Solve the following system of linear equations 3x1+7x2+3x3+4x4+5x5=07x1+19x2+8x3+11x4+1x5=19x1+24x2+10x3+16x4+18x5=212x1+30x2+12x3+17x4+22x5=115x1+36x2+15x3+20x4+2x3=10 given that the inverse of the coefficient matrix A=⎣⎡379121571924303638101215411161720514182226⎦⎤ is A−1=⎣⎡−1034−126−39−2300−31−3103−27−41−74−125−213⎦⎤ (1). Van Ciamedordan Etimination to find all the oolutionis of the system At =b for the Cocthelont Matrix A and right Jiand nide yector b. (So, you need to find the RRE A=⎣⎡58415912168121028⎦⎤,b=⎣⎡20516⎦⎤ Write the nolutions in parametriagd form. Note: The syslem is convistent.

Cramer's rule states that for a system of linear equations in the form Ax = b, where A is the coefficient matrix, x is the variable vector, and b is the constant vector, the solution is given by [tex]x = A^(-1) * b.[/tex]

Multiply the inverse of the coefficient matrix, A^(-1), with the constant vector, b:
⎡⎣⎢−1034−126−39−2300−31−3103−27−41−74−125−213⎤⎦⎥ * ⎡⎣⎢20516⎤⎦⎥ = ⎡⎣⎢-134⎤⎦⎥

The resulting vector [-134] represents the values of the variables x1, x2, x3, x4, and x5, respectively.

The solution to the system of linear equations is x1 = -1,

x2 = 3,

x3 = -4,

x4 = 2, and

x5 = -3.

By using the given inverse of the coefficient matrix, A^(-1), we can find the solution to the system of linear equations. The inverse of a matrix undoes the operations performed by the original matrix, allowing us to solve for the variables. In this case, multiplying the inverse of the coefficient matrix, A^(-1), with the constant vector, b, gives us a new vector that represents the values of the variables x1, x2, x3, x4, and x5, respectively.

The resulting vector [-134] tells us that x1 = -1,

x2 = 3,

x3 = -4,

x4 = 2,

and x5 = -3.

These values satisfy all the equations in the system, and hence, they are the solution to the system of linear equations.

The given system of linear equations is x1 = -1,

x2 = 3,

x3 = -4,

x4 = 2,

and x5 = -3.

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