1. **State the problem:** Calculate the determinant of the 4x4 matrix $$A=\begin{pmatrix} -1 & 3 & -5 & -1 \\ -6 & 4 & 3 & 2 \\ -4 & -4 & -6 & -4 \\ -5 & 3 & 2 & 2 \end{pmatrix}$$. 2. **Formula and method:** The determinant of a 4x4 matrix can be found by expanding along a row or column using cofactors. The determinant is given by: $$\det(A) = \sum_{j=1}^4 (-1)^{1+j} a_{1j} M_{1j}$$ where $a_{1j}$ are elements of the first row and $M_{1j}$ are the determinants of the 3x3 minors obtained by removing the first row and $j$-th column. 3. **Calculate cofactors for the first row:** - For $a_{11} = -1$: $$M_{11} = \det\begin{pmatrix}4 & 3 & 2 \\ -4 & -6 & -4 \\ 3 & 2 & 2 \end{pmatrix}$$ Calculate this 3x3 determinant: $$=4((-6)(2) - (-4)(2)) - 3((-4)(2) - (-4)(3)) + 2((-4)(2) - (-6)(3))$$ $$=4(-12 + 8) - 3(-8 + 12) + 2(-8 + 18)$$ $$=4(-4) - 3(4) + 2(10) = -16 - 12 + 20 = -8$$ - For $a_{12} = 3$: $$M_{12} = \det\begin{pmatrix} -6 & 3 & 2 \\ -4 & -6 & -4 \\ -5 & 2 & 2 \end{pmatrix}$$ Calculate: $$= -6((-6)(2) - (-4)(2)) - 3((-4)(2) - (-4)(-5)) + 2((-4)(2) - (-6)(-5))$$ $$= -6(-12 + 8) - 3(-8 - 20) + 2(-8 - 30)$$ $$= -6(-4) - 3(-28) + 2(-38) = 24 + 84 - 76 = 32$$ - For $a_{13} = -5$: $$M_{13} = \det\begin{pmatrix} -6 & 4 & 2 \\ -4 & -4 & -4 \\ -5 & 3 & 2 \end{pmatrix}$$ Calculate: $$= -6((-4)(2) - (-4)(3)) - 4((-4)(2) - (-4)(-5)) + 2((-4)(3) - (-4)(-5))$$ $$= -6(-8 + 12) - 4(-8 - 20) + 2(-12 - 20)$$ $$= -6(4) - 4(-28) + 2(-32) = -24 + 112 - 64 = 24$$ - For $a_{14} = -1$: $$M_{14} = \det\begin{pmatrix} -6 & 4 & 3 \\ -4 & -4 & -6 \\ -5 & 3 & 2 \end{pmatrix}$$ Calculate: $$= -6((-4)(2) - (-6)(3)) - 4((-4)(2) - (-6)(-5)) + 3((-4)(3) - (-4)(-5))$$ $$= -6(-8 + 18) - 4(-8 - 30) + 3(-12 - 20)$$ $$= -6(10) - 4(-38) + 3(-32) = -60 + 152 - 96 = -4$$ 4. **Apply signs and sum:** $$\det(A) = (-1)^{1+1}(-1)(-8) + (-1)^{1+2}(3)(32) + (-1)^{1+3}(-5)(24) + (-1)^{1+4}(-1)(-4)$$ $$= 1 \times (-1) \times (-8) - 1 \times 3 \times 32 + 1 \times (-5) \times 24 - 1 \times (-1) \times (-4)$$ $$= 8 - 96 - 120 - 4 = -212$$ 5. **Final answer:** $$\boxed{\det(A) = -212}$$