answer:To solve the system of linear equations using Gaussian elimination method, we first write the augmented matrix: | 2 3 4 | 11 | | 3 4 5 | 14 | | 4 5 6 | 17 | Step 1: Eliminate the first element of the second and third rows. To do this, we can subtract 1.5 times the first row from the second row and 2 times the first row from the third row. | 2 3 4 | 11 | | 0 -0.5 -1 | -2.5 | | 0 -1 -2 | -5 | Step 2: Eliminate the second element of the third row. To do this, we can subtract 2 times the second row from the third row. | 2 3 4 | 11 | | 0 -0.5 -1 | -2.5 | | 0 0 0 | 0 | Now, we have reached a row of all zeros in the last row, which means the system is dependent and has infinitely many solutions. We can't find unique values for x, y, and z. To comment on whether the given system is ill-conditioned or not, we need to calculate the determinant of the coefficient matrix. | 2 3 4 | | 3 4 5 | | 4 5 6 | Determinant = 2((4*6)-(5*5)) - 3((3*6)-(4*5)) + 4((3*5)-(4*4)) Determinant = 2(24-25) - 3(18-20) + 4(15-16) Determinant = 2(-1) + 3(-2) + 4(-1) Determinant = -2 - 6 - 4 Determinant = -12 Since the determinant is non-zero, the system is not ill-conditioned. However, it is a dependent system with infinitely many solutions, so we cannot find unique values for x, y, and z.

answer:To solve the given system of linear equations using Gaussian elimination method, we first write the augmented matrix: | 2 3 | 7 | | 10 15.001 | 35 | Step 1: Divide the first row by 2 to get a leading 1: | 1 1.5 | 3.5 | | 10 15.001 | 35 | Step 2: Replace the second row with the second row minus 10 times the first row: | 1 1.5 | 3.5 | | 0 -0.499 | -0.5 | Step 3: Divide the second row by -0.499 to get a leading 1: | 1 1.5 | 3.5 | | 0 1 | 1.002004 | Step 4: Replace the first row with the first row minus 1.5 times the second row: | 1 0 | 1.497006 | | 0 1 | 1.002004 | The resulting matrix represents the system of equations: x = 1.497006 y = 1.002004 So, the values of x and y that satisfy the given system of equations are approximately x = 1.497 and y = 1.002. Now, let's discuss the conditioning of the matrix. The condition number of a matrix is a measure of how sensitive the solution is to small changes in the input. A matrix is considered ill-conditioned if its condition number is large, which means that small changes in the input can lead to large changes in the solution. In this case, the original matrix is: | 2 3 | | 10 15.001 | The determinant of this matrix is (2 * 15.001) - (3 * 10) = 0.002, which is very close to zero. This indicates that the matrix is nearly singular, and thus, it is ill-conditioned. This means that the solution we found may be sensitive to small changes in the input, and the system of equations is not well-suited for numerical computations.

answer:First, let's write the augmented matrix for the given system of linear equations: | 3 2 1 | 11 | | 6 4 2 | 22 | | 9 6 3 | 33 | Now, we'll perform Gaussian elimination: Step 1: Divide the first row by 3 to get a leading 1: | 1 2/3 1/3 | 11/3 | | 6 4 2 | 22 | | 9 6 3 | 33 | Step 2: Eliminate the first column below the first row by subtracting multiples of the first row from the other rows: Row 2 = Row 2 - 6 * Row 1 Row 3 = Row 3 - 9 * Row 1 | 1 2/3 1/3 | 11/3 | | 0 0 0 | 0 | | 0 0 0 | 0 | The second and third rows are all zeros, which means the system is dependent. This implies that there are infinitely many solutions for this system of linear equations. To find the general solution, we can express the variables in terms of a free variable. In this case, let's use z as the free variable: x = 11/3 - 2/3 * y - 1/3 * z y = y z = z The general solution is given by: x = 11/3 - 2/3 * y - 1/3 * z y = y z = z Now, let's discuss why the coefficient matrix is ill-conditioned. The coefficient matrix is: | 3 2 1 | | 6 4 2 | | 9 6 3 | We can see that the rows of the matrix are linearly dependent, as the second row is just the first row multiplied by 2, and the third row is the first row multiplied by 3. This means that the matrix is singular (i.e., it has a determinant of 0) and is not invertible. In the context of solving linear systems, an ill-conditioned matrix is one that is close to being singular, which can lead to numerical instability and inaccurate results when using certain algorithms. In this case, the matrix is not just ill-conditioned, but it is actually singular, which leads to a dependent system with infinitely many solutions.

answer:First, we set up the augmented matrix for the given system of equations: | 1 2 | 5 | 1.001 1.999 | 4.999 Next, we perform Gaussian elimination with partial pivoting. Since the second row has a larger coefficient for x (1.001 > 1), we will swap the two rows: | 1.001 1.999 | 4.999 | 1 2 | 5 Now, we will eliminate the x term in the second row. To do this, we will subtract (1/1.001) times the first row from the second row: | 1.001 1.999 | 4.999 | 0 0.001 | 0.001 Now, we can solve for y by dividing the second row by 0.001: | 1.001 1.999 | 4.999 | 0 1 | 1 Finally, we will eliminate the y term in the first row by subtracting 1.999 times the second row from the first row: | 1 0 | 3 | 0 1 | 1 Now, we have the solution in the augmented matrix: x = 3 y = 1 So, the solution to the system of equations is x = 3 and y = 1.