a. Write the linear system as a matrix equation
in the form AX= B.
b. Solve the system using the inverse that is given for the coefficient matrix.
PLEASE HELP WILL GIVE A LOT OF POINTS!!

a) The linear system as a matrix equation in the form AX= B as

[tex]\left[\begin{array}{ccc}-1&1&1\\0&-2&-1\\2&1&0\end{array}\right][/tex] [tex]\left[\begin{array}{c}x\\y\\z\end{array}\right][/tex] = [tex]\left[\begin{array}{c}0\\-2\\4\end{array}\right][/tex]

b) The solution set is {x = 2, y = -4, z = -2}.

Given that A⁻¹ = [tex]\left[\begin{array}{ccc}-1&1&1\\2&-2&-1\\-4&3&2\end{array}\right][/tex],

To find the matrix equation in the form AX = B, multiply both sides of the equation by A⁻¹:

A⁻¹(A⁻¹X) = A⁻¹B

X = A⁻¹B

Therefore, the linear system as a matrix equation in the form AX = B is:

[tex]\left[\begin{array}{ccc}-1&1&1\\0&-2&-1\\2&1&0\end{array}\right][/tex] [tex]\left[\begin{array}{c}x\\y\\z\end{array}\right][/tex] = [tex]\left[\begin{array}{c}0\\-2\\4\end{array}\right][/tex]

2. The given system of equations can be represented as:

                                    A = [tex]\left[\begin{array}{ccc}-1&1&1\\0&-2&-1\\2&1&0\end{array}\right][/tex]

                                    X  = [tex]\left[\begin{array}{c}x\\y\\z\end{array}\right][/tex]

                                    B = [tex]\left[\begin{array}{c}0\\-2\\4\end{array}\right][/tex]

Using the given inverse of A:

A⁻¹ = [tex]\left[\begin{array}{ccc}-1&1&1\\2&-2&-1\\-4&3&2\end{array}\right][/tex]

Multiply A⁻¹ by B to solve for X:

[tex]X = A^{(-1)} B[/tex]

[tex]\left[\begin{array}{c}x\\y\\z\end{array}\right][/tex]   = [tex]\left[\begin{array}{ccc}-1&1&1\\2&-2&-1\\-4&3&2\end{array}\right][/tex]  [tex]\left[\begin{array}{c}0\\-2\\4\end{array}\right][/tex]

Performing the matrix multiplication:

|  x  | =  | (-1)(0) + (1)(-2) + (1)(4)|   |  2  |

|  y  | = | (-2)(0) + (3)(-2) + (2)(4)|  | -4  |

|  z  |   | (2)(0) + (-2)(-2) + (-1)(4)|   | -2  |

Therefore, the solution set is {x = 2, y = -4, z = -2}.

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