answer:To solve the given system of linear equations using Cramer's rule, we first need to find the determinant of the coefficient matrix (D), and then the determinants of the matrices obtained by replacing each column with the constant terms (Dx, Dy, Dz). Finally, we'll find the values of x, y, and z by dividing the respective determinants by D. The coefficient matrix is: | 2 3 4 | | 5 -2 1 | | 1 1 -3 | D = | 2 3 4 | = 2((-2)(-3) - (1)(1)) - 3(5(-3) - (1)(1)) + 4(5(1) - (-2)(1)) | 5 -2 1 | = 2(6 - 1) - 3(-15 - 1) + 4(5 + 2) | 1 1 -3 | = 2(5) + 3(16) + 4(7) = 10 + 48 + 28 = 86 Now, let's find Dx, Dy, and Dz. Dx is obtained by replacing the first column of the coefficient matrix with the constant terms: | 25 3 4 | | 10 -2 1 | | -4 1 -3 | Dx = | 25 3 4 | = 25((-2)(-3) - (1)(1)) - 3(10(-3) - (1)(-4)) + 4(10(1) - (-2)(-4)) | 10 -2 1 | = 25(6 - 1) - 3(-30 + 4) + 4(10 - 8) | -4 1 -3 | = 25(5) + 3(26) + 4(2) = 125 + 78 + 8 = 211 Dy is obtained by replacing the second column of the coefficient matrix with the constant terms: | 2 25 4 | | 5 10 1 | | 1 -4 -3 | Dy = | 2 25 4 | = 2(10(-3) - (1)(-4)) - 25(5(-3) - (1)(-4)) + 4(5(-4) - (10)(1)) | 5 10 1 | = 2(-30 + 4) - 25(-15 + 4) + 4(-20 - 10) | 1 -4 -3 | = 2(-26) + 25(11) - 4(30) = -52 + 275 - 120 = 103 Dz is obtained by replacing the third column of the coefficient matrix with the constant terms: | 2 3 25 | | 5 -2 10 | | 1 1 -4 | Dz = | 2 3 25 | = 2((-2)(-4) - (10)(1)) - 3(5(-4) - (10)(1)) + 25(5(1) - (-2)(1)) | 5 -2 10 | = 2(8 - 10) - 3(-20 - 10) + 25(5 + 2) | 1 1 -4 | = 2(-2) + 3(30) + 25(7) = -4 + 90 + 175 = 261 Now, we can find the values of x, y, and z: x = Dx / D = 211 / 86 y = Dy / D = 103 / 86 z = Dz / D = 261 / 86 So, the solution to the given system of linear equations is: x = 211/86 y = 103/86 z = 261/86

answer:To solve the given system of linear equations using Cramer's rule, we first need to find the determinant of the coefficient matrix (D), and then the determinants of the matrices obtained by replacing the columns with the constant terms (Dx, Dy, Dz). The coefficient matrix is: | 3 2 -1 | | 5 -3 2 | | 1 4 -7 | To find the determinant D: D = 3((-3)(-7) - (2)(4)) - 2((5)(-7) - (2)(1)) - 1((5)(4) - (-3)(1)) D = 3(21 - 8) - 2(-35 - 2) - 1(20 + 3) D = 3(13) + 2(37) + 23 D = 39 + 74 + 23 D = 136 Now, we'll find Dx by replacing the first column with the constant terms: | 5 -6 0 | | 2 -3 2 | | 4 4 -7 | Dx = 5((-3)(-7) - (2)(4)) - (-6)((2)(-7) - (2)(1)) + 0 Dx = 5(21 - 8) + 6(14 - 2) Dx = 5(13) + 6(12) Dx = 65 + 72 Dx = 137 Now, we'll find Dy by replacing the second column with the constant terms: | 3 5 -1 | | 5 -6 2 | | 1 0 -7 | Dy = 3((-6)(-7) - (2)(0)) - 5((5)(-7) - (2)(1)) - 1((5)(0) - (-6)(1)) Dy = 3(42) - 5(-35 - 2) - 6 Dy = 126 + 5(37) - 6 Dy = 126 + 185 - 6 Dy = 305 Now, we'll find Dz by replacing the third column with the constant terms: | 3 2 5 | | 5 -3 -6 | | 1 4 0 | Dz = 3((-3)(0) - (-6)(4)) - 2((5)(0) - (-6)(1)) - 5((5)(4) - (-3)(1)) Dz = 3(24) - 2(-6) - 5(20 + 3) Dz = 72 + 12 - 5(23) Dz = 84 - 115 Dz = -31 Now, we can find the values of x, y, and z using the determinants: x = Dx / D = 137 / 136 y = Dy / D = 305 / 136 z = Dz / D = -31 / 136 So, the solution is: x = 137/136 y = 305/136 z = -31/136

answer:To solve the given system of linear equations using Cramer's rule, we first need to find the determinant of the coefficient matrix (D), and then the determinants of the matrices obtained by replacing the columns with the constant terms (Dx, Dy, Dz). The coefficient matrix is: | 2 1 -1 | | 1 -1 3 | | 3 4 2 | Let's find the determinant of the coefficient matrix (D): D = 2((-1)(2) - (3)(4)) - 1((1)(2) - (3)(3)) + (-1)((1)(4) - (-1)(3)) D = 2(-2 - 12) - 1(2 - 9) - (4 + 3) D = 2(-14) + 1(7) - 7 D = -28 + 7 - 7 D = -28 Now, let's find the determinant of the matrix obtained by replacing the first column with the constant terms (Dx): | 5 1 -1 | | 4 -1 3 | | 7 4 2 | Dx = 5((-1)(2) - (3)(4)) - 1((4)(2) - (3)(7)) + (-1)((4)(4) - (-1)(7)) Dx = 5(-2 - 12) - 1(8 - 21) + (16 + 7) Dx = 5(-14) + 1(13) + 23 Dx = -70 + 13 + 23 Dx = -34 Now, let's find the determinant of the matrix obtained by replacing the second column with the constant terms (Dy): | 2 5 -1 | | 1 4 3 | | 3 7 2 | Dy = 2((4)(2) - (3)(7)) - 5((1)(2) - (3)(3)) + (-1)((1)(7) - (4)(3)) Dy = 2(8 - 21) - 5(2 - 9) + (3 - 12) Dy = 2(-13) + 5(7) - 9 Dy = -26 + 35 - 9 Dy = 0 Now, let's find the determinant of the matrix obtained by replacing the third column with the constant terms (Dz): | 2 1 5 | | 1 -1 4 | | 3 4 7 | Dz = 2((-1)(7) - (4)(4)) - 1((1)(7) - (-1)(3)) + 5((1)(4) - (-1)(3)) Dz = 2(-7 - 16) - 1(7 + 3) + 5(4 + 3) Dz = 2(-23) - 1(10) + 5(7) Dz = -46 - 10 + 35 Dz = -21 Now, we can find the values of x, y, and z using the determinants: x = Dx / D = -34 / -28 = 17/14 y = Dy / D = 0 / -28 = 0 z = Dz / D = -21 / -28 = 3/4 So, the solution to the given system of linear equations is: x = 17/14, y = 0, z = 3/4

answer:To solve the system of linear equations using Cramer's rule, we first need to find the determinant of the coefficient matrix (D), and then the determinants of the matrices obtained by replacing the columns with the constants (Dx, Dy, Dz). Finally, we will find the values of x, y, and z by dividing the respective determinants by D. The coefficient matrix is: | 2 3 5 | | 4 2 3 | | 5 4 2 | To find the determinant D, we use the formula: D = 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. D = 2(2*2 - 3*4) - 3(4*2 - 5*3) + 5(4*4 - 5*2) D = 2(-4) - 3(-7) + 5(6) D = -8 + 21 + 30 D = 43 Now, we will find the determinants Dx, Dy, and Dz by replacing the columns with the constants. For Dx, replace the first column with the constants: | 1 3 5 | | -2 2 3 | | -1 4 2 | Dx = 1(2*2 - 3*4) - 3(-2*2 - (-1)*3) + 5(-2*4 - (-1)*2) Dx = 1(-4) - 3(4) + 5(-6) Dx = -4 - 12 - 30 Dx = -46 For Dy, replace the second column with the constants: | 2 1 5 | | 4 -2 3 | | 5 -1 2 | Dy = 2(-2*2 - 3*(-1)) - 1(4*2 - 5*3) + 5(4*(-1) - 5*(-2)) Dy = 2(-4 + 3) - 1(-2) + 5(-4 + 10) Dy = 2(-1) + 2 + 5(6) Dy = -2 + 2 + 30 Dy = 30 For Dz, replace the third column with the constants: | 2 3 1 | | 4 2 -2 | | 5 4 -1 | Dz = 2(2*(-1) - 4*(-2)) - 3(4*(-1) - 5*(-2)) + 1(4*4 - 5*2) Dz = 2(-2 + 8) - 3(-4 + 10) + 1(16 - 10) Dz = 2(6) - 3(6) + 1(6) Dz = 12 - 18 + 6 Dz = 0 Now, we can find the values of x, y, and z: x = Dx / D = -46 / 43 y = Dy / D = 30 / 43 z = Dz / D = 0 / 43 = 0 So, the solution for the system of linear equations is: x = -46/43 y = 30/43 z = 0