问题

You are supposed to provide a solution to a given problem.

Problem:
For matrix A = [[5, 4], [1, 2]], what are its eigen values?
Solution:

答案

模型 模型答案 正确答案
金声玉亮2.0 To find the eigenvalues of a 2x2 matrix A = [[a, b], [c, d]], we need to solve the characteristic equation, which is given by:

det(A - λI) = 0

where λ represents the eigenvalues, I is the identity matrix, and det() is the determinant of the matrix.

For matrix A = [[5, 4], [1, 2]], the characteristic equation is:

det([[5 - λ, 4], [1, 2 - λ]]) = 0

Expanding the determinant, we get:

(5 - λ)(2 - λ) - (4)(1) = 0

Simplifying the equation, we get:

λ^2 - 7λ + 6 = 0

Solving the quadratic equation, we get:

λ = (7 ± √(49 - 24))/2 = (7 ± √25)/2 = (7 ± 5)/2

So, the eigenvalues of matrix A are:

λ₁ = (7 + 5)/2 = 6

λ₂ = (7 - 5)/2 = 1

Therefore, the eigenvalues of matrix A = [[5, 4], [1, 2]] are 6 and 1.		 [1, 6]
deepseekr1

The eigenvalues of the matrix \( A = \begin{bmatrix} 5 & 4 \\ 1 & 2 \end{bmatrix} \) are found by solving the characteristic equation \( \det(A - \lambda I) = 0 \).

1. Compute \( A - \lambda I \):
\[
\begin{bmatrix}
5 - \lambda & 4 \\
1 & 2 - \lambda
\end{bmatrix}
\]

2. Calculate the determinant:
\[
(5 - \lambda)(2 - \lambda) - (4)(1) = \lambda^2 - 7\lambda + 6 = 0
\]

3. Solve the quadratic equation \( \lambda^2 - 7\lambda + 6 = 0 \):
\[
\lambda = \frac{7 \pm \sqrt{49 - 24}}{2} = \frac{7 \pm 5}{2}
\]
\[
\lambda_1 = \frac{12}{2} = 6, \quad \lambda_2 = \frac{2}{2} = 1
\]

**Eigenvalues:** \( \boxed{6} \) and \( \boxed{1} \)		 [1, 6]
gpto1 To find the eigenvalues of the matrix \( A = \begin{bmatrix} 5 & 4 \\ 1 & 2 \end{bmatrix} \), we need to solve the characteristic equation obtained from the determinant of \( (A - \lambda I) \), where \( \lambda \) represents the eigenvalues and \( I \) is the identity matrix.

### Step 1: Write the Matrix \( A - \lambda I \)

Subtract \( \lambda \) times the identity matrix \( I \) from matrix \( A \):

\[
A - \lambda I = \begin{bmatrix} 5 - \lambda & 4 \\ 1 & 2 - \lambda \end{bmatrix}
\]

### Step 2: Compute the Determinant of \( A - \lambda I \)

Set the determinant equal to zero to find the eigenvalues:

\[
\det(A - \lambda I) = 0
\]

Compute the determinant:

\[
\begin{align*}
\det(A - \lambda I) &= (5 - \lambda)(2 - \lambda) - (4)(1) \\
&= (5 - \lambda)(2 - \lambda) - 4
\end{align*}
\]

### Step 3: Expand the Determinant

Expand the expression:

\[
\begin{align*}
(5 - \lambda)(2 - \lambda) &= 5 \cdot 2 - 5 \lambda - 2 \lambda + \lambda^2 \\
&= 10 - 7 \lambda + \lambda^2
\end{align*}
\]

Subtract 4 from the result:

\[
\det(A - \lambda I) = [10 - 7 \lambda + \lambda^2] - 4 = 6 - 7 \lambda + \lambda^2
\]

### Step 4: Form the Characteristic Equation

Set the determinant equal to zero:

\[
\lambda^2 - 7 \lambda + 6 = 0
\]

### Step 5: Solve the Quadratic Equation

Solve for \( \lambda \):

\[
\lambda^2 - 7 \lambda + 6 = 0
\]

Factor the quadratic:

\[
(\lambda - 6)(\lambda - 1) = 0
\]

Set each factor equal to zero:

1. \( \lambda - 6 = 0 \quad \Rightarrow \quad \lambda = 6 \)
2. \( \lambda - 1 = 0 \quad \Rightarrow \quad \lambda = 1 \)

### **Answer:**

The eigenvalues of the matrix \( A \) are:

\[
\lambda_1 = 6 \quad \text{and} \quad \lambda_2 = 1
\]		 [1, 6]
GPT-4o To find the eigenvalues of a matrix \( A \), we need to solve the characteristic equation \( \text{det}(A - \lambda I) = 0 \), where \( \lambda \) represents the eigenvalues and \( I \) is the identity matrix of the same dimension as \( A \).

Given matrix:
\[ A = \begin{bmatrix} 5 & 4 \\ 1 & 2 \end{bmatrix} \]

First, we set up the matrix \( A - \lambda I \):
\[ A - \lambda I = \begin{bmatrix} 5 & 4 \\ 1 & 2 \end{bmatrix} - \lambda \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 5 - \lambda & 4 \\ 1 & 2 - \lambda \end{bmatrix} \]

Next, we find the determinant of \( A - \lambda I \):
\[ \text{det}(A - \lambda I) = \begin{vmatrix} 5 - \lambda & 4 \\ 1 & 2 - \lambda \end{vmatrix} \]

The determinant of a 2x2 matrix \( \begin{bmatrix} a & b \\ c & d \end{bmatrix} \) is calculated as \( ad - bc \). Applying this to our matrix:
\[ \text{det}(A - \lambda I) = (5 - \lambda)(2 - \lambda) - (1 \cdot 4) \]
\[ \text{det}(A - \lambda I) = (5 - \lambda)(2 - \lambda) - 4 \]
\[ \text{det}(A - \lambda I) = 10 - 5\lambda - 2\lambda + \lambda^2 - 4 \]
\[ \text{det}(A - \lambda I) = \lambda^2 - 7\lambda + 6 \]

Now, we solve the characteristic equation:
\[ \lambda^2 - 7\lambda + 6 = 0 \]

We can factor this quadratic equation:
\[ \lambda^2 - 7\lambda + 6 = (\lambda - 1)(\lambda - 6) = 0 \]

Setting each factor equal to zero gives us the eigenvalues:
\[ \lambda - 1 = 0 \quad \text{or} \quad \lambda - 6 = 0 \]
\[ \lambda = 1 \quad \text{or} \quad \lambda = 6 \]

Thus, the eigenvalues of the matrix \( A \) are:
\[ \boxed{\lambda = 1 \text{ and } \lambda = 6} \]		 [1, 6]
llama3.1-70B-Instruct

To find the eigenvalues of matrix A = [[5, 4], [1, 2]], we need to solve the characteristic equation det(A - λI) = 0, where λ is the eigenvalue and I is the identity matrix.

First, we need to find the characteristic matrix A - λI:

A - λI = [[5 - λ, 4], [1, 2 - λ]]

Next, we need to find the determinant of this matrix:

det(A - λI) = (5 - λ)(2 - λ) - 4

Expanding and simplifying, we get:

det(A - λI) = λ^2 - 7λ + 6

Now, we need to solve the equation det(A - λI) = 0:

λ^2 - 7λ + 6 = 0

Factoring the quadratic equation, we get:

(λ - 1)(λ - 6) = 0

This gives us two possible values for λ:

λ = 1 or λ = 6

Therefore, the eigenvalues of matrix A = [[5, 4], [1, 2]] are λ = 1 and λ = 6.		 [1, 6]
llama3.1-405B-Instruct

To find the eigenvalues of a matrix A, we need to solve the characteristic equation det(A - λI) = 0, where λ is the eigenvalue and I is the identity matrix.

For the given matrix A = [[5, 4], [1, 2]], we can calculate the characteristic equation as follows:

det(A - λI) = det([[5 - λ, 4], [1, 2 - λ]])
= (5 - λ)(2 - λ) - 4
= λ^2 - 7λ + 6

Now, we need to solve the quadratic equation λ^2 - 7λ + 6 = 0 to find the eigenvalues.

Using the quadratic formula, we get:

λ = (7 ± √(49 - 24)) / 2
= (7 ± √25) / 2
= (7 ± 5) / 2

Therefore, the eigenvalues of the matrix A are:

λ1 = (7 + 5) / 2 = 6
λ2 = (7 - 5) / 2 = 1

So, the eigenvalues of the matrix A = [[5, 4], [1, 2]] are 6 and 1.		 [1, 6]
DeepSeek-Coder-V2 To find the eigenvalues of the matrix A = [[5, 4], [1, 2]], we need to solve the characteristic equation, which is given by the determinant of (A - λI) = 0, where λ is the eigenvalue and I is the identity matrix.

Step 1: Form the matrix (A - λI)
A - λI = [[5, 4], [1, 2]] - λ [[1, 0], [0, 1]]
A - λI = [[5 - λ, 4], [1, 2 - λ]]

Step 2: Calculate the determinant of (A - λI)
det(A - λI) = (5 - λ)(2 - λ) - (4)(1)
det(A - λI) = (5 - λ)(2 - λ) - 4
det(A - λI) = 10 - 5λ - 2λ + λ^2 - 4
det(A - λI) = λ^2 - 7λ + 6

Step 3: Solve the characteristic equation λ^2 - 7λ + 6 = 0
To solve this quadratic equation, we can factor it:
λ^2 - 7λ + 6 = (λ - 1)(λ - 6) = 0

Step 4: Find the eigenvalues by setting each factor equal to zero
λ - 1 = 0 => λ = 1
λ - 6 = 0 => λ = 6

So, the eigenvalues of the matrix A = [[5, 4], [1, 2]] are λ = 1 and λ = 6.		 [1, 6]
Qwen2-72B-Instruct To find the eigenvalues of a 2x2 matrix A = [[a, b], [c, d]], we need to solve the characteristic equation, which is given by:

det(A - λI) = 0

where λ represents the eigenvalues, I is the identity matrix, and det() is the determinant of the matrix.

For matrix A = [[5, 4], [1, 2]], the characteristic equation is:

det([[5 - λ, 4], [1, 2 - λ]]) = 0

Expanding the determinant, we get:

(5 - λ)(2 - λ) - (4)(1) = 0

Simplifying the equation, we get:

λ^2 - 7λ + 6 = 0

Solving the quadratic equation, we get:

λ = (7 ± √(49 - 24))/2 = (7 ± √25)/2 = (7 ± 5)/2

So, the eigenvalues of matrix A are:

λ₁ = (7 + 5)/2 = 6

λ₂ = (7 - 5)/2 = 1

Therefore, the eigenvalues of matrix A = [[5, 4], [1, 2]] are 6 and 1.		 [1, 6]
baichuan4 To find the eigenvalues of matrix A = [[5, 4], [1, 2]], we need to solve the characteristic equation |A - λI| = 0, where λ represents the eigenvalues and I is the identity matrix.

First, we construct the matrix A - λI:

A - λI = [[5 - λ, 4], [1, 2 - λ]]

Next, we calculate the determinant of A - λI:

|A - λI| = (5 - λ)(2 - λ) - (4)(1) = λ^2 - 7λ + 6

Now, we set the determinant equal to zero and solve for λ:

λ^2 - 7λ + 6 = 0

(λ - 6)(λ - 1) = 0

This gives us two solutions for λ:

λ = 6 and λ = 1

Therefore, the eigenvalues of matrix A are 6 and 1.		 [1, 6]
doubao The characteristic equation of a 2x2 matrix [[a,b],[c,d]] is given by λ² - (a+d)λ + (ad-bc) = 0.

For the given matrix A = [[5, 4], [1, 2]], a = 5, d = 2, b = 4, c = 1.

The characteristic equation is λ² - (5+2)λ + (5*2 - 4*1) = 0, which simplifies to λ² - 7λ + 6 = 0.

Factoring this equation gives (λ - 1)(λ - 6) = 0.

So the eigenvalues are λ = 1 and λ = 6.		 [1, 6]
minimax To find the eigenvalues of a matrix A, we need to solve the characteristic equation given by:

det(A - λI) = 0

where A is the matrix, λ is the eigenvalue, and I is the identity matrix of the same size as A.

Given the matrix A = [[5, 4], [1, 2]], let's find its eigenvalues.

First, we construct the matrix (A - λI):

A - λI = [[5 - λ, 4], [1, 2 - λ]]

Now, we find the determinant of this matrix:

det(A - λI) = (5 - λ)(2 - λ) - (4)(1)

Expanding this, we get:

det(A - λI) = λ^2 - 7λ + 10 - 4
det(A - λI) = λ^2 - 7λ + 6

Now, we solve for λ by setting the determinant to zero:

λ^2 - 7λ + 6 = 0

This is a quadratic equation, and we can solve for λ by factoring:

(λ - 6)(λ - 1) = 0

Setting each factor equal to zero gives us the eigenvalues:

λ - 6 = 0 => λ = 6
λ - 1 = 0 => λ = 1

Therefore, the eigenvalues of matrix A are λ1 = 6 and λ2 = 1.		 [1, 6]
kimi To find the eigenvalues of a 2x2 matrix A = [[a, b], [c, d]], you can use the characteristic equation, which is derived from the determinant of (A - λI), where λ represents the eigenvalues and I is the identity matrix.

For the matrix A = [[5, 4], [1, 2]], the characteristic equation is:

|A - λI| = |(5 - λ) 4 | = (5 - λ)(2 - λ) - (4)(1) = λ^2 - 7λ + 6

| 1 (2 - λ) |

Expanding the determinant, we get:

λ^2 - 7λ + 6 = 0

This is a quadratic equation, which can be solved using the quadratic formula:

λ = [-b ± sqrt(b^2 - 4ac)] / (2a)

In this case, a = 1, b = -7, and c = 6. Plugging these values into the quadratic formula gives:

λ = [7 ± sqrt((-7)^2 - 4*1*6)] / (2*1)
λ = [7 ± sqrt(49 - 24)] / 2
λ = [7 ± sqrt(25)] / 2
λ = [7 ± 5] / 2

So the eigenvalues are:

λ1 = (7 + 5) / 2 = 12 / 2 = 6
λ2 = (7 - 5) / 2 = 2 / 2 = 1

Therefore, the eigenvalues of the matrix A = [[5, 4], [1, 2]] are λ1 = 6 and λ2 = 1.		 [1, 6]
wenxinyiyan To find the eigenvalues of matrix A = [[5, 4], [1, 2]], we need to solve the characteristic equation det(A - λI) = 0, where λ is the eigenvalue and I is the identity matrix.

First, we form the matrix A - λI by subtracting λ from each diagonal element of A:

A - λI = [[5 - λ, 4], [1, 2 - λ]]

Next, we find the determinant of A - λI:

det(A - λI) = (5 - λ)(2 - λ) - 4*1 = λ^2 - 7λ + 6

Now, we solve the characteristic equation det(A - λI) = 0:

λ^2 - 7λ + 6 = 0

This is a quadratic equation in λ, which we can solve using the quadratic formula:

λ = [-b ± sqrt(b^2 - 4ac)] / 2a

where a = 1, b = -7, and c = 6. Solving this equation, we get two eigenvalues:

λ1 = [-(-7) + sqrt((-7)^2 - 4*1*6)] / 2*1 = (7 + sqrt(49 - 24)) / 2 = (7 + sqrt(25)) / 2 = (7 + 5) / 2 = 6

λ2 = [-(-7) - sqrt((-7)^2 - 4*1*6)] / 2*1 = (7 - sqrt(49 - 24)) / 2 = (7 - sqrt(25)) / 2 = (7 - 5) / 2 = 1

So, the eigenvalues of matrix A = [[5, 4], [1, 2]] are λ1 = 6 and λ2 = 1.		 [1, 6]
zhipu To find the eigenvalues of a matrix, we need to solve the characteristic equation, which is given by:

\[ \text{det}(A - \lambda I) = 0 \]

where \( A \) is the matrix, \( \lambda \) represents the eigenvalues, \( I \) is the identity matrix of the same size as \( A \), and \( \text{det} \) denotes the determinant.

Given the matrix \( A = \begin{bmatrix} 5 & 4 \\ 1 & 2 \end{bmatrix} \), let's calculate the determinant of \( A - \lambda I \):

\[ A - \lambda I = \begin{bmatrix} 5 & 4 \\ 1 & 2 \end{bmatrix} - \begin{bmatrix} \lambda & 0 \\ 0 & \lambda \end{bmatrix} = \begin{bmatrix} 5 - \lambda & 4 \\ 1 & 2 - \lambda \end{bmatrix} \]

Now, we find the determinant of this matrix:

\[ \text{det}(A - \lambda I) = (5 - \lambda)(2 - \lambda) - (4)(1) \]

\[ = (5 - \lambda)(2 - \lambda) - 4 \]

\[ = 10 - 5\lambda - 2\lambda + \lambda^2 - 4 \]

\[ = \lambda^2 - 7\lambda + 6 \]

Set the determinant equal to zero to get the characteristic equation:

\[ \lambda^2 - 7\lambda + 6 = 0 \]

Now, solve this quadratic equation for \( \lambda \):

\[ \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

where \( a = 1 \), \( b = -7 \), and \( c = 6 \).

\[ \lambda = \frac{7 \pm \sqrt{(-7)^2 - 4(1)(6)}}{2(1)} \]

\[ \lambda = \frac{7 \pm \sqrt{49 - 24}}{2} \]

\[ \lambda = \frac{7 \pm \sqrt{25}}{2} \]

\[ \lambda = \frac{7 \pm 5}{2} \]

This gives us two solutions:

\[ \lambda_1 = \frac{7 + 5}{2} = 6 \]

\[ \lambda_2 = \frac{7 - 5}{2} = 1 \]

Therefore, the eigenvalues of the matrix \( A \) are \( \lambda_1 = 6 \) and \( \lambda_2 = 1 \).		 [1, 6]
gpto1 To find the eigenvalues of the matrix \( A = \begin{bmatrix} 5 & 4 \\ 1 & 2 \end{bmatrix} \), we need to solve the characteristic equation obtained from the determinant of \( (A - \lambda I) \), where \( \lambda \) represents the eigenvalues and \( I \) is the identity matrix.

### Step 1: Write the Matrix \( A - \lambda I \)

Subtract \( \lambda \) times the identity matrix \( I \) from matrix \( A \):

\[
A - \lambda I = \begin{bmatrix} 5 - \lambda & 4 \\ 1 & 2 - \lambda \end{bmatrix}
\]

### Step 2: Compute the Determinant of \( A - \lambda I \)

Set the determinant equal to zero to find the eigenvalues:

\[
\det(A - \lambda I) = 0
\]

Compute the determinant:

\[
\begin{align*}
\det(A - \lambda I) &= (5 - \lambda)(2 - \lambda) - (4)(1) \\
&= (5 - \lambda)(2 - \lambda) - 4
\end{align*}
\]

### Step 3: Expand the Determinant

Expand the expression:

\[
\begin{align*}
(5 - \lambda)(2 - \lambda) &= 5 \cdot 2 - 5 \lambda - 2 \lambda + \lambda^2 \\
&= 10 - 7 \lambda + \lambda^2
\end{align*}
\]

Subtract 4 from the result:

\[
\det(A - \lambda I) = [10 - 7 \lambda + \lambda^2] - 4 = 6 - 7 \lambda + \lambda^2
\]

### Step 4: Form the Characteristic Equation

Set the determinant equal to zero:

\[
\lambda^2 - 7 \lambda + 6 = 0
\]

### Step 5: Solve the Quadratic Equation

Solve for \( \lambda \):

\[
\lambda^2 - 7 \lambda + 6 = 0
\]

Factor the quadratic:

\[
(\lambda - 6)(\lambda - 1) = 0
\]

Set each factor equal to zero:

1. \( \lambda - 6 = 0 \quad \Rightarrow \quad \lambda = 6 \)
2. \( \lambda - 1 = 0 \quad \Rightarrow \quad \lambda = 1 \)

### **Answer:**

The eigenvalues of the matrix \( A \) are:

\[
\lambda_1 = 6 \quad \text{and} \quad \lambda_2 = 1
\]		 [1, 6]
deepseekv3 To find the eigenvalues of the matrix \( A = \begin{bmatrix} 5 & 4 \\ 1 & 2 \end{bmatrix} \), follow these steps:

### Step 1: Write down the characteristic equation
The eigenvalues of a matrix \( A \) are found by solving the characteristic equation:
\[
\det(A - \lambda I) = 0
\]
where \( \lambda \) is the eigenvalue and \( I \) is the identity matrix.

### Step 2: Compute \( A - \lambda I \)
\[
A - \lambda I = \begin{bmatrix} 5 & 4 \\ 1 & 2 \end{bmatrix} - \lambda \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 5 - \lambda & 4 \\ 1 & 2 - \lambda \end{bmatrix}
\]

### Step 3: Compute the determinant of \( A - \lambda I \)
\[
\det(A - \lambda I) = (5 - \lambda)(2 - \lambda) - (4)(1)
\]
\[
\det(A - \lambda I) = (10 - 5\lambda - 2\lambda + \lambda^2) - 4
\]
\[
\det(A - \lambda I) = \lambda^2 - 7\lambda + 6
\]

### Step 4: Set the determinant equal to zero and solve for \( \lambda \)
\[
\lambda^2 - 7\lambda + 6 = 0
\]
This is a quadratic equation, which can be factored as:
\[
(\lambda - 1)(\lambda - 6) = 0
\]
Thus, the eigenvalues are:
\[
\lambda_1 = 1, \quad \lambda_2 = 6
\]

### Final Answer:
The eigenvalues of the matrix \( A \) are \( \lambda_1 = 1 \) and \( \lambda_2 = 6 \).		 [1, 6]