begin{array}{ccc} 1 & 3 & 1 \\ 2 & -8 & -3 \\ 5 & 9 & 1 \end{array} \right| = \\
= 1 \cdot (-8) \cdot 1 + 3 \cdot (-3) \cdot 5 + 2 \cdot 9 \cdot 1 - 5 \cdot (-8) \cdot 1 - 1 \cdot 9 \cdot (-3) - 2 \cdot 3 \cdot 1 = \\
= -8 - 45 + 18 + 40 + 27 - 6 = 26 \\
\Delta_{c} = \left| \begin{array}{ccc} 1 & 3 & 3 \\ 2 & -1 & -8 \\ 5 & 2 & 9 \end{array} \right| = \\
= 1 \cdot (-1) \cdot 9 + 3 \cdot (-8) \cdot 5 + 2 \cdot 2 \cdot 3 - 5 \cdot (-1) \cdot 3 - 1 \cdot 2 \cdot (-8) - 2 \cdot 3 \cdot 9 = \\
= -9 - 120 + 12 + 15 + 16 - 54 = -140 \\
a = \frac{\Delta_{a}}{\Delta} = \frac{-49}{-37} = 1 \frac{12}{37} \\
b = \frac{\Delta_{b}}{\Delta} = \frac{26}{-37} \\
c = \frac{\Delta_{c}}{\Delta} = \frac{-140}{-37} = 3 \frac{29}{37} \\
\end{array} = 3 ⋅ ( − 1 ) ⋅ 1 + 3 ⋅ ( − 3 ) ⋅ 9 + ( − 8 ) ⋅ 2 ⋅ 1 − 9 ⋅ ( − 1 ) ⋅ 1 − 3 ⋅ 2 ⋅ ( − 3 ) − ( − 8 ) ⋅ 3 ⋅ 1 = = − 3 − 81 − 16 + 9 + 18 + 24 = − 49 Δ b = ∣ ∣ 1 2 5 3 − 8 9 1 − 3 1 ∣ ∣ = = 1 ⋅ ( − 8 ) ⋅ 1 + 3 ⋅ ( − 3 ) ⋅ 5 + 2 ⋅ 9 ⋅ 1 − 5 ⋅ ( − 8 ) ⋅ 1 − 1 ⋅ 9 ⋅ ( − 3 ) − 2 ⋅ 3 ⋅ 1 = = − 8 − 45 + 18 + 40 + 27 − 6 = 26 Δ c = ∣ ∣ 1 2 5 3 − 1 2 3 − 8 9 ∣ ∣ = = 1 ⋅ ( − 1 ) ⋅ 9 + 3 ⋅ ( − 8 ) ⋅ 5 + 2 ⋅ 2 ⋅ 3 − 5 ⋅ ( − 1 ) ⋅ 3 − 1 ⋅ 2 ⋅ ( − 8 ) − 2 ⋅ 3 ⋅ 9 = = − 9 − 120 + 12 + 15 + 16 − 54 = − 140 a = Δ Δ a = − 37 − 49 = 1 37 12 b = Δ Δ b = − 37 26 c = Δ Δ c = − 37 − 140 = 3 37 29 a = 1 12 37 , b = 26 − 37 , c = 3 29 37 a = 1 \frac{12}{37}, b = \frac{26}{-37}, c = 3 \frac{29}{37} a = 1 37 12 , b = − 37 26 , c = 3 37 29
Answer: c = 3 29 37 c = 3 \frac{29}{37} c = 3 37 29
5. If A . x = λ x A.x = \lambda x A . x = λ x A = ∥ ∥ 211232 − 212 ∥ ∥ A = \|\|211232 - 212\|\| A = ∥∥211232 − 212∥∥ A A A λ = 1 \lambda = 1 λ = 1
If
λ = 1 \lambda = 1 λ = 1
, what is a
Solution
A ^ x = λ x \hat{A}x = \lambda x A ^ x = λ x A ^ x = [ 2 1 1 2 3 2 − 2 1 2 ] \hat{A}x = \left[ \begin{array}{ccc} 2 & 1 & 1 \\ 2 & 3 & 2 \\ -2 & 1 & 2 \end{array} \right] A ^ x = ⎣ ⎡ 2 2 − 2 1 3 1 1 2 2 ⎦ ⎤
Let's determine the eigen values of matrix A
A ^ x = λ E ^ x \hat{A}x = \lambda \hat{E}x A ^ x = λ E ^ x A ^ x − λ E ^ x = 0 \hat{A}x - \lambda \hat{E}x = 0 A ^ x − λ E ^ x = 0 ( A ^ − λ E ^ ) x = 0 (\hat{A} - \lambda \hat{E})x = 0 ( A ^ − λ E ^ ) x = 0 det [ 2 − λ 1 1 2 3 − λ 2 − 2 1 2 − λ ] = 0 \det \begin{bmatrix}
2 - \lambda & 1 & 1 \\
2 & 3 - \lambda & 2 \\
-2 & 1 & 2 - \lambda
\end{bmatrix} = 0 det ⎣ ⎡ 2 − λ 2 − 2 1 3 − λ 1 1 2 2 − λ ⎦ ⎤ = 0 ∣ 2 − λ 1 1 2 3 − λ 2 − 2 1 2 − λ ∣ = 0 \begin{vmatrix}
2 - \lambda & 1 & 1 \\
2 & 3 - \lambda & 2 \\
-2 & 1 & 2 - \lambda
\end{vmatrix} = 0 ∣ ∣ 2 − λ 2 − 2 1 3 − λ 1 1 2 2 − λ ∣ ∣ = 0 ( 2 − λ ) ⋅ ( 3 − λ ) ⋅ ( 2 − λ ) + 1 ⋅ 2 ⋅ ( − 2 ) + 2 ⋅ 1 ⋅ 1 − ( − 2 ) ⋅ ( 3 − λ ) ⋅ 1 − 1 ⋅ 2 ⋅ ( 2 − λ ) − 1 ⋅ 2 ⋅ ( 2 − λ ) = 0 (2 - \lambda) \cdot (3 - \lambda) \cdot (2 - \lambda) + 1 \cdot 2 \cdot (-2) + 2 \cdot 1 \cdot 1 - (-2) \cdot (3 - \lambda) \cdot 1 - 1 \cdot 2 \cdot (2 - \lambda) - 1 \cdot 2 \cdot (2 - \lambda) = 0 ( 2 − λ ) ⋅ ( 3 − λ ) ⋅ ( 2 − λ ) + 1 ⋅ 2 ⋅ ( − 2 ) + 2 ⋅ 1 ⋅ 1 − ( − 2 ) ⋅ ( 3 − λ ) ⋅ 1 − 1 ⋅ 2 ⋅ ( 2 − λ ) − 1 ⋅ 2 ⋅ ( 2 − λ ) = 0 ( 2 − λ ) 2 ⋅ ( 3 − λ ) − 4 + 2 + 6 − 2 λ + 4 λ − 8 = 0 (2 - \lambda)^2 \cdot (3 - \lambda) - 4 + 2 + 6 - 2\lambda + 4\lambda - 8 = 0 ( 2 − λ ) 2 ⋅ ( 3 − λ ) − 4 + 2 + 6 − 2 λ + 4 λ − 8 = 0 ( 2 − λ ) 2 ⋅ ( 3 − λ ) + 2 λ − 4 = 0 (2 - \lambda)^2 \cdot (3 - \lambda) + 2\lambda - 4 = 0 ( 2 − λ ) 2 ⋅ ( 3 − λ ) + 2 λ − 4 = 0 ( 2 − λ ) 2 ⋅ ( 3 − λ ) − 2 ( 2 − λ ) = 0 (2 - \lambda)^2 \cdot (3 - \lambda) - 2(2 - \lambda) = 0 ( 2 − λ ) 2 ⋅ ( 3 − λ ) − 2 ( 2 − λ ) = 0 ( 2 − λ ) ⋅ ( ( 2 − λ ) ⋅ ( 3 − λ ) − 2 ) = 0 (2 - \lambda) \cdot ((2 - \lambda) \cdot (3 - \lambda) - 2) = 0 ( 2 − λ ) ⋅ (( 2 − λ ) ⋅ ( 3 − λ ) − 2 ) = 0 ( 2 − λ ) ⋅ ( λ 2 − 5 λ + 6 − 2 ) = 0 (2 - \lambda) \cdot (\lambda^2 - 5\lambda + 6 - 2) = 0 ( 2 − λ ) ⋅ ( λ 2 − 5 λ + 6 − 2 ) = 0 ( 2 − λ ) ⋅ ( λ 2 − 5 λ + 4 ) = 0 (2 - \lambda) \cdot (\lambda^2 - 5\lambda + 4) = 0 ( 2 − λ ) ⋅ ( λ 2 − 5 λ + 4 ) = 0 λ 1 = 2 \lambda_1 = 2 λ 1 = 2 λ 2 − 5 λ + 4 = 0 \lambda^2 - 5\lambda + 4 = 0 λ 2 − 5 λ + 4 = 0 D = 25 − 16 = 9 D = 25 - 16 = 9 D = 25 − 16 = 9 λ 2 = 4 λ 3 = 1 \lambda_2 = 4 \quad \lambda_3 = 1 λ 2 = 4 λ 3 = 1 λ = 1 , λ = 2 , λ = 4 \lambda = 1, \lambda = 2, \lambda = 4 λ = 1 , λ = 2 , λ = 4
Determine the eigen vector corresponding to each eigen value.
λ = 1 : \lambda = 1: λ = 1 : ∣ 2 − 1 1 1 0 2 3 − 1 2 0 − 2 1 2 − 1 0 ∣ = 0 \begin{vmatrix}
2 - 1 & 1 & 1 & 0 \\
2 & 3 - 1 & 2 & 0 \\
-2 & 1 & 2 - 1 & 0
\end{vmatrix} = 0 ∣ ∣ 2 − 1 2 − 2 1 3 − 1 1 1 2 2 − 1 0 0 0 ∣ ∣ = 0 ∣ 1 1 1 0 2 2 2 0 − 2 1 1 0 ∣ = ∣ 1 1 1 0 0 0 0 0 0 3 3 0 ∣ = ∣ 1 1 1 0 0 3 3 0 ∣ \begin{vmatrix}
1 & 1 & 1 & 0 \\
2 & 2 & 2 & 0 \\
-2 & 1 & 1 & 0
\end{vmatrix} = \begin{vmatrix}
1 & 1 & 1 & 0 \\
0 & 0 & 0 & 0 \\
0 & 3 & 3 & 0
\end{vmatrix} = \begin{vmatrix}
1 & 1 & 1 & 0 \\
0 & 3 & 3 & 0
\end{vmatrix} ∣ ∣ 1 2 − 2 1 2 1 1 2 1 0 0 0 ∣ ∣ = ∣ ∣ 1 0 0 1 0 3 1 0 3 0 0 0 ∣ ∣ = ∣ ∣ 1 0 1 3 1 3 0 0 ∣ ∣ { x 1 + x 2 + x 3 = 0 0 + x 2 + x 3 = 0 = = > { x 1 = 0 x 2 + x 3 = 0 \begin{cases}
x_1 + x_2 + x_3 = 0 \\
0 + x_2 + x_3 = 0
\end{cases}
= = > \begin{cases}
x_1 & = 0 \\
x_2 + x_3 = 0
\end{cases} { x 1 + x 2 + x 3 = 0 0 + x 2 + x 3 = 0 ==> { x 1 x 2 + x 3 = 0 = 0 x 1 = 0 x_1 = 0 x 1 = 0 x 2 = − x 3 x_2 = -x_3 x 2 = − x 3 X = [ x 1 x 2 x 3 ] = [ 0 − x 3 x 3 ] = x 3 [ 0 − 1 1 ] , x 3 ≠ 0. X = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 0 \\ -x_3 \\ x_3 \end{bmatrix} = x_3 \begin{bmatrix} 0 \\ -1 \\ 1 \end{bmatrix}, x_3 \neq 0. X = ⎣ ⎡ x 1 x 2 x 3 ⎦ ⎤ = ⎣ ⎡ 0 − x 3 x 3 ⎦ ⎤ = x 3 ⎣ ⎡ 0 − 1 1 ⎦ ⎤ , x 3 = 0. λ = 2 : \lambda = 2: λ = 2 : ∣ 2 − 2 1 1 0 2 3 − 2 2 0 − 2 1 2 − 2 0 ∣ = 0 \begin{vmatrix}
2 - 2 & 1 & 1 & 0 \\
2 & 3 - 2 & 2 & 0 \\
-2 & 1 & 2 - 2 & 0
\end{vmatrix}
= 0 ∣ ∣ 2 − 2 2 − 2 1 3 − 2 1 1 2 2 − 2 0 0 0 ∣ ∣ = 0 ∣ 0 1 1 0 2 1 2 0 − 2 1 0 0 ∣ = ∣ 0 1 1 0 1 1 2 1 0 0 2 2 0 ∣ = ∣ 0 1 1 0 1 1 2 1 0 ∣ \begin{vmatrix}
0 & 1 & 1 & 0 \\
2 & 1 & 2 & 0 \\
-2 & 1 & 0 & 0
\end{vmatrix}
= \begin{vmatrix}
0 & 1 & 1 & 0 \\
1 & \frac{1}{2} & 1 & 0 \\
0 & 2 & 2 & 0
\end{vmatrix}
= \begin{vmatrix}
0 & 1 & 1 & 0 \\
1 & \frac{1}{2} & 1 & 0
\end{vmatrix} ∣ ∣ 0 2 − 2 1 1 1 1 2 0 0 0 0 ∣ ∣ = ∣ ∣ 0 1 0 1 2 1 2 1 1 2 0 0 0 ∣ ∣ = ∣ ∣ 0 1 1 2 1 1 1 0 0 ∣ ∣ { 0 + x 2 + x 3 = 0 x 1 + 1 2 x 2 + x 3 = 0 = ⇒ { x 1 = − 1 2 x 3 x 2 = − x 3 \begin{cases}
0 + x_2 + x_3 = 0 \\
x_1 + \frac{1}{2}x_2 + x_3 = 0
\end{cases}
= \Rightarrow \begin{cases}
x_1 = -\frac{1}{2}x_3 \\
x_2 = -x_3
\end{cases} { 0 + x 2 + x 3 = 0 x 1 + 2 1 x 2 + x 3 = 0 =⇒ { x 1 = − 2 1 x 3 x 2 = − x 3 x 1 = − 1 2 x 3 x_1 = -\frac{1}{2}x_3 x 1 = − 2 1 x 3 x 2 = − x 3 x_2 = -x_3 x 2 = − x 3 X = [ x 1 x 2 x 3 ] = [ − 1 2 x 3 − x 3 x 3 ] = x 3 [ − 1 2 − 1 1 ] X = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} -\frac{1}{2}x_3 \\ -x_3 \\ x_3 \end{bmatrix} = x_3 \begin{bmatrix} -\frac{1}{2} \\ -1 \\ 1 \end{bmatrix} X = ⎣ ⎡ x 1 x 2 x 3 ⎦ ⎤ = ⎣ ⎡ − 2 1 x 3 − x 3 x 3 ⎦ ⎤ = x 3 ⎣ ⎡ − 2 1 − 1 1 ⎦ ⎤ λ = 4 : \lambda = 4: λ = 4 : ∣ 2 − 4 1 1 0 2 3 − 4 2 0 − 2 1 2 − 4 0 ∣ = 0 \begin{vmatrix}
2 - 4 & 1 & 1 & 0 \\
2 & 3 - 4 & 2 & 0 \\
-2 & 1 & 2 - 4 & 0
\end{vmatrix}
= 0 ∣ ∣ 2 − 4 2 − 2 1 3 − 4 1 1 2 2 − 4 0 0 0 ∣ ∣ = 0 ∣ − 2 1 1 0 2 − 1 2 0 − 2 1 − 2 0 ∣ = ∣ 1 − 1 2 − 1 2 1 − 1 2 1 1 − 1 2 1 ∣ = ∣ 1 − 1 2 − 1 2 1 − 1 2 1 ∣ \begin{vmatrix}
-2 & 1 & 1 & 0 \\
2 & -1 & 2 & 0 \\
-2 & 1 & -2 & 0
\end{vmatrix}
= \begin{vmatrix}
1 & -\frac{1}{2} & -\frac{1}{2} \\
1 & -\frac{1}{2} & 1 \\
1 & -\frac{1}{2} & 1
\end{vmatrix}
= \begin{vmatrix}
1 & -\frac{1}{2} & -\frac{1}{2} \\
1 & -\frac{1}{2} & 1
\end{vmatrix} ∣ ∣ − 2 2 − 2 1 − 1 1 1 2 − 2 0 0 0 ∣ ∣ = ∣ ∣ 1 1 1 − 2 1 − 2 1 − 2 1 − 2 1 1 1 ∣ ∣ = ∣ ∣ 1 1 − 2 1 − 2 1 − 2 1 1 ∣ ∣ { x 1 − 1 2 x 2 − 1 2 x 3 = 0 x 1 − 1 2 x 2 + x 3 = 0 = ⇒ { x 1 = 1 2 x 2 x 3 = 0 \begin{cases}
x_1 - \frac{1}{2}x_2 - \frac{1}{2}x_3 = 0 \\
x_1 - \frac{1}{2}x_2 + x_3 = 0
\end{cases}
= \Rightarrow \begin{cases}
x_1 = \frac{1}{2}x_2 \\
x_3 = 0
\end{cases} { x 1 − 2 1 x 2 − 2 1 x 3 = 0 x 1 − 2 1 x 2 + x 3 = 0 =⇒ { x 1 = 2 1 x 2 x 3 = 0 x 1 = 1 2 x 2 x_1 = \frac{1}{2}x_2 x 1 = 2 1 x 2 x 3 = 0 x_3 = 0 x 3 = 0 X = [ x 1 x 2 x 3 ] = [ 1 2 x 2 x 2 0 ] = x 2 [ 1 2 1 0 ] , x 2 ≠ 0. X = \left[ \begin{array}{c} x _ {1} \\ x _ {2} \\ x _ {3} \end{array} \right] = \left[ \begin{array}{c} \frac {1}{2} x _ {2} \\ x _ {2} \\ 0 \end{array} \right] = x _ {2} \left[ \begin{array}{c} \frac {1}{2} \\ 1 \\ 0 \end{array} \right], x _ {2} \neq 0. X = ⎣ ⎡ x 1 x 2 x 3 ⎦ ⎤ = ⎣ ⎡ 2 1 x 2 x 2 0 ⎦ ⎤ = x 2 ⎣ ⎡ 2 1 1 0 ⎦ ⎤ , x 2 = 0.
If λ = 1 \lambda = 1 λ = 1
X = [ x 1 x 2 x 3 ] = [ 0 − x 3 x 3 ] = x 3 [ 0 − 1 1 ] , x 3 ≠ 0. X = \left[ \begin{array}{c} x _ {1} \\ x _ {2} \\ x _ {3} \end{array} \right] = \left[ \begin{array}{c} 0 \\ - x _ {3} \\ x _ {3} \end{array} \right] = x _ {3} \left[ \begin{array}{c} 0 \\ - 1 \\ 1 \end{array} \right], x _ {3} \neq 0. X = ⎣ ⎡ x 1 x 2 x 3 ⎦ ⎤ = ⎣ ⎡ 0 − x 3 x 3 ⎦ ⎤ = x 3 ⎣ ⎡ 0 − 1 1 ⎦ ⎤ , x 3 = 0.
Answer:
λ = 1 : a = [ 0 − x 3 x 3 ] \lambda = 1: a = \left[ \begin{array}{c} 0 \\ - x _ {3} \\ x _ {3} \end{array} \right] λ = 1 : a = ⎣ ⎡ 0 − x 3 x 3 ⎦ ⎤ λ = 2 b = [ − 1 2 x 3 − x 3 x 3 ] \lambda = 2 \quad b = \left[ \begin{array}{c} - \frac {1}{2} x _ {3} \\ - x _ {3} \\ x _ {3} \end{array} \right] λ = 2 b = ⎣ ⎡ − 2 1 x 3 − x 3 x 3 ⎦ ⎤ λ = 4 c = [ 1 2 x 2 x 2 0 ] \lambda = 4 \quad c = \left[ \begin{array}{c} \frac {1}{2} x _ {2} \\ x _ {2} \\ 0 \end{array} \right] λ = 4 c = ⎣ ⎡ 2 1 x 2 x 2 0 ⎦ ⎤
If λ = 1 \lambda = 1 λ = 1 a = [ 0 − x 3 x 3 ] a = \left[ \begin{array}{c}0\\ -x_3\\ x_3 \end{array} \right] a = ⎣ ⎡ 0 − x 3 x 3 ⎦ ⎤
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#339914 on Dec 2023