answer:To find the determinant of a 3x3 matrix, we can use the following formula: det(A) = a(ei - fh) - b(di - fg) + c(dh - eg) where A is the given matrix, and the elements are represented as follows: begin{bmatrix} a & b & c d & e & f g & h & i end{bmatrix} In our case, the matrix is: begin{bmatrix} 3 & -2 & 4 1 & 5 & -3 2 & 0 & 2 end{bmatrix} So, we have: a = 3, b = -2, c = 4 d = 1, e = 5, f = -3 g = 2, h = 0, i = 2 Now, we can plug these values into the formula: det(A) = 3(5 cdot 2 - (-3) cdot 0) - (-2)(1 cdot 2 - (-3) cdot 2) + 4(1 cdot 0 - 5 cdot 2) det(A) = 3(10 - 0) + 2(2 + 6) - 4(0 - 10) det(A) = 30 + 16 + 40 det(A) = 86 So, the determinant of the given matrix is 86.

answer:To find the determinant of a 3x3 matrix, we will use the following formula: det(A) = a(ei - fh) - b(di - fg) + c(dh - eg) where a, b, c, d, e, f, g, h, and i are the elements of the matrix: | a b c | | d e f | | g h i | In our case, the matrix is: | 2 5 8 | | 1 0 -3 | | 4 -1 2 | So, a = 2, b = 5, c = 8, d = 1, e = 0, f = -3, g = 4, h = -1, and i = 2. Now, we will plug these values into the formula: det(A) = 2(0*2 - (-3)*(-1)) - 5(1*2 - (-3)*4) + 8(1*(-1) - 0*4) det(A) = 2(0 - 3) - 5(2 + 12) + 8(-1) det(A) = 2(-3) - 5(14) + (-8) det(A) = -6 - 70 - 8 det(A) = -84 So, the determinant of the given 3x3 matrix is -84.

answer:To find the determinant of a 3x3 matrix, we can use the following formula: det(A) = a(ei - fh) - b(di - fg) + c(dh - eg) where the matrix A is: begin{bmatrix}a & b & c d & e & f g & h & iend{bmatrix} In our case, the matrix is: begin{bmatrix}2 & 1 & -3 -1 & 4 & 2 0 & 5 & 1end{bmatrix} So, we have: a = 2, b = 1, c = -3, d = -1, e = 4, f = 2, g = 0, h = 5, i = 1 Now, we can plug these values into the formula: det(A) = 2(4cdot1 - 2cdot5) - 1(-1cdot1 - 2cdot0) + (-3)(-1cdot5 - 4cdot0) det(A) = 2(4 - 10) - 1(1) - 3(5) det(A) = 2(-6) - 1 - 15 det(A) = -12 - 1 - 15 det(A) = -28 So, the determinant of the given matrix is -28.

answer:To find the determinant of a 3x3 matrix, we can use the following formula: det(A) = a(ei - fh) - b(di - fg) + c(dh - eg) where a, b, and c are the elements of the first row, d, e, and f are the elements of the second row, and g, h, and i are the elements of the third row. For the given matrix: begin{bmatrix} 3 & 2 & 1 0 & 4 & 2 1 & 5 & 2 end{bmatrix} a = 3, b = 2, c = 1, d = 0, e = 4, f = 2, g = 1, h = 5, and i = 2. Now, we can plug these values into the formula: det(A) = 3(4*2 - 2*5) - 2(0*2 - 2*1) + 1(0*5 - 4*1) det(A) = 3(8 - 10) - 2(0 - 2) + 1(0 - 4) det(A) = 3(-2) - 2(-2) + 1(-4) det(A) = -6 + 4 - 4 det(A) = -6 So, the determinant of the matrix is -6.