QuestionFinding the determinant of a $3 \times 3$ matrix
Evaluate the following determinant. $$\left|\begin{array}{ccc} 0 & -4 & 6 \\ 3 & 3 & -3 \\ -3 & 3 & 5 \end{array}\right|$$

Studdy Solution
Calculate the determinant using the cofactor expansion formula along the first row:
The determinant of a $3 \times 3$ matrix is given by: $\text{det}(A) = a_{11}C_{11} + a_{12}C_{12} + a_{13}C_{13}$ where $C_{ij}$ is the cofactor of element $a_{ij}$.
For the given matrix: $\text{det}(A) = 0 \cdot C_{11} + (-4) \cdot C_{12} + 6 \cdot C_{13}$
Calculate each cofactor: - $C_{11}$ is not needed since it is multiplied by 0. - $C_{12}$ is the determinant of the submatrix obtained by removing the first row and second column: C_{12} = \left|\begin{array}{cc} 3 & -3 \\ -3 & 5 \end{array}\right| \] Calculate \(C_{12}\): C_{12} = (3)(5) - (-3)(-3) = 15 - 9 = 6 \]
- $C_{13}$ is the determinant of the submatrix obtained by removing the first row and third column: C_{13} = \left|\begin{array}{cc} 3 & 3 \\ -3 & 3 \end{array}\right| \] Calculate \(C_{13}\): C_{13} = (3)(3) - (3)(-3) = 9 + 9 = 18 \]
Substitute back into the determinant formula: $\text{det}(A) = 0 \cdot C_{11} + (-4) \cdot 6 + 6 \cdot 18$ $\text{det}(A) = 0 - 24 + 108$ $\text{det}(A) = 84$
The determinant of the matrix is:
$$\boxed{84}$$