answer:To solve this system of linear equations, we can use the method of substitution or elimination. In this case, we will use the elimination method. First, let's eliminate x from the second and third equations. To do this, we can multiply the first equation by 3 and the second equation by 2, then subtract the first equation from the second equation: (3)(2x + y - 3z) = (3)(7) => 6x + 3y - 9z = 21 (2)(3x - 2y + 5z) = (2)(-4) => 6x - 4y + 10z = -8 Now subtract the first equation from the second equation: (6x - 4y + 10z) - (6x + 3y - 9z) = -8 - 21 -7y + 19z = -29 Now, let's eliminate x from the first and third equations. To do this, we can multiply the first equation by -1 and the third equation by 2, then add the first equation to the third equation: (-1)(2x + y - 3z) = (-1)(7) => -2x - y + 3z = -7 (2)(x + 4y + z) = (2)(6) => 2x + 8y + 2z = 12 Now add the first equation to the second equation: (-2x - y + 3z) + (2x + 8y + 2z) = -7 + 12 7y + 5z = 5 Now we have a system of two linear equations in two variables: -7y + 19z = -29 7y + 5z = 5 Add the two equations together to eliminate y: (-7y + 19z) + (7y + 5z) = -29 + 5 24z = -24 Now solve for z: z = -24 / 24 z = -1 Now that we have z, we can substitute it back into one of the equations with y: 7y + 5z = 5 7y + 5(-1) = 5 7y - 5 = 5 7y = 10 y = 10 / 7 Now that we have y, we can substitute y and z back into one of the equations with x: 2x + y - 3z = 7 2x + (10 / 7) - 3(-1) = 7 2x + 10 / 7 + 3 = 7 2x = 7 - 10 / 7 - 3 2x = (49 - 10 - 21) / 7 2x = 18 / 7 x = 9 / 7 So the solution to the system of linear equations is: x = 9 / 7 y = 10 / 7 z = -1,To solve the given system of linear equations, we can use the method of substitution or elimination. In this case, I'll use the elimination method. First, let's multiply the first equation by 3 and the second equation by 2 to make the coefficients of x in both equations equal: (1) 6x + 3y - 9z = 21 (2) 6x - 4y + 10z = -8 Now, subtract the second equation from the first to eliminate x: (3) 7y - 19z = 29 Next, let's multiply the first equation by 4 and the third equation by 1 to make the coefficients of y in both equations equal: (1) 8x + 4y - 12z = 28 (4) x + 4y + z = 6 Now, subtract the third equation from the first to eliminate y: (5) 7y - 11z = 22 Now we have two equations with two variables: (3) 7y - 19z = 29 (5) 7y - 11z = 22 Subtract equation (5) from equation (3) to eliminate y: (-8z) = -7 z = 7/8 Now substitute the value of z in equation (5) to find y: 7y - 11(7/8) = 22 y = (22 + 11/8)/7 y = 49/4 Finally, substitute the values of y and z in the third equation to find x: x + 4(49/4) + (7/8) = 6 x = 6 - 4(49/4) - 7/8 x = -27/8 So, the solution to the system of linear equations is: x = -27/8, y = 49/4, z = 7/8, We can use theGaussian elimination method to solve this system. First, let's write it in matrix form: 2 1 -3 x 7 A = 3 -2 5 , X = y -4 1 4 1 z 6 To eliminate the variable x, we can add -3/2 times the first equation to the second equation and add -1 times the first equation to the third equation: 2 1 -3 x 7 A ≈ 0 -3.5 6.5, X ≈ y -7.5 0 3.5 -2.5 z 3.5 Now, we can divide the second and third rows by -3.5 and 3.5 respectively to normalize them: 2 1 -3 x 7 A ≈ 0 -1 2 , X ≈ y -2.1429 0 1 -0.7143, Z ≈ z 1 Next, we'll eliminate the variable y by adding the second row to the third row: 2 1 -3 x 7 A ≈ 0 -1 2 , X ≈ y -2.1429 0 0 -0.2857, Z ≈ z 0.8571 Since we have no more variables to eliminate, we can now use back substitution to solve for the variables. Starting with z: z ≈ 0.8571 Next, we can substitute z back into the equation for y: -y ≈ 0.8571 y ≈ -0.8571 Finally, we can substitute both y and z back into the equation for x: 2x + y - 3z ≈ 7 2x ≈ 7 - (-0.8571) + 3(0.8571) 2x ≈ 9.381 x ≈ 4.6905 So the solution to the system is approximately (x, y, z) = (4.6905, -0.8571, 0.8571).

answer:To solve the non-homogeneous system of linear equations, we can use the Gaussian elimination method. First, we write the augmented matrix for the given system: | 1 2 -1 | 7 | | 2 -1 1 | 3 | | 3 1 2 | 13 | Step 1: Eliminate x from the second and third rows. To do this, we'll subtract 2 times the first row from the second row, and add 3 times the first row to the third row. | 1 2 -1 | 7 | | 0 -5 3 | -13| | 0 -5 5 | 2 | Step 2: Divide the second row by -5 to make the leading coefficient of y equal to 1. | 1 2 -1 | 7 | | 0 1 -0.6| 2.6 | | 0 -5 5 | 2 | Step 3: Eliminate y from the third row. To do this, we'll add 5 times the second row to the third row. | 1 2 -1 | 7 | | 0 1 -0.6| 2.6 | | 0 0 2.4 | 17 | Step 4: Divide the third row by 2.4 to make the leading coefficient of z equal to 1. | 1 2 -1 | 7 | | 0 1 -0.6| 2.6 | | 0 0 1 | 7.08 | Step 5: Back-substitute to find the values of x, y, and z. From the third row, we have z = 7.08. Plugging this value into the second row, we get: y - 0.6(7.08) = 2.6 y = 2.6 + 0.6(7.08) ≈ 6.888 Finally, plugging z and y into the first row, we get: x + 2(6.888) - 7.08 = 7 , To find the solution to a non-homogeneous system of linear equations, we can use either elimination or substitution method. In this case, I will use the elimination method. First, add the first two equations to eliminate x: (x + 2y - z = 7) + (2x - y + z = 3) This gives us: 3x + y = 10 Rightarrow y = 10 - 3x Now, substitute y in the third equation: 3x + (10 - 3x) + 2z = 13 Rightarrow 2z = 3 Rightarrow z = frac{3}{2} Now that we have z, we can find y: y = 10 - 3x Rightarrow y = 10 - 3left(frac{2}{3}right) = 10 - 2 = 8 Finally, find x using the first equation: x + 2(8) - frac{3}{2} = 7 Rightarrow x = 7 - 16 + frac{3}{2} = -frac{5}{2} So the solution to the non-homogeneous system is: boxed{left(-frac{5}{2}, 8, frac{3}{2}right)},To solve the non-homogeneous system of linear equations, we can use the method of substitution or elimination. Here, we will use the elimination method. First, let's eliminate z from the first two equations: 1) x + 2y - z = 7 2) 2x - y + z = 3 Add equations 1 and 2: 3x + y = 10 (call this equation 4) Now, let's eliminate z from the second and third equations: 2) 2x - y + z = 3 3) 3x + y + 2z = 13 Multiply equation 2 by 2: 4x - 2y + 2z = 6 Add equations 3 and the modified equation 2: 7x = 19 Now, solve for x: x = 19/7 Now, substitute x back into equation 4: 3(19/7) + y = 10 Solve for y: y = 10 - 3(19/7) y = (70 - 57) / 7 y = 13/7 Now, substitute x and y back into equation 1: (19/7) + 2(13/7) - z = 7 Solve for z: z = (19 + 26) / 7 - 7 z = 45/7 - 49/7 z = -4/7 So, the solution to the non-homogeneous system of linear equations is: x = 19/7, y = 13/7, and z = -4/7.

answer:To solve the given system of linear equations using matrix methods, we first write the system in the matrix form AX = B, where A is the coefficient matrix, X is the column matrix of variables, and B is the column matrix of constants. A = begin{bmatrix} 2 & 3 & -1 4 & 1 & 2 1 & -2 & 3 end{bmatrix}, X = begin{bmatrix} x_1 x_2 x_3 end{bmatrix}, B = begin{bmatrix} 10 14 15 end{bmatrix} Now, we will find the inverse of matrix A, denoted as A_inv. If A_inv exists, we can multiply both sides of the equation AX = B by A_inv to get X = A_inv * B, which will give us the solution to the system of linear equations. First, we find the determinant of A: det(A) = 2(1 cdot 3 - 2 cdot (-2)) - 3(4 cdot 3 - 2 cdot 1) - (-1)(4 cdot (-2) - 1 cdot 1) det(A) = 2(3 + 4) - 3(12 - 2) - (-1)(8 - 1) det(A) = 2(7) - 3(10) + 7 det(A) = 14 - 30 + 7 det(A) = -9 Since det(A) ≠ 0, A_inv exists. Now, we find A_inv using the formula A_inv = (1/det(A)) * adj(A), where adj(A) is the adjugate of A. adj(A) = begin{bmatrix} (1 cdot 3 - 2 cdot (-2)) & -(4 cdot 3 - 2 cdot 1) & (4 cdot (-2) - 1 cdot 1) -(3 cdot 3 - (-1) cdot (-2)) & (2 cdot 3 - (-1) cdot 1) & -(2 cdot (-2) - 1 cdot 1) (3 cdot 2 - 1 cdot 1) & -(2 cdot 2 - 1 cdot 4) & (2 cdot 1 - 3 cdot 4) end{bmatrix} adj(A) = begin{bmatrix} 7 & -10 & 7 -7 & 7 & 3 5 & 0 & -10 end{bmatrix} Now, we find A_inv: A_inv = frac{1}{-9} begin{bmatrix} 7 & -10 & 7 -7 & 7 & 3 5 & 0 & -10 end{bmatrix} A_inv = begin{bmatrix} -frac{7}{9} & frac{10}{9} & -frac{7}{9} frac{7}{9} & -frac{7}{9} & -frac{1}{3} -frac{5}{9} & 0 & frac{10}{9} end{bmatrix} Now, we can find the solution X by multiplying A_inv with B: X = A_inv * B = begin{bmatrix} -frac{7}{9} & frac{10}{9} & -frac{7}{9} frac{7}{9} & -frac{7}{9} & -frac{1}{3} -frac{5}{9} & 0 & frac{10}{9} end{bmatrix} * begin{bmatrix} 10 14 15 end{bmatrix} X = begin{bmatrix} (-frac{7}{9} cdot 10) + (frac{10}{9} cdot 14) + (-frac{7}{9} cdot 15) (frac{7}{9} cdot 10) + (-frac{7}{9} cdot 14) + (-frac{1}{3} cdot 15) (-frac{5}{9} cdot 10) + (0 cdot 14) + (frac{10}{9} cdot 15) end{bmatrix} X = begin{bmatrix} x_1 x_2 x_3 end{bmatrix} = begin{bmatrix} 2 -1 3 end{bmatrix} So, the solution to the system of linear equations is x_1 = 2, x_2 = -1, and x_3 = 3., To solve the system of linear equations, we first write it in matrix form: AX = B where A = begin{bmatrix} 2 & 3 & -1 4 & 1 & 2 1 & -2 & 3 end{bmatrix}, X = begin{bmatrix} x_1 x_2 x_3 end{bmatrix}, and B = begin{bmatrix} 10 14 15 end{bmatrix} First, we find the inverse of matrix A. We can use any method to find the inverse, but here we will use the formula for a 3 times 3 matrix: A^{-1} = dfrac{1}{text{det}(A)} cdot text{adj}(A) First, let's compute the determinant of A: text{det}(A) = 2(1 cdot 3 - 2(-2)) - 3(4 cdot 3 - 2 cdot 1) - 1(4(-2) - 1 cdot 1) = 2(3 + 4) - 3(12 - 2) - 1(-8 - 1) = 2 cdot 7 - 3 cdot 10 + 1 cdot 9 = 14 - 30 + 9 = -7 Now let's compute the adjugate of A: text{adj}(A) = begin{bmatrix} 1 cdot 3 - (-2) cdot 1 & -(1 cdot 2 - 4 cdot 1) & 1 cdot (-2) - 3 cdot 4 -(2 cdot 3 - 1 cdot 1) & 2 cdot 1 - 3 cdot (-2) & -2 cdot (-2) - 1 cdot 1 1 cdot (1 - 4) & -1 cdot (2 - 4) & 1 cdot 2 - 3 cdot 1 end{bmatrix} = begin{bmatrix} 7 & -6 & -14 -5 & 7 & -7 -3 & 2 & -5 end{bmatrix} Now we have: A^{-1} = -dfrac{1}{7} cdot begin{bmatrix} 7 & -6 & -14 -5 & 7 & -7 -3 & 2 & -5 end{bmatrix} = begin{bmatrix} -dfrac{1}{1} & dfrac{6}{7} & dfrac{2}{1} dfrac{5}{7} & -dfrac{1}{1} & dfrac{1}{1} dfrac{3}{7} & -dfrac{2}{7} & dfrac{5}{1} end{bmatrix} Now we can find the solution by multiplying A^{-1} and B: X = A^{-1}B = begin{bmatrix} -dfrac{1}{1} & dfrac{6}{7} & dfrac{2}{1} dfrac{5}{7} & -dfrac{1}{1} & dfrac{1}{1} dfrac{3}{7} & -dfrac{2}{7} & dfrac{5}{1} end{bmatrix} cdot begin{bmatrix} 10 14 15 end{bmatrix} = begin{bmatrix} -dfrac{1}{1} cdot 10 + dfrac{6}{7} cdot 14 + dfrac{2}{1} cdot 15 dfrac{5}{7} cdot 10 - dfrac{1}{1} cdot 14 + dfrac{1}{1} cdot 15 dfrac{3}{7} cdot 10 - dfrac{2}{7} cdot 14 + dfrac{5}{1} cdot 15 end{bmatrix} = begin{bmatrix} -10 + 12 + 30 5 - 14 + 15 3 - 4 + 75 end{bmatrix} = begin{bmatrix} 22 6 74 end{bmatrix} So the solution to the system of linear equations is boxed{begin{bmatrix} x_1, x_2, x_3 end{bmatrix} = begin{bmatrix} 22, 6, 74 end{bmatrix}}.,To solve the given system of linear equations using matrix methods, we first need to represent the system in the form AX = B, where A is a matrix of coefficients, X is a column matrix of variable values, and B is a column matrix of constants. The given system is: begin{cases} 2x_1 + 3x_2 - x_3 = 10 4x_1 + x_2 + 2x_3 = 14 x_1 - 2x_2 + 3x_3 = 15 end{cases} So, we have the following matrix equation: begin{bmatrix} 2 & 3 & -1 4 & 1 & 2 1 & -2 & 3 end{bmatrix} begin{bmatrix} x_1 x_2 x_3 end{bmatrix} = begin{bmatrix} 10 14 15 end{bmatrix} Now, we need to find the inverse of matrix A, denoted as A^{-1}. Then, we can find the solution matrix X by multiplying the inverse of matrix A with matrix B: X = A^{-1}B First, let's find the determinant of matrix A: |A| = 2(1)(3) - 3(4)(3) + (-1)(1)(-2) = -21 Since the determinant is non-zero, the inverse of matrix A exists. Now, let's find the adjoint of matrix A: adj(A) = begin{bmatrix} (1)(3) - (2)(-2) & -(4)(3) - (1)(-2) & (4)(1) - (1)(-2) -(1)(2) + (3)(-2) & (2)(2) + (1)(1) & -(2)(1) + (3)(1) (2)(2) - (3)(1) & (3

answer:To find the eigenvalues of the matrix A, we need to solve the characteristic equation, which is given by the determinant of (A - λI), where λ is the eigenvalue and I is the identity matrix. A - λI = 4 - λ, 2, 1, 3 - λ Now, we find the determinant: det(A - λI) = (4 - λ)(3 - λ) - (2)(1) = λ^2 - 7λ + 10 To find the eigenvalues, we need to solve the quadratic equation: λ^2 - 7λ + 10 = 0 By factoring, we get: (λ - 2)(λ - 5) = 0 So, the eigenvalues are λ1 = 2 and λ2 = 5. Now, we will find the eigenvectors corresponding to each eigenvalue. For λ1 = 2: (A - 2I)v1 = 0 2, 2, 1, 1v1 = 0 From the first row, we get the equation: 2v1_1 + 2v1_2 = 0 v1_1 = -v1_2 Let v1_2 = 1, then v1_1 = -1. So, the eigenvector v1 = -1, 1. For λ2 = 5: (A - 5I)v2 = 0 -1, 2, 1, -2v2 = 0 From the first row, we get the equation: -v2_1 + 2v2_2 = 0 v2_1 = 2v2_2 Let v2_2 = 1, then v2_1 = 2. So, the eigenvector v2 = 2, 1. In summary, the eigenvalues are λ1 = 2 and λ2 = 5, and their corresponding eigenvectors are v1 = -1, 1 and v2 = 2, 1.,To find the eigenvalues and eigenvectors of the given matrix A, we will follow these steps: 1. Find the eigenvalues by solving the characteristic equation. 2. For each eigenvalue, find the eigenvectors by solving the corresponding linear system. Matrix A: 4, 2, 1, 3 Step 1: Find the eigenvalues by solving the characteristic equation det(A - λI) = 0, where λ is the eigenvalue and I is the identity matrix. A - λI = 4 - λ, 2, 1, 3 - λ The characteristic equation is given by the determinant of this matrix: (4 - λ)(3 - λ) - (2)(1) = λ^2 - 7λ + 10 Now, solve for λ: λ^2 - 7λ + 10 = 0 (λ - 2)(λ - 5) = 0 So, the eigenvalues are λ1 = 2 and λ2 = 5. Step 2: For each eigenvalue, find the eigenvectors by solving the corresponding linear system. For λ1 = 2: (A - λ1I)X = 0 (2, 2, 1, 1 * x = 0) Solving the system, we get the eigenvector: v1 = 1, -1 For λ2 = 5: (A - λ2I)X = 0 (-1, 2, 1, -2 * x = 0) Solving the system, we get the eigenvector: v2 = 2, 1 So, the eigenvalues and eigenvectors of matrix A are: Eigenvalue λ1 = 2, Eigenvector v1 = 1, -1 Eigenvalue λ2 = 5, Eigenvector v2 = 2, 1, To find the eigenvalues, we need to solve the characteristic equation of matrix A, which is given by: |A - lambda I| = 0 where lambda is an eigenvalue and I is the identity matrix. For our given matrix A, this equation becomes: begin{vmatrix} 4-lambda & 2 1 & 3-lambda end{vmatrix} = (4-lambda)(3-lambda) - (1)(2) = 0 Expanding and simplifying, we get: lambda^2 - 7lambda + 10 = (lambda-5)(lambda-2) = 0 So the eigenvalues are lambda_1 = 5 and lambda_2 = 2. Now let's find the eigenvectors. For lambda_1 = 5, we need to solve (A - 5I)X = 0: begin{bmatrix} -1 & 2 1 & -2 end{bmatrix} begin{bmatrix} x_1 x_2 end{bmatrix} = begin{bmatrix} 0 0 end{bmatrix} The first row gives us x_1 = 2x_2. Let x_2 = 1, then x_1 = 2. Thus, the eigenvector corresponding to lambda_1 = 5 is mathbf{v}_1 = begin{bmatrix} 2 1 end{bmatrix}. For lambda_2 = 2, we need to solve (A - 2I)X = 0: begin{bmatrix} 2 & 2 1 & 1 end{bmatrix} begin{bmatrix} x_1 x_2 end{bmatrix} = begin{bmatrix} 0 0 end{bmatrix} The second row gives us x_1 = -x_2. Let x_2 = 1, then x_1 = -1. Thus, the eigenvector corresponding to lambda_2 = 2 is mathbf{v}_2 = begin{bmatrix} -1 1 end{bmatrix}. In summary, the eigenvalues are lambda_1 = 5 and lambda_2 = 2, and their corresponding eigenvectors are mathbf{v}_1 = begin{bmatrix} 2 1 end{bmatrix} and mathbf{v}_2 = begin{bmatrix} -1 1 end{bmatrix}.