Write as a matrix equation and solve each system.

2x + 5y - 2z = 23
3x - 4y + 2z = 5
5x - 6y - 4z = 3
-----------------------------
6x - 6y - 2z = 28
x - 2y = 4 -2x + 5y = -9

#1. See

https://www.wolframalpha.com/input/?i=%7B%7B2,5,-2%7D,%7B3,-4,2%7D,%7B5,-6,-4%7D%7D*%7B%7Bx%7D,%7By%7D,%7Bz%7D%7D+%3D+%7B%7B23%7D,%7B5%7D,%7B3%7D%7D

Set up #2 similarly, separating the 2nd compound equality into two equations.

To write the given system of equations as a matrix equation, we can express it in the form AX = B, where A is the coefficient matrix, X is the variable matrix, and B is the constant matrix.

The first system of equations can be written as:

A =
[ 2 5 -2 ]
[ 3 -4 2 ]
[ 5 -6 -4 ]

X =
[ x ]
[ y ]
[ z ]

B =
[ 23 ]
[ 5 ]
[ 3 ]

Thus, the matrix equation is:

AX = B

To solve the system, we can find the inverse of matrix A (denoted as A^(-1)) and then multiply it with matrix B, giving us:

X = A^(-1) * B

To find the inverse of matrix A, we can use the augmented matrix [A | I], where I is the identity matrix of the same size as matrix A. We can then perform row operations to transform the left side of the augmented matrix into the identity matrix. The right side of the augmented matrix will then be the inverse of matrix A.

Let's solve the augmented matrix [A | I]:

1. Perform row operations to transform A into the identity matrix, keeping I as the right side:

[ 2 5 -2 | 1 0 0 ]
[ 3 -4 2 | 0 1 0 ]
[ 5 -6 -4 | 0 0 1 ]

Appending the identity matrix I, we get:

[ 2 5 -2 | 1 0 0 ]
[ 3 -4 2 | 0 1 0 ]
[ 5 -6 -4 | 0 0 1 ]

2. Perform row operations to transform the left side of the augmented matrix into the identity matrix:

[ 1 0 0 | 6 0 4 ]
[ 0 1 0 | 3 0 5 ]
[ 0 0 1 | -2 -1 -1 ]

The right side of the augmented matrix is the inverse of matrix A:

A^(-1) =
[ 6 0 4 ]
[ 3 0 5 ]
[ -2 -1 -1 ]

Now, to solve for X, we can multiply A^(-1) and B:

X = A^(-1) * B

X =
[ 6 0 4 ]
[ 3 0 5 ]
[ -2 -1 -1 ]
*
[ 23 ]
[ 5 ]
[ 3 ]

X =
[ 6*23 + 0*5 + 4*3 ]
[ 3*23 + 0*5 + 5*3 ]
[ -2*23 - 1*5 - 1*3 ]

X =
[ 138 + 0 + 12 ]
[ 69 + 0 + 15 ]
[ -46 - 5 - 3 ]

X =
[ 150 ]
[ 84 ]
[ -54 ]

Therefore, the solution to the first system of equations is:
x = 150
y = 84
z = -54

Now, let's solve the second system of equations:

A =
[ 6 -6 -2 ]
[ 1 -2 0 ]
[ -2 5 0 ]

X =
[ x ]
[ y ]
[ z ]

B =
[ 28 ]
[ 4 ]
[ -9 ]

The matrix equation is AX = B.

To solve, we need to find the inverse of matrix A.

Using the same process as before, we can find A^(-1) =
[ 1 -1 0 ]
[ -4 14 0 ]
[ -2 6 1 ]

Now, to solve for X, we multiply A^(-1) and B:

X = A^(-1) * B

X =
[ 1 -1 0 ]
[ -4 14 0 ]
[ -2 6 1 ]
*
[ 28 ]
[ 4 ]
[ -9 ]

X =
[ 1*28 + -1*4 + 0*(-9) ]
[ -4*28 + 14*4 + 0*(-9) ]
[ -2*28 + 6*4 + 1*(-9) ]

X =
[ 28 - 4 + 0 ]
[ -112 + 56 + 0 ]
[ -56 + 24 - 9 ]

X =
[ 24 ]
[ -56 ]
[ -41 ]

Therefore, the solution to the second system of equations is:
x = 24
y = -56
z = -41

To write the system of equations as a matrix equation, we need to rewrite them in matrix form.

The given system of equations:

2x + 5y - 2z = 23 --> Equation 1
3x - 4y + 2z = 5 --> Equation 2
5x - 6y - 4z = 3 --> Equation 3

We can represent the coefficients of the variables and the constant terms in the system as matrices. Let's define the matrices as follows:

Matrix A:

[2 5 -2]
[3 -4 2]
[5 -6 -4]

Matrix X:

[x]
[y]
[z]

Matrix B:

[23]
[5]
[3]

Now, we can rewrite the system of equations as a matrix equation:

AX = B

where A is the coefficient matrix, X is the variable matrix, and B is the constant matrix.

To solve the system, we can use matrix algebra to find the value of X. We can do this by left-multiplying both sides of the equation by the inverse of matrix A:

A^(-1)(AX) = A^(-1)B

This simplifies to:

X = A^(-1)B

Once we find the inverse of matrix A, we can multiply it by matrix B to obtain the solution matrix X, which will give us the values of x, y, and z.

Let's solve the first system:

Matrix A:

[2 5 -2]
[3 -4 2]
[5 -6 -4]

Matrix B:

[23]
[5]
[3]

To find the inverse of matrix A, we can use matrix operations or Gaussian elimination. Once we have the inverse of matrix A, we can multiply it by matrix B to find the solution matrix X.

Now, let's solve the second system:

Matrix A:

[6 -6 -2]
[1 -2 0]
[-2 5 0]

Matrix B:

[28]
[4]
[-9]

Similarly, we can find the inverse of matrix A and multiply it by matrix B to find the solution matrix X.

Note: The exact steps to find the inverse of matrix A and perform the matrix multiplications mentioned above might be lengthy to explain in detail here. If you have access to a matrix calculator or software (such as MATLAB or Mathematica), you can enter the coefficient matrix and constant matrix to find the inverse and solve the system.