1. **State the problem:** Calculate the determinant of the given 5x5 matrix: $$\begin{bmatrix} 1 & 2 & 3 & 1 & 5 \\ 0 & 1 & 0 & 5 & 1 \\ 2 & 1 & 2 & 3 & 2 \\ 0 & 3 & 0 & 1 & 3 \\ 3 & 2 & 1 & 3 & 4 \end{bmatrix}$$ 2. **Formula and rules:** The determinant of a 5x5 matrix can be calculated by expansion along a row or column using minors and cofactors: $$\det(A) = \sum_{j=1}^5 (-1)^{i+j} a_{ij} M_{ij}$$ where $a_{ij}$ is the element in row $i$, column $j$, and $M_{ij}$ is the determinant of the $(5-1) \times (5-1)$ submatrix formed by deleting row $i$ and column $j$. 3. **Choose row or column for expansion:** We choose the second row because it has two zeros, simplifying calculations: Row 2: $[0, 1, 0, 5, 1]$ 4. **Calculate cofactors for nonzero elements in row 2:** - For $a_{2,2} = 1$: Delete row 2 and column 2: $$\begin{bmatrix} 1 & 3 & 1 & 5 \\ 2 & 2 & 3 & 2 \\ 0 & 0 & 1 & 3 \\ 3 & 1 & 3 & 4 \end{bmatrix}$$ - For $a_{2,4} = 5$: Delete row 2 and column 4: $$\begin{bmatrix} 1 & 2 & 3 & 5 \\ 2 & 1 & 2 & 2 \\ 0 & 3 & 0 & 3 \\ 3 & 2 & 1 & 4 \end{bmatrix}$$ - For $a_{2,5} = 1$: Delete row 2 and column 5: $$\begin{bmatrix} 1 & 2 & 3 & 1 \\ 2 & 1 & 2 & 3 \\ 0 & 3 & 0 & 1 \\ 3 & 2 & 1 & 3 \end{bmatrix}$$ 5. **Calculate each 4x4 determinant:** We calculate each 4x4 determinant by expanding along the first row. **First 4x4 determinant:** $$D_1 = \begin{vmatrix} 1 & 3 & 1 & 5 \\ 2 & 2 & 3 & 2 \\ 0 & 0 & 1 & 3 \\ 3 & 1 & 3 & 4 \end{vmatrix}$$ Expand along first row: $$D_1 = 1 \cdot C_{11} - 3 \cdot C_{12} + 1 \cdot C_{13} - 5 \cdot C_{14}$$ Calculate minors $C_{1j}$ (3x3 determinants): - $C_{11} = \begin{vmatrix} 2 & 3 & 2 \\ 0 & 1 & 3 \\ 1 & 3 & 4 \end{vmatrix} = 2(1\cdot4 - 3\cdot3) - 3(0\cdot4 - 3\cdot1) + 2(0\cdot3 - 1\cdot1) = 2(4-9) - 3(0-3) + 2(0-1) = 2(-5) + 9 - 2 = -10 + 9 - 2 = -3$ - $C_{12} = \begin{vmatrix} 2 & 3 & 2 \\ 0 & 1 & 3 \\ 3 & 3 & 4 \end{vmatrix} = 2(1\cdot4 - 3\cdot3) - 3(0\cdot4 - 3\cdot3) + 2(0\cdot3 - 1\cdot3) = 2(4-9) - 3(0-9) + 2(0-3) = 2(-5) + 27 - 6 = -10 + 27 - 6 = 11$ - $C_{13} = \begin{vmatrix} 2 & 2 & 2 \\ 0 & 0 & 3 \\ 3 & 1 & 4 \end{vmatrix} = 2(0\cdot4 - 3\cdot1) - 2(0\cdot4 - 3\cdot3) + 2(0\cdot1 - 0\cdot3) = 2(0-3) - 2(0-9) + 2(0-0) = -6 + 18 + 0 = 12$ - $C_{14} = \begin{vmatrix} 2 & 2 & 3 \\ 0 & 0 & 1 \\ 3 & 1 & 3 \end{vmatrix} = 2(0\cdot3 - 1\cdot1) - 2(0\cdot3 - 1\cdot3) + 3(0\cdot1 - 0\cdot3) = 2(0-1) - 2(0-3) + 3(0-0) = -2 + 6 + 0 = 4$ Calculate $D_1$: $$D_1 = 1(-3) - 3(11) + 1(12) - 5(4) = -3 - 33 + 12 - 20 = -44$$ **Second 4x4 determinant:** $$D_2 = \begin{vmatrix} 1 & 2 & 3 & 5 \\ 2 & 1 & 2 & 2 \\ 0 & 3 & 0 & 3 \\ 3 & 2 & 1 & 4 \end{vmatrix}$$ Expand along first row: $$D_2 = 1 \cdot C_{11} - 2 \cdot C_{12} + 3 \cdot C_{13} - 5 \cdot C_{14}$$ Calculate minors: - $C_{11} = \begin{vmatrix} 1 & 2 & 2 \\ 3 & 0 & 3 \\ 2 & 1 & 4 \end{vmatrix} = 1(0\cdot4 - 3\cdot1) - 2(3\cdot4 - 3\cdot2) + 2(3\cdot1 - 0\cdot2) = 1(0-3) - 2(12-6) + 2(3-0) = -3 - 12 + 6 = -9$ - $C_{12} = \begin{vmatrix} 2 & 2 & 2 \\ 0 & 0 & 3 \\ 3 & 1 & 4 \end{vmatrix} = 2(0\cdot4 - 3\cdot1) - 2(0\cdot4 - 3\cdot3) + 2(0\cdot1 - 0\cdot3) = 2(0-3) - 2(0-9) + 2(0-0) = -6 + 18 + 0 = 12$ - $C_{13} = \begin{vmatrix} 2 & 1 & 2 \\ 0 & 3 & 3 \\ 3 & 2 & 4 \end{vmatrix} = 2(3\cdot4 - 3\cdot2) - 1(0\cdot4 - 3\cdot3) + 2(0\cdot2 - 3\cdot3) = 2(12-6) - 1(0-9) + 2(0-9) = 12 + 9 - 18 = 3$ - $C_{14} = \begin{vmatrix} 2 & 1 & 2 \\ 0 & 3 & 0 \\ 3 & 2 & 1 \end{vmatrix} = 2(3\cdot1 - 0\cdot2) - 1(0\cdot1 - 0\cdot3) + 2(0\cdot2 - 3\cdot3) = 2(3-0) - 1(0-0) + 2(0-9) = 6 + 0 - 18 = -12$ Calculate $D_2$: $$D_2 = 1(-9) - 2(12) + 3(3) - 5(-12) = -9 - 24 + 9 + 60 = 36$$ **Third 4x4 determinant:** $$D_3 = \begin{vmatrix} 1 & 2 & 3 & 1 \\ 2 & 1 & 2 & 3 \\ 0 & 3 & 0 & 1 \\ 3 & 2 & 1 & 3 \end{vmatrix}$$ Expand along first row: $$D_3 = 1 \cdot C_{11} - 2 \cdot C_{12} + 3 \cdot C_{13} - 1 \cdot C_{14}$$ Calculate minors: - $C_{11} = \begin{vmatrix} 1 & 2 & 3 \\ 3 & 0 & 1 \\ 2 & 1 & 3 \end{vmatrix} = 1(0\cdot3 - 1\cdot1) - 2(3\cdot3 - 1\cdot2) + 3(3\cdot1 - 0\cdot2) = 1(0-1) - 2(9-2) + 3(3-0) = -1 - 14 + 9 = -6$ - $C_{12} = \begin{vmatrix} 2 & 2 & 3 \\ 0 & 0 & 1 \\ 3 & 1 & 3 \end{vmatrix} = 2(0\cdot3 - 1\cdot1) - 2(0\cdot3 - 1\cdot3) + 3(0\cdot1 - 0\cdot3) = 2(0-1) - 2(0-3) + 3(0-0) = -2 + 6 + 0 = 4$ - $C_{13} = \begin{vmatrix} 2 & 1 & 3 \\ 0 & 3 & 1 \\ 3 & 2 & 3 \end{vmatrix} = 2(3\cdot3 - 1\cdot2) - 1(0\cdot3 - 1\cdot3) + 3(0\cdot2 - 3\cdot3) = 2(9-2) - 1(0-3) + 3(0-9) = 14 + 3 - 27 = -10$ - $C_{14} = \begin{vmatrix} 2 & 1 & 2 \\ 0 & 3 & 0 \\ 3 & 2 & 1 \end{vmatrix} = 2(3\cdot1 - 0\cdot2) - 1(0\cdot1 - 0\cdot3) + 2(0\cdot2 - 3\cdot3) = 2(3-0) - 1(0-0) + 2(0-9) = 6 + 0 - 18 = -12$ Calculate $D_3$: $$D_3 = 1(-6) - 2(4) + 3(-10) - 1(-12) = -6 - 8 - 30 + 12 = -32$$ 6. **Calculate determinant of original matrix using row 2 expansion:** $$\det = \sum_{j=1}^5 (-1)^{2+j} a_{2j} M_{2j} = (-1)^{2+2} \cdot 1 \cdot D_1 + (-1)^{2+4} \cdot 5 \cdot D_2 + (-1)^{2+5} \cdot 1 \cdot D_3$$ Calculate signs: - $(-1)^{4} = 1$ - $(-1)^{6} = 1$ - $(-1)^{7} = -1$ So: $$\det = 1 \cdot (-44) + 1 \cdot 5 \cdot 36 - 1 \cdot (-32) = -44 + 180 + 32 = 168$$ **Final answer:** $$\boxed{168}$$