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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].
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