answer:To solve the system of linear equations with complex coefficients, we can treat the real and imaginary parts separately as if they were parts of separate real-valued equations. Let's start by solving for x in one of the equations and y in the other: First, let's solve for x in the first equation and y in the second equation: (1) 3x + 4yi = 7 + 2i (2) 2x - 5yi = -2i From equation (1), we have: 3x = 7 + 2i - 4yi Now, multiply both sides of equation (1) by 1/3: x = (7 + 2i - 4yi) / 3 From equation (2), we have: 2x = -2i + 5yi Now, multiply both sides of equation (2) by 1/2: x = (-2i + 5yi) / 2 Since x should be the same on both sides of the equation, we can equate the expressions for x from both equations: (7 + 2i - 4yi) / 3 = (-2i + 5yi) / 2 Cross-multiplying gives us: 2(7 + 2i - 4yi) = 3(-2i + 5yi) Expanding and simplifying: 14 - 8yi + 4y - 20y^2 = -6i + 15y Rearranging the terms: 28yi - 5y^2 = -20yi - 15y^2 - 28y + 14i Combine like terms: (28yi - 20yi) + (5y^2 - 15y^2) = 14i - 28y Combining: 8yi - 10y^2 = 14i - 28y 8yi - 10y^2 = 14i - 14y Factor out y and y^2: y(8i - 10y) = (14 - 14y)i Now, equate the real and imaginary,To solve the system of linear equations with complex coefficients, we can treat the real and imaginary parts separately. Let's rewrite the given equations as: (1) 3x + 4yi = 7 + 2i (2) 2x - 5yi = -2i Now, separate the real and imaginary parts: (1a) 3x = 7 (1b) 4y = 2 (2a) 2x = 0 (2b) -5y = -2 Now, we can solve for x and y: From (1a), we have: x = 7/3 From (1b), we have: y = 2/4 = 1/2 Now, we can check if these values satisfy the second equation: (2a) 2x = 2(7/3) = 14/3 ≠ 0 (2b) -5y = -5(1/2) = -5/2 = -2 Since the values of x and y do not satisfy the second equation, there is no solution to the given system of linear equations with complex coefficients., To solve the system of linear equations, we first separate the real and imaginary parts: For the real part: 3x = 7 x = 7/3 For the imaginary part: 4yi = 2i - 5yi 4yi + 5yi = 2i 9yi = 2i Now, we solve for y by dividing both sides by 9: y = (2i)/9 So the solution is: x = 7/3 y = (2/9)i

answer:First, let's rewrite the third equation to make it easier to read: 3x + (2 + 4i)y - 5z = 3 - 4i Now, let's start the elimination process. We will eliminate x from the second and third equations. To do this, we will multiply the first equation by -1/2 and the second equation by 2: -1x - 3/2y + i/2 z = -2 2x - 4y + 6i z = 4 Now, add the modified first and second equations: (-1x - 3/2y + i/2 z) + (2x - 4y + 6i z) = -2 + 4 x - 5/2y + (13i/2)z = 2 Now, we will eliminate x from the third equation. To do this, we will multiply the first equation by -3/2 and the third equation by 2: -3x - 9/2y + 3i/2 z = -6 6x + (4 + 8i)y - 10z = 6 - 8i Now, add the modified first and third equations: (-3x - 9/2y + 3i/2 z) + (6x + (4 + 8i)y - 10z) = -6 + 6 - 8i 3x + (-1/2 + 8i)y - (19/2 - 3i/2)z = -8i Now we have two equations with two variables (y and z): x - 5/2y + (13i/2)z = 2 (-1/2 + 8i)y - (19/2 - 3i/2)z = -8i To eliminate y, we will multiply the first equation by (1/2 - 8i) and the second equation by 5: (1/2 - 8i)x - (25/4 - 20i)y + (13i/4 - 52)z = 2(1/2 - 8i) (-5/2 + 40i)y + (95/4 + 15i/4)z = -40i Now, add the modified first and second equations: (1/2 - 8i)x - (25/4 - 20i)y + (13i/4 - 52)z + (-5/2 + 40i)y + (95/4 + 15i/4)z = 2(1/2 - 8i) - 40i Now, simplify the equation: (1/2 - 8i)x + (15i/4 + 60)z = -64i + 1 Now, we can solve for x: x = (64i - 1)/(1/2 - 8i) Now, substitute x back into the equation x - 5/2y + (13i/2)z = 2: (64i - 1)/(1/2 - 8i) - 5/2y + (13i/2)z = 2 Now, solve for y: y = (64i - 1)/(1/2 - 8i) - 2 - (13i/2)z / (-5/2) Finally, substitute x and y back into the original first equation: 2x + 3y - i z = 4 Now, we have: 2(64i - 1)/(1/2 - 8i) + 3(64i - 1)/(1/2 - 8i) - 2 - (13i/2)z / (-5/2) - i z = 4 Solve for z: z = (2(64i - 1) + 3(64i - 1 - 2(1/2 - 8i) - 13i(-5/2)))/(1/2 - 8i) - 4/i Now, we have the expressions for x, y, and z in terms of i: x = (64i - 1)/(1/2 - 8i) y = (64i - 1)/(1/2 - 8i) - 2 - (13i/2)z / (-5/2) z = (2(64i - 1) + 3(64i - 1 - 2(1/2 - 8i) - 13i(-5/2)))/(1/2 - 8i) - 4/i These expressions represent the solutions for x, y, and z in terms of the imaginary unit i.,To solve the system of linear equations using the elimination method, we first need to convert the imaginary constants into real numbers by multiplying the equations by their complex conjugates. The complex conjugate of i is -i, and the complex conjugate of 4i is -4i. 1. Multiply the first equation by -i: -i(2x + 3y - i z) = -i(4) -2x + 3y + i*z = -4i 2. Multiply the second equation by 2i: 2i(x - 2y + 3i z) = 2i(2) 2x - 4y - 6zi = 4i Now, we combine the two equations: First, subtract the second equation from the first equation to eliminate the x term: (-2x + 3y + i*z) - (2x - 4y - 6zi) = -4i - 4i y + 7zi = 0 Next, add the first equation and the modified third equation to eliminate the x term: (2x + 3y - i z) + (3x + (2 + 4i) y - 5z) = 4 + (3 - 4i) 5x + 7y - 4zi = 7 - 4i Now we have two equations with two variables: 1. y + 7zi = 0 2. 5x + 7y - 4zi = 7 - 4i To eliminate y, add 7 times the first equation to the second equation: 7(y + 7zi) + (5x + 7y - 4zi) = 7(0) + (7 - 4i) 35y + 7xy + 35zi - 4zi = 7 - 4i Now we have an equation with only one variable, x: 35x - 4zi = 7 - 4i To isolate x, divide both sides of the equation by 35: x - (4/35)zi = (1/5) - (4/35)zi Now, To solve the given system of linear equations with complex coefficients, we can use the elimination method. First, let's write the augmented matrix for the system: 2 3 -i | 4 1 -2 3i | 2 3 2+4i -5 | 3-4i Now, let's apply row operations to eliminate variables and find the solution. 1. Interchange rows 1 and 2 to get a leading coefficient of 1 for x: 1 -2 3i | 2 2 3 -i | 4 3 2+4i -5 | 3-4i 2. Subtract 2 times row 1 from row 2 and then subtract 3 times row 1 from row 3: 1 -2 3i | 2 0 7 -7i | 0 0 8+4i -14 | -3 3. Now, we'll eliminate y by multiplying row 2 by -frac{1}{7} and adding it to row 3: 1 -2 3i | 2 0 1 1-i/7 | 0 0 0 +20i/7 | -3 4. Next, multiply row 2 by 2 + i and add it to row 1: 1 0 -2i + 2 + i | 4 + 2i 0 1 1-i/7 | 0 0 0 +20i/7 | -3 5. Finally, divide row 1 by -i + 2 + i to obtain the solution: 1 0 -1 | (4 - 2i)/5 0 1 1 - i/7 | 0 0 0 20i/7 | -3 The system has infinitely many solutions, where: x = frac{4-2i}{5} - z y = frac{i}{7}z for any complex number z.

answer:To solve this system of linear equations with complex coefficients, we can use the method of substitution or elimination. We will use the elimination method in this case. First, let's rewrite the given equations: Equation 1: 3x + 5yi = 7 + 2zi Equation 2: 2x - 4yi = 2 + 3zi We can eliminate the yi term by multiplying Equation 1 by 4 and Equation 2 by 5, and then adding the two equations: 4 * (3x + 5yi) = 4 * (7 + 2zi) 5 * (2x - 4yi) = 5 * (2 + 3zi) Which gives us: 12x + 20yi = 28 + 8zi 10x - 20yi = 10 + 15zi Now, add the two equations: (12x + 20yi) + (10x - 20yi) = (28 + 8zi) + (10 + 15zi) This simplifies to: 22x = 38 + 23zi Now, divide by 22: x = (38 + 23zi) / 22 x = 19/11 + (23/22)zi Now that we have the value of x, we can substitute it into either Equation 1 or Equation 2 to find the value of yi. We will use Equation 1: 3x + 5yi = 7 + 2zi 3(19/11 + (23/22)zi) + 5yi = 7 + 2zi (57/11 + (69/22)zi) + 5yi = 7 + 2zi Now, equate the real and imaginary parts separately: Real part: 57/11 + 5yi = 7 Imaginary part: 69/22zi = 2zi For the imaginary part, we have: 69/22 = 2 69 = 44 yi = 69/44 Now we have the values of x and yi: x = 19/11 + (23/22)zi yi = 69/44 So the solution to the system of linear equations with complex coefficients is: x = 19/11 + (23/22)zi yi = 69/44, To solve the system of linear equations, we first separate the real and imaginary parts: For the real parts: 3x = 7 + 2zr 2x = 2 + 3zr For the imaginary parts: 5y = 2zi -4y = 3zi Now, let's solve for x and y. From the real parts equation, we can express zr in terms of x: zr = (3x - 7) / 2 ....(1) zr = (2x - 2) / 3 ....(2) Equating (1) and (2), we get: (3x - 7) / 2 = (2x - 2) / 3 9x - 21 = 4x - 4 5x = 17 x = 17/5 Now, using this value of x, we can find the value of zr by substituting x into either (1) or (2): zr = (3 * 17/5 - 7) / 2 zr = (51/5 - 35/5) / 2 zr = 16/10 zr = 4/5 Next, let's find the value of y using the imaginary parts equation: From 5y = 2zi, we have: y = (2zi) / 5 And from -4y = 3zi, we have: y = -(3zi) / 4 Equating these two expressions for y, we get: (2zi) / 5 = -(3zi) / 4 8zi = -15zi z = -8/15 Now we can find the values of x, y, and z: x = 17/5, y = 2zi / 5 = -(3zi) / 4, and z = 4/5 - 8/15 i So, the solution to the system of linear equations is: (x, y, z) = (17/5, -3/20 i, 4/5 - 8/15 i) or in complex form, x + yi + z(ji) = 17/5 - 3/20 i + (4/5 - 8/15 i)j, where j is another imaginary unit perpendicular to i, such that j^2 = -1 and ij = -ji.,To solve the given system of linear equations, we can use the Gaussian elimination method. First, let's write the given system of equations in the form of an augmented matrix: | 3 5i | 7 + 2zi | | 2 -4i | 2 + 3zi | Step 1: Eliminate the "x" term from the second equation. To do this, multiply the first equation by -2/3 and add it to the second equation: | 3 5i | 7 + 2zi | | 0 -11i | 2 - 5zi | Step 2: Divide the second equation by -11i: | 3 5i | 7 + 2zi | | 0 i | 2zi + 3 | Step 3: Eliminate the "5i" term from the first equation. To do this, multiply the second equation by -3i and add it to the first equation: | 3 0 | 7 - 5zi | | 0 1 | 2zi + 3 | Now, we have an upper triangular matrix, and we can find the solution by back-substitution: From the second equation, we have: 2zi + 3 = 0 2zi = -3 zi = -3/2 Now, substitute the value of "zi" into the first equation: 3x + 5yi = 7 - 5zi 3x + 5yi = 7 + 5/2 3x + 5yi = 19/2 Since we have no more equations to solve for y or x, we can express the solution in terms of the constants in the equations: x = (19/6 - 5/2i) / 3 x = (19 - 15i) / 6 So, the solution to the given system of linear equations is: x = (19 - 15i) / 6 y = (-3/2) / 1 z = -3/2

answer:To solve this system of linear equations, we can use the method of substitution or elimination. Here, we will use the elimination method. First, let's multiply the first equation by 2 and the second equation by 1 to make the coefficients of x in both equations equal: (1) 2(x + 2y - z) = 2(6) (2) 1(2x - 3y + 4z) = 1(-7) Now we have: (1) 2x + 4y - 2z = 12 (2) 2x - 3y + 4z = -7 Next, subtract equation (2) from equation (1): (1) - (2): (2x + 4y - 2z) - (2x - 3y + 4z) = 12 - (-7) This simplifies to: 7y - 6z = 19 Now, let's eliminate x from the second and third equations. Multiply the first equation by 3 and the third equation by 2: (1) 3(x + 2y - z) = 3(6) (3) 2(3x + y + 2iz) = 2(4) Now we have: (1) 3x + 6y - 3z = 18 (3) 6x + 2y + 4iz = 8 Subtract equation (1) from equation (3): (3) - (1): (6x + 2y + 4iz) - (3x + 6y - 3z) = 8 - 18 This simplifies to: 3x - 4y + 4iz + 3z = -10 Now we have two equations with two variables (y and z): 7y - 6z = 19 3x - 4y + 4iz + 3z = -10 Let's solve the first equation for y: y = (19 + 6z) / 7 Now substitute this expression for y in the second equation: 3x - 4((19 + 6z) / 7) + 4iz + 3z = -10 Clear the fraction by multiplying both sides by 7: 21x - 4(19 + 6z) + 28iz + 21z = -70 Distribute the -4: 21x - 76 - 24z + 28iz + 21z = -70 Combine like terms: 21x - 76 - 3z + 28iz = -70 Now, let's solve for x: 21x = 70 - 28iz + 76 + 3z x = (146 + 3z - 28iz) / 21 Now we have expressions for x and y in terms of z: x = (146 + 3z - 28iz) / 21 y = (19 + 6z) / 7 Substitute these expressions for x and y in any of the original equations. We'll use the first one: x + 2y - z = 6 (146 + 3z - 28iz) / 21 + 2(19 + 6z) / 7 - z = 6 Clear the fractions by multiplying both sides by 21: (146 + 3z - 28iz) + 6(19 + 6z) - 21z = 126 Distribute the 6: 146 + 3z - 28iz + 114 + 36z - 21z = 126 Combine like terms: 18z - 28iz = -134 Now, we have a single equation with the variable z: 18z - 28iz = -134 To solve for z, we can rewrite the equation as: z(18 - 28i) = -134 Now, divide both sides by (18 - 28i): z = -134 / (18 - 28i) To rationalize the denominator, multiply the numerator and denominator by the conjugate of the denominator, (18 + 28i): z = (-134 * (18 + 28i)) / ((18 - 28i) * (18 + 28i)) Multiply the numerator: z = (-2412 - 3752i) / (18^2 + 28^2) Divide by the denominator: z = (-2412 - 3752i) / (324 + 784) z = (-2412 - 3752i) / 1108 Simplify: z = -2412/1108 - 3752i/1108 z = -2 - 3.4i Now that we have z, we can find x and y using the expressions we found earlier: x = (146 + 3z - 28iz) / 21 x = (146 + 3(-2 - 3.4i) - 28i(-2 - 3.4i)) / 21 x = (146 - 6 - 10.2i + 56i + 95.2i^2) / 21 x = (140 + 45.8i - 95.2) / 21 x = (44.8 + 45.8i) / 21 x = 2.4 + 2.2i y = (19 + 6z) / 7 y = (19 + 6(-2 - 3.4i)) / 7 y = (19 - 12 - 20.4i) / 7 y = (7 - 20.4i) / 7 y = 1 - 2.9i So the solution to the system of linear equations is: x = 2.4 + 2.2i y = 1 - 2.9i z = -2 - 3.4i, We can solve this system of linear equations using Gaussian elimination. First, we will convert the third equation to have real coefficients by multiplying both sides by -i: -3ix + iy - 2z = -4i Now, we have the following system of equations: x + 2y - z = 6 2x - 3y + 4z = -7 -3x + iy - 2z = -4i To eliminate the x term in the second and third equations, add -2 times the first equation to the second equation and add frac{3}{2} times the first equation to the third equation: (2x - 3y + 4z) - 2(x + 2y - z) = -7 - 2(6) left(-frac{3}{2}x + iy - 2zright) + frac{3}{2}(x + 2y - z) = -4i + frac{3}{2}(6) This gives us -5y + 6z = -19 left(-frac{1}{2}iright)y + z = frac{9}{2} - 4i Next, multiply the second equation by 2i to get rid of the imaginary coefficient: y + 2iz = 9 - 8i Now, we can use back substitution to solve for y and z: z = frac{1}{2}(y + 9 - 8i) Substitute this expression for z into the equation for y: y + 2ileft(frac{1}{2}(y + 9 - 8i)right) = 9 - 8i Multiplying through, we get y + iy + 9i - 16i^2 = 9 - 8i Since i^2 = -1, this simplifies to y + iy = 25 - 8i Now, equate the real and imaginary parts: y + Re(iy) = 25 0 = -8 The second equation has no solution, so the system of equations has no solution. This makes sense since we started with three equations in three variables but only two of them were independent (the third one could be written as a linear combination of the first two).,To solve the system of linear equations with complex coefficients, we will use the Gaussian elimination method. First, let's write the augmented matrix for the given system: | 1 2 -1 | 6 | 2 -3 4 | -7 | 3 1 2i | 4 Step 1: Eliminate the coefficients below the leading 1 in the first column. Row2 = Row2 - 2 * Row1 Row3 = Row3 - 3 * Row1 | 1 2 -1 | 6 | 0 -7 6 | -19 | 0 -5 5i | -14 Step 2: Divide the second row by -7 to get a leading 1 in the second column. Row2 = Row2 / -7 | 1 2 -1 | 6 | 0 1 -6/7 | 19/7 | 0 -5 5i | -14 Step 3: Eliminate the coefficients below the leading 1 in the second column. Row3 = Row3 + 5 * Row2 | 1 2 -1 | 6 | 0 1 -6/7 | 19/7 | 0 0 5i - 30i/7 | -14 + 30i/7 Step 4: Divide the third row by 5i to get a leading 1 in the third column. Row3 = Row3 / (5i) | 1 2 -1 | 6 | 0 1 -6/7 | 19/7 | 0 0 1 15/7 | -14/5 - 30/5i Now we can find the solutions for x, y, and z. From the third row, we have z = (15/7) - 30i/7. Substituting this into the second row: y - 6/7 * ((15/7) - 30i/7) = 19/7 y - 15/7 + 15i/7 = 19/7 y = 1