Use the inverse matrix to solve the following simultaneous equations:

x − 2z = 4

2x + 2y + 3z = 15

x + 3y + 2z = 12

Step-by-Step

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**Step 1:** Write all the equations in the same order: variables (in order) = RHS. Write down the matrices A and B. Thematrix A is the matrix of coefficients of the three equations, all arranged in the same format; B is the column matrix consisting of the constants from the RHS of the equations.

A = \left(\begin{matrix} 1 & 0 & -2 \\ 2 & 2 & 3 \\ 1 & 3 & 2 \end{matrix} \right) B = \left(\begin{matrix} 4 \\ 15\\ 12 \end{matrix} \right)

Step 2: Since X = A^{−1} B, determine the inverse of A. However, if you look carefully at A, you will see that this matrix is identical to the matrix D in Worked Examples 9.17 and 9.18 in which the inverse of D was determined; therefore

A^{−1} =\frac{1}{-13} \left(\begin{matrix} -5 & -6 & 4 \\ -1 & 4 & -7 \\ 4 & -3 & 2 \end{matrix} \right)

So proceed straight away to the next step.

**Step 3:** Premultiply the matrix B by the inverse of A:

X = A^{−1} B

\left(\begin{matrix} x \\ y\\ z \end{matrix} \right) =\frac{1}{-13} \left(\begin{matrix} -5 & -6 & 4 \\ -1 & 4 & -7 \\ 4 & -3 & 2 \end{matrix} \right) \left(\begin{matrix} 4 \\ 15\\ 12 \end{matrix}\right) = -\frac{1}{13} \left(\begin{matrix} -62 \\ -28\\ -5 \end{matrix}\right) = \left(\begin{matrix} 4.77 \\ 2.15\\ 0.38 \end{matrix}\right)

\left(\begin{matrix} x \\ y\\ z \end{matrix} \right) = \left(\begin{matrix} 4.77 \\ 2.15\\ 0.38 \end{matrix}\right)

The solutions may simply be read off (correct to two decimal places), x = 4.77,

y = 2.15, z = 0.38.

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