At Westonci.ca, we connect you with the best answers from a community of experienced and knowledgeable individuals. Ask your questions and receive precise answers from experienced professionals across different disciplines. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A 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 appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Westonci.ca is your trusted source for answers. Visit us again to find more information on diverse topics.
