1. **Stating the problem:** Calculate the determinant of the matrix $$\begin{pmatrix} 1 & 1 & -2 & 0 \\ 3 & 6 & -2 & 5 \\ 1 & 0 & 6 & 4 \\ 2 & 3 & 5 & -1 \end{pmatrix}$$ using three methods: a) Expansion along the $i$-th row ($i=4$). b) Expansion along the $j$-th column ($j=1$). c) Make all elements except one in the $j$-th column zero and expand along that column. 2. **Formula and rules:** The determinant of a matrix can be expanded along any row or column using the cofactor expansion: $$\det(A) = \sum_{k=1}^n (-1)^{i+k} a_{ik} M_{ik}$$ for expansion along row $i$, or $$\det(A) = \sum_{k=1}^n (-1)^{k+j} a_{kj} M_{kj}$$ for expansion along column $j$, where $M_{ik}$ or $M_{kj}$ is the minor determinant after removing the $i$-th row and $k$-th column or $k$-th row and $j$-th column. 3. **Matrix given:** $$A = \begin{pmatrix} 1 & 1 & -2 & 0 \\ 3 & 6 & -2 & 5 \\ 1 & 0 & 6 & 4 \\ 2 & 3 & 5 & -1 \end{pmatrix}$$ --- ### a) Expansion along the 4th row ($i=4$): Row 4 elements: $2, 3, 5, -1$ Calculate cofactors: - For element $a_{4,1} = 2$: $$M_{4,1} = \det \begin{pmatrix} 1 & -2 & 0 \\ 6 & -2 & 5 \\ 0 & 6 & 4 \end{pmatrix}$$ Calculate this minor: $$= 1 \times \det \begin{pmatrix} -2 & 5 \\ 6 & 4 \end{pmatrix} - (-2) \times \det \begin{pmatrix} 6 & 5 \\ 0 & 4 \end{pmatrix} + 0 \times \det(...)$$ $$= 1((-2)(4) - 6(5)) + 2(6 \times 4 - 0) + 0 = 1(-8 - 30) + 2(24) = -38 + 48 = 10$$ - For element $a_{4,2} = 3$: $$M_{4,2} = \det \begin{pmatrix} 1 & -2 & 0 \\ 3 & -2 & 5 \\ 1 & 6 & 4 \end{pmatrix}$$ Calculate this minor: $$= 1 \times \det \begin{pmatrix} -2 & 5 \\ 6 & 4 \end{pmatrix} - (-2) \times \det \begin{pmatrix} 3 & 5 \\ 1 & 4 \end{pmatrix} + 0 \times \det(...)$$ $$= 1((-2)(4) - 6(5)) + 2(3 \times 4 - 1 \times 5) + 0 = 1(-8 - 30) + 2(12 - 5) = -38 + 14 = -24$$ - For element $a_{4,3} = 5$: $$M_{4,3} = \det \begin{pmatrix} 1 & 1 & 0 \\ 3 & 6 & 5 \\ 1 & 0 & 4 \end{pmatrix}$$ Calculate this minor: $$= 1 \times \det \begin{pmatrix} 6 & 5 \\ 0 & 4 \end{pmatrix} - 1 \times \det \begin{pmatrix} 3 & 5 \\ 1 & 4 \end{pmatrix} + 0 \times \det(...)$$ $$= 1(6 \times 4 - 0) - 1(3 \times 4 - 1 \times 5) + 0 = 24 - (12 - 5) = 24 - 7 = 17$$ - For element $a_{4,4} = -1$: $$M_{4,4} = \det \begin{pmatrix} 1 & 1 & -2 \\ 3 & 6 & -2 \\ 1 & 0 & 6 \end{pmatrix}$$ Calculate this minor: $$= 1 \times \det \begin{pmatrix} 6 & -2 \\ 0 & 6 \end{pmatrix} - 1 \times \det \begin{pmatrix} 3 & -2 \\ 1 & 6 \end{pmatrix} + (-2) \times \det \begin{pmatrix} 3 & 6 \\ 1 & 0 \end{pmatrix}$$ $$= 1(6 \times 6 - 0) - 1(3 \times 6 - 1 \times (-2)) + (-2)(3 \times 0 - 1 \times 6)$$ $$= 36 - (18 + 2) + (-2)(0 - 6) = 36 - 20 + 12 = 28$$ Now apply signs $(-1)^{4+k}$: - $k=1$: $(-1)^{5} = -1$ - $k=2$: $(-1)^{6} = 1$ - $k=3$: $(-1)^{7} = -1$ - $k=4$: $(-1)^{8} = 1$ Calculate determinant: $$\det(A) = 2(-1)(10) + 3(1)(-24) + 5(-1)(17) + (-1)(1)(28) = -20 - 72 - 85 - 28 = -205$$ --- ### b) Expansion along the 1st column ($j=1$): Column 1 elements: $1, 3, 1, 2$ Calculate cofactors: - For $a_{1,1} = 1$: $$M_{1,1} = \det \begin{pmatrix} 6 & -2 & 5 \\ 0 & 6 & 4 \\ 3 & 5 & -1 \end{pmatrix}$$ Calculate minor: $$= 6 \times \det \begin{pmatrix} 6 & 4 \\ 5 & -1 \end{pmatrix} - (-2) \times \det \begin{pmatrix} 0 & 4 \\ 3 & -1 \end{pmatrix} + 5 \times \det \begin{pmatrix} 0 & 6 \\ 3 & 5 \end{pmatrix}$$ $$= 6(6 \times (-1) - 5 \times 4) + 2(0 \times (-1) - 3 \times 4) + 5(0 \times 5 - 3 \times 6)$$ $$= 6(-6 - 20) + 2(0 - 12) + 5(0 - 18) = 6(-26) + 2(-12) + 5(-18) = -156 - 24 - 90 = -270$$ - For $a_{2,1} = 3$: $$M_{2,1} = \det \begin{pmatrix} 1 & -2 & 0 \\ 0 & 6 & 4 \\ 3 & 5 & -1 \end{pmatrix}$$ Calculate minor: $$= 1 \times \det \begin{pmatrix} 6 & 4 \\ 5 & -1 \end{pmatrix} - (-2) \times \det \begin{pmatrix} 0 & 4 \\ 3 & -1 \end{pmatrix} + 0 \times \det(...)$$ $$= 1(6 \times (-1) - 5 \times 4) + 2(0 \times (-1) - 3 \times 4) + 0 = 1(-6 - 20) + 2(0 - 12) = -26 - 24 = -50$$ - For $a_{3,1} = 1$: $$M_{3,1} = \det \begin{pmatrix} 1 & -2 & 0 \\ 6 & -2 & 5 \\ 3 & 5 & -1 \end{pmatrix}$$ Calculate minor: $$= 1 \times \det \begin{pmatrix} -2 & 5 \\ 5 & -1 \end{pmatrix} - (-2) \times \det \begin{pmatrix} 6 & 5 \\ 3 & -1 \end{pmatrix} + 0 \times \det(...)$$ $$= 1((-2)(-1) - 5 \times 5) + 2(6 \times (-1) - 3 \times 5) + 0 = 1(2 - 25) + 2(-6 - 15) = -23 + 2(-21) = -23 - 42 = -65$$ - For $a_{4,1} = 2$: $$M_{4,1} = \det \begin{pmatrix} 1 & -2 & 0 \\ 6 & -2 & 5 \\ 0 & 6 & 4 \end{pmatrix}$$ Calculate minor: $$= 1 \times \det \begin{pmatrix} -2 & 5 \\ 6 & 4 \end{pmatrix} - (-2) \times \det \begin{pmatrix} 6 & 5 \\ 0 & 4 \end{pmatrix} + 0 \times \det(...)$$ $$= 1((-2)(4) - 6(5)) + 2(6 \times 4 - 0) + 0 = 1(-8 - 30) + 2(24) = -38 + 48 = 10$$ Apply signs $(-1)^{k+1}$ for column 1: - $k=1$: $(-1)^2 = 1$ - $k=2$: $(-1)^3 = -1$ - $k=3$: $(-1)^4 = 1$ - $k=4$: $(-1)^5 = -1$ Calculate determinant: $$\det(A) = 1(1)(-270) + 3(-1)(-50) + 1(1)(-65) + 2(-1)(10) = -270 + 150 - 65 - 20 = -205$$ --- ### c) Make all elements except one in the 1st column zero and expand along that column: We can use row operations to zero out elements in column 1 except for the first element. - $R_2 \to R_2 - 3R_1$: $$\begin{pmatrix} 1 & 1 & -2 & 0 \\ 0 & 3 & 4 & 5 \\ 1 & 0 & 6 & 4 \\ 2 & 3 & 5 & -1 \end{pmatrix}$$ - $R_3 \to R_3 - R_1$: $$\begin{pmatrix} 1 & 1 & -2 & 0 \\ 0 & 3 & 4 & 5 \\ 0 & -1 & 8 & 4 \\ 2 & 3 & 5 & -1 \end{pmatrix}$$ - $R_4 \to R_4 - 2R_1$: $$\begin{pmatrix} 1 & 1 & -2 & 0 \\ 0 & 3 & 4 & 5 \\ 0 & -1 & 8 & 4 \\ 0 & 1 & 9 & -1 \end{pmatrix}$$ Now column 1 has only one nonzero element at $a_{1,1} = 1$. Expand along column 1: $$\det(A) = 1 \times (-1)^{1+1} \times \det \begin{pmatrix} 3 & 4 & 5 \\ -1 & 8 & 4 \\ 1 & 9 & -1 \end{pmatrix} = \det \begin{pmatrix} 3 & 4 & 5 \\ -1 & 8 & 4 \\ 1 & 9 & -1 \end{pmatrix}$$ Calculate this 3x3 determinant: $$= 3 \times \det \begin{pmatrix} 8 & 4 \\ 9 & -1 \end{pmatrix} - 4 \times \det \begin{pmatrix} -1 & 4 \\ 1 & -1 \end{pmatrix} + 5 \times \det \begin{pmatrix} -1 & 8 \\ 1 & 9 \end{pmatrix}$$ $$= 3(8 \times (-1) - 9 \times 4) - 4((-1)(-1) - 1 \times 4) + 5((-1)(9) - 1 \times 8)$$ $$= 3(-8 - 36) - 4(1 - 4) + 5(-9 - 8) = 3(-44) - 4(-3) + 5(-17) = -132 + 12 - 85 = -205$$ --- **Final answer:** $$\boxed{-205}$$ All three methods yield the same determinant value $-205$.