Matrix Determinants 2Ad1Cc

1. **Problem Statement:** Find the determinant of each given matrix. 2. **Formula and Rules:** - For a 2x2 matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$, determinant is $ad - bc$. - For 3x3 and 4x4 matrices, use cofactor expansion or row reduction. --- ### (a) Matrix $\begin{bmatrix} 1 & -4 \\ 2 & 8 \end{bmatrix}$ 3. Calculate determinant: $$\det = (1)(8) - (2)(-4) = 8 + 8 = 16$$ --- ### (b) Matrix $\begin{bmatrix} 2 & -1 \\ 1 & 4 \end{bmatrix}$ 4. Calculate determinant: $$\det = (2)(4) - (1)(-1) = 8 + 1 = 9$$ --- ### (c) Matrix $\begin{bmatrix} 0 & 1 & 1 \\ 2 & 3 & 0 \\ 1 & 2 & -1 \end{bmatrix}$ 5. Use cofactor expansion along first row: $$\det = 0 \cdot C_{11} - 1 \cdot C_{12} + 1 \cdot C_{13}$$ 6. Calculate cofactors: - $C_{12} = (-1)^{1+2} \det \begin{bmatrix} 2 & 0 \\ 1 & -1 \end{bmatrix} = -1 \times (2 \times -1 - 1 \times 0) = -1 \times (-2) = 2$ - $C_{13} = (-1)^{1+3} \det \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix} = 1 \times (2 \times 2 - 1 \times 3) = 1 \times (4 - 3) = 1$ 7. Substitute: $$\det = 0 - 1 \times 2 + 1 \times 1 = -2 + 1 = -1$$ --- ### (d) Matrix $\begin{bmatrix} 1 & -1 & 2 \\ 1 & 0 & -1 \\ 3 & -2 & 0 \end{bmatrix}$ 8. Use cofactor expansion along first row: $$\det = 1 \cdot C_{11} - (-1) \cdot C_{12} + 2 \cdot C_{13}$$ 9. Calculate cofactors: - $C_{11} = \det \begin{bmatrix} 0 & -1 \\ -2 & 0 \end{bmatrix} = 0 \times 0 - (-1) \times (-2) = 0 - 2 = -2$ - $C_{12} = \det \begin{bmatrix} 1 & -1 \\ 3 & 0 \end{bmatrix} = 1 \times 0 - (-1) \times 3 = 0 + 3 = 3$ - $C_{13} = \det \begin{bmatrix} 1 & 0 \\ 3 & -2 \end{bmatrix} = 1 \times (-2) - 0 \times 3 = -2$ 10. Substitute: $$\det = 1 \times (-2) + 1 \times 3 + 2 \times (-2) = -2 + 3 - 4 = -3$$ --- ### (e) Matrix $\begin{bmatrix} 1 & 2 & 1 \\ -1 & 1 & 3 \\ 1 & 0 & 2 \end{bmatrix}$ 11. Use cofactor expansion along first row: $$\det = 1 \cdot C_{11} - 2 \cdot C_{12} + 1 \cdot C_{13}$$ 12. Calculate cofactors: - $C_{11} = \det \begin{bmatrix} 1 & 3 \\ 0 & 2 \end{bmatrix} = 1 \times 2 - 3 \times 0 = 2$ - $C_{12} = \det \begin{bmatrix} -1 & 3 \\ 1 & 2 \end{bmatrix} = (-1) \times 2 - 3 \times 1 = -2 - 3 = -5$ - $C_{13} = \det \begin{bmatrix} -1 & 1 \\ 1 & 0 \end{bmatrix} = (-1) \times 0 - 1 \times 1 = -1$ 13. Substitute: $$\det = 1 \times 2 - 2 \times (-5) + 1 \times (-1) = 2 + 10 - 1 = 11$$ --- ### (f) Matrix $\begin{bmatrix} 1 & 1 & 2 & -1 \\ 0 & 1 & 3 & 0 \\ 2 & 1 & -1 & 1 \\ 0 & 0 & 1 & -1 \end{bmatrix}$ 14. Use cofactor expansion along first row: $$\det = 1 \cdot C_{11} - 1 \cdot C_{12} + 2 \cdot C_{13} - (-1) \cdot C_{14}$$ 15. Calculate cofactors (3x3 determinants): - $C_{11} = \det \begin{bmatrix} 1 & 3 & 0 \\ 1 & -1 & 1 \\ 0 & 1 & -1 \end{bmatrix}$ 16. Calculate $C_{11}$: Expand along first row: $$= 1 \times \det \begin{bmatrix} -1 & 1 \\ 1 & -1 \end{bmatrix} - 3 \times \det \begin{bmatrix} 1 & 1 \\ 0 & -1 \end{bmatrix} + 0$$ 17. Calculate minors: $$\det \begin{bmatrix} -1 & 1 \\ 1 & -1 \end{bmatrix} = (-1)(-1) - (1)(1) = 1 - 1 = 0$$ $$\det \begin{bmatrix} 1 & 1 \\ 0 & -1 \end{bmatrix} = 1 \times (-1) - 1 \times 0 = -1$$ 18. Substitute: $$C_{11} = 1 \times 0 - 3 \times (-1) + 0 = 3$$ 19. Calculate $C_{12} = \det \begin{bmatrix} 0 & 3 & 0 \\ 2 & -1 & 1 \\ 0 & 1 & -1 \end{bmatrix}$ 20. Expand along first row: $$= 0 - 3 \times \det \begin{bmatrix} 2 & 1 \\ 0 & -1 \end{bmatrix} + 0 = -3 \times (2 \times -1 - 0) = -3 \times (-2) = 6$$ 21. Calculate $C_{13} = \det \begin{bmatrix} 0 & 1 & 0 \\ 2 & 1 & 1 \\ 0 & 0 & -1 \end{bmatrix}$ 22. Expand along first row: $$= 0 - 1 \times \det \begin{bmatrix} 2 & 1 \\ 0 & -1 \end{bmatrix} + 0 = -1 \times (2 \times -1 - 0) = -1 \times (-2) = 2$$ 23. Calculate $C_{14} = \det \begin{bmatrix} 0 & 1 & 3 \\ 2 & 1 & -1 \\ 0 & 0 & 1 \end{bmatrix}$ 24. Expand along first row: $$= 0 - 1 \times \det \begin{bmatrix} 2 & -1 \\ 0 & 1 \end{bmatrix} + 3 \times \det \begin{bmatrix} 2 & 1 \\ 0 & 0 \end{bmatrix}$$ 25. Calculate minors: $$\det \begin{bmatrix} 2 & -1 \\ 0 & 1 \end{bmatrix} = 2 \times 1 - 0 = 2$$ $$\det \begin{bmatrix} 2 & 1 \\ 0 & 0 \end{bmatrix} = 2 \times 0 - 0 = 0$$ 26. Substitute: $$C_{14} = 0 - 1 \times 2 + 3 \times 0 = -2$$ 27. Substitute all cofactors: $$\det = 1 \times 3 - 1 \times 6 + 2 \times 2 - (-1) \times (-2) = 3 - 6 + 4 - 2 = -1$$ --- **Final answers:** - (a) $16$ - (b) $9$ - (c) $-1$ - (d) $-3$ - (e) $11$ - (f) $-1$