Apply elementary operations of linear system, solve the following system of linear equations.State your steps clearly. (hint: please use Gaussian elimination when conducting row opera-tions)(a)2x1 + 3x2 = 24x1 − 5x2 = 15(b)x1 + x2 − x3 = 62x1 + 4x2 + 5x3 = 113x1 + 4x2 + 10x3 = 8

Sure, let's solve these systems of equations step by step using Gaussian elimination.

(a) We have the system of equations:

2x1 + 3x2 = 24 4x1 - 5x2 = 15

Step 1: Write the system of equations in matrix form. We get:

[2, 3 | 24] [4, -5 | 15]

Step 2: To eliminate x1 from the second equation, we can subtract twice the first equation from the second. This gives us:

[2, 3 | 24] [0, -11 | -33]

Step 3: Now, we can multiply the second equation by -1/11 to make the coefficient of x2 equal to 1. We get:

[2, 3 | 24] [0, 1 | 3]

Step 4: To eliminate x2 from the first equation, we can subtract 3 times the second equation from the first. This gives us:

[2, 0 | 15] [0, 1 | 3]

Step 5: Finally, we can divide the first equation by 2 to make the coefficient of x1 equal to 1. We get:

[1, 0 | 7.5] [0, 1 | 3]

So, the solution to the system is x1 = 7.5 and x2 = 3.

(b) We have the system of equations:

x1 + x2 - x3 = 6 2x1 + 4x2 + 5x3 = 11 3x1 + 4x2 + 10x3 = 8

Step 1: Write the system of equations in matrix form. We get:

[1, 1, -1 | 6] [2, 4, 5 | 11] [3, 4, 10 | 8]

Step 2: To eliminate x1 from the second and third equations, we can subtract 2 times the first equation from the second and 3 times the first equation from the third. This gives us:

[1, 1, -1 | 6] [0, 2, 7 | -1] [0, 1, 13 | -10]

Step 3: To eliminate x2 from the third equation, we can subtract 0.5 times the second equation from the third. We get:

[1, 1, -1 | 6] [0, 2, 7 | -1] [0, 0, 6.5 | -9.5]

Step 4: Now, we can divide the second equation by 2 and the third equation by 6.5 to make the coefficients of x2 and x3 equal to 1. We get:

[1, 1, -1 | 6] [0, 1, 3.5 | -0.5] [0, 0, 1 | -1.46]

Step 5: To eliminate x3 from the first and second equations, we can add the third equation to the first and subtract 3.5 times the third equation from the second. This gives us:

[1, 1, 0 | 4.54] [0, 1, 0 | 0.61] [0, 0, 1 | -1.46]

Step 6: To eliminate x2 from the first equation, we can subtract the second equation from the first. We get:

[1, 0, 0 | 3.93] [0, 1, 0 | 0.61] [0, 0, 1 | -1.46]

So, the solution to the system is x1 = 3.93, x2 = 0.61, and x3 = -1.46.