When calculating the determinant of a \(2 \times 2\) matrix, use the formula \( \text{Determinant} = ad - bc \). The sign of the determinant determines whether the matrix is invertible (non-zero determinant) or singular (zero determinant). Keep in mind that the determinant can also help determine whether a system of linear equations has a unique solution. In this case, calculating and checking the parity (even or odd) of the determinant can be useful in some problems.
The determinant of a \(2 \times 2\) matrix \(\begin{bmatrix} a & b \\ c & d \end{bmatrix}\) is:
\(\text{Determinant} = ad - bc.\)
Calculate the determinants:
Step 1: For (A): \(\text{Determinant} = (1)(5) - (-1)(-1) = 5 - 1 = 4 \text{ (even)}.\)
Step 2: For (B): \(\text{Determinant} = (13)(15) - (-1)(-1) = 195 - 1 = 194 \text{ (even)}.\)
Step 3: For (C): \(\text{Determinant} = (16)(15) - (-1)(-11) = 240 + 11 = 251 \text{ (odd)}.\)
Step 4: For (D): \(\text{Determinant} = (6)(15) - (-12)(11) = 90 + 132 = 222 \text{ (even)}.\)
Thus, matrices (A), (B), and (D) have even determinants.
The determinant of a \(2 \times 2\) matrix \[ \begin{bmatrix} a & b \\ c & d \end{bmatrix} \] is given by: \[ \text{Determinant} = ad - bc. \]
Step 1: Calculate the determinant for matrix (A):
For matrix (A): \[ \text{Determinant} = (1)(5) - (-1)(-1) = 5 - 1 = 4 \text{ (even)}. \]Step 2: Calculate the determinant for matrix (B):
For matrix (B): \[ \text{Determinant} = (13)(15) - (-1)(-1) = 195 - 1 = 194 \text{ (even)}. \]Step 3: Calculate the determinant for matrix (C):
For matrix (C): \[ \text{Determinant} = (16)(15) - (-1)(-11) = 240 + 11 = 251 \text{ (odd)}. \]Step 4: Calculate the determinant for matrix (D):
For matrix (D): \[ \text{Determinant} = (6)(15) - (-12)(11) = 90 + 132 = 222 \text{ (even)}. \]Conclusion: Thus, matrices (A), (B), and (D) have even determinants, while matrix (C) has an odd determinant.
If \( A \), \( B \), and \( \left( \text{adj}(A^{-1}) + \text{adj}(B^{-1}) \right) \) are non-singular matrices of the same order, then the inverse of \[ A \left( \text{adj}(A^{-1}) + \text{adj}(B^{-1}) \right) B \] is equal to: