Welcome to Westonci.ca, your ultimate destination for finding answers to a wide range of questions from experts. Discover the answers you need from a community of experts ready to help you with their knowledge and experience in various fields. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform.
Sagot :
To determine the value of [tex]\( k \)[/tex] such that the rank of the matrix [tex]\( A \)[/tex] is maximized, we follow these steps:
First, we define the matrix [tex]\( A \)[/tex]:
[tex]\[ A = \begin{bmatrix} 1 & 1 & -1 \\ 2 & -3 & 4 \\ 3 & -2 & k \end{bmatrix} \][/tex]
### Step 1: Evaluate the 2x2 Submatrix Determinants
1. Select the submatrix formed by the first two rows and the first two columns:
[tex]\[ A_{1} = \begin{bmatrix} 1 & 1 \\ 2 & -3 \end{bmatrix} \][/tex]
Calculate the determinant of [tex]\( A_{1} \)[/tex]:
[tex]\[ \text{det}(A_{1}) = (1)(-3) - (1)(2) = -3 - 2 = -5 \][/tex]
2. Select the submatrix formed by the first two rows and the first and third columns:
[tex]\[ A_{2} = \begin{bmatrix} 1 & -1 \\ 2 & 4 \end{bmatrix} \][/tex]
Calculate the determinant of [tex]\( A_{2} \)[/tex]:
[tex]\[ \text{det}(A_{2}) = (1)(4) - (-1)(2) = 4 + 2 = 6 \][/tex]
3. Select the submatrix formed by the first two rows and the second and third columns:
[tex]\[ A_{3} = \begin{bmatrix} 1 & -1 \\ -3 & 4 \end{bmatrix} \][/tex]
Calculate the determinant of [tex]\( A_{3} \)[/tex]:
[tex]\[ \text{det}(A_{3}) = (1)(4) - (-1)(-3) = 4 - 3 = 1 \][/tex]
Given these calculations, we confirm that the 2x2 submatrix determinants are [tex]\( -5 \)[/tex], [tex]\( 6 \)[/tex], and [tex]\( 1 \)[/tex]. Since there are non-zero determinants among the 2x2 submatrices, the rank is at least 2.
### Step 2: Evaluate the 3x3 Determinant
Next, we need to evaluate if the rank of [tex]\( A \)[/tex] can be 3, which requires the determinant of the entire 3x3 matrix [tex]\( A \)[/tex] to be non-zero.
Calculate the determinant of [tex]\( A \)[/tex]:
[tex]\[ \text{det}(A) = \begin{vmatrix} 1 & 1 & -1 \\ 2 & -3 & 4 \\ 3 & -2 & k \end{vmatrix} \][/tex]
Use cofactor expansion along the first row:
[tex]\[ \text{det}(A) = 1 \cdot \begin{vmatrix} -3 & 4 \\ -2 & k \end{vmatrix} - 1 \cdot \begin{vmatrix} 2 & 4 \\ 3 & k \end{vmatrix} - 1 \cdot \begin{vmatrix} 2 & -3 \\ 3 & -2 \end{vmatrix} \][/tex]
Calculate each 2x2 determinant:
[tex]\[ \begin{vmatrix} -3 & 4 \\ -2 & k \end{vmatrix} = (-3)(k) - (4)(-2) = -3k + 8 \][/tex]
[tex]\[ \begin{vmatrix} 2 & 4 \\ 3 & k \end{vmatrix} = (2)(k) - (4)(3) = 2k - 12 \][/tex]
[tex]\[ \begin{vmatrix} 2 & -3 \\ 3 & -2 \end{vmatrix} = (2)(-2) - (-3)(3) = -4 + 9 = 5 \][/tex]
Combine these results:
[tex]\[ \text{det}(A) = 1(-3k + 8) - 1(2k - 12) - 5 \][/tex]
[tex]\[ = -3k + 8 - 2k + 12 - 5 \][/tex]
[tex]\[ = -5k + 15 \][/tex]
To maximize the rank of [tex]\( A \)[/tex], we need this determinant to be non-zero:
[tex]\[ -5k + 15 \neq 0 \][/tex]
[tex]\[ -5k \neq -15 \][/tex]
[tex]\[ k \neq 3 \][/tex]
Thus, [tex]\( k = 3 \)[/tex] is the value at which the matrix [tex]\( A \)[/tex] has a rank of 2. For the matrix [tex]\( A \)[/tex] to have a rank of 3, [tex]\( k \)[/tex] must take any value other than 3. Therefore, the rank of the matrix is maximized when [tex]\( k \neq 3 \)[/tex].
First, we define the matrix [tex]\( A \)[/tex]:
[tex]\[ A = \begin{bmatrix} 1 & 1 & -1 \\ 2 & -3 & 4 \\ 3 & -2 & k \end{bmatrix} \][/tex]
### Step 1: Evaluate the 2x2 Submatrix Determinants
1. Select the submatrix formed by the first two rows and the first two columns:
[tex]\[ A_{1} = \begin{bmatrix} 1 & 1 \\ 2 & -3 \end{bmatrix} \][/tex]
Calculate the determinant of [tex]\( A_{1} \)[/tex]:
[tex]\[ \text{det}(A_{1}) = (1)(-3) - (1)(2) = -3 - 2 = -5 \][/tex]
2. Select the submatrix formed by the first two rows and the first and third columns:
[tex]\[ A_{2} = \begin{bmatrix} 1 & -1 \\ 2 & 4 \end{bmatrix} \][/tex]
Calculate the determinant of [tex]\( A_{2} \)[/tex]:
[tex]\[ \text{det}(A_{2}) = (1)(4) - (-1)(2) = 4 + 2 = 6 \][/tex]
3. Select the submatrix formed by the first two rows and the second and third columns:
[tex]\[ A_{3} = \begin{bmatrix} 1 & -1 \\ -3 & 4 \end{bmatrix} \][/tex]
Calculate the determinant of [tex]\( A_{3} \)[/tex]:
[tex]\[ \text{det}(A_{3}) = (1)(4) - (-1)(-3) = 4 - 3 = 1 \][/tex]
Given these calculations, we confirm that the 2x2 submatrix determinants are [tex]\( -5 \)[/tex], [tex]\( 6 \)[/tex], and [tex]\( 1 \)[/tex]. Since there are non-zero determinants among the 2x2 submatrices, the rank is at least 2.
### Step 2: Evaluate the 3x3 Determinant
Next, we need to evaluate if the rank of [tex]\( A \)[/tex] can be 3, which requires the determinant of the entire 3x3 matrix [tex]\( A \)[/tex] to be non-zero.
Calculate the determinant of [tex]\( A \)[/tex]:
[tex]\[ \text{det}(A) = \begin{vmatrix} 1 & 1 & -1 \\ 2 & -3 & 4 \\ 3 & -2 & k \end{vmatrix} \][/tex]
Use cofactor expansion along the first row:
[tex]\[ \text{det}(A) = 1 \cdot \begin{vmatrix} -3 & 4 \\ -2 & k \end{vmatrix} - 1 \cdot \begin{vmatrix} 2 & 4 \\ 3 & k \end{vmatrix} - 1 \cdot \begin{vmatrix} 2 & -3 \\ 3 & -2 \end{vmatrix} \][/tex]
Calculate each 2x2 determinant:
[tex]\[ \begin{vmatrix} -3 & 4 \\ -2 & k \end{vmatrix} = (-3)(k) - (4)(-2) = -3k + 8 \][/tex]
[tex]\[ \begin{vmatrix} 2 & 4 \\ 3 & k \end{vmatrix} = (2)(k) - (4)(3) = 2k - 12 \][/tex]
[tex]\[ \begin{vmatrix} 2 & -3 \\ 3 & -2 \end{vmatrix} = (2)(-2) - (-3)(3) = -4 + 9 = 5 \][/tex]
Combine these results:
[tex]\[ \text{det}(A) = 1(-3k + 8) - 1(2k - 12) - 5 \][/tex]
[tex]\[ = -3k + 8 - 2k + 12 - 5 \][/tex]
[tex]\[ = -5k + 15 \][/tex]
To maximize the rank of [tex]\( A \)[/tex], we need this determinant to be non-zero:
[tex]\[ -5k + 15 \neq 0 \][/tex]
[tex]\[ -5k \neq -15 \][/tex]
[tex]\[ k \neq 3 \][/tex]
Thus, [tex]\( k = 3 \)[/tex] is the value at which the matrix [tex]\( A \)[/tex] has a rank of 2. For the matrix [tex]\( A \)[/tex] to have a rank of 3, [tex]\( k \)[/tex] must take any value other than 3. Therefore, the rank of the matrix is maximized when [tex]\( k \neq 3 \)[/tex].
We hope this information was helpful. Feel free to return anytime for more answers to your questions and concerns. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Keep exploring Westonci.ca for more insightful answers to your questions. We're here to help.
