Question

question 23 points 2

solve the system of equations using the matrix inverse method.
\\x + y + z = 3\\
\\2x - y + z = -1\\
\\2x + y - 3z = -9\\

\\(x = 1/5, y = 3/5, z = 8/7\\)
\\(x = -8/7, y = 10/7, z = 19/7\\)

Explanation:

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<reasoning>

Write the system in matrix form

Using the Coefficient Matrix knowledge point
\[
A =

$$\begin{bmatrix} 1 & 1 & 1 \\ 2 & -1 & 1 \\ 2 & 1 & -3 \end{bmatrix}$$

, \quad X =

$$\begin{bmatrix} x \\ y \\ z \end{bmatrix}$$

, \quad B =

$$\begin{bmatrix} 3 \\ -1 \\ -9 \end{bmatrix}$$

\]

Calculate the determinant of A

Using the Matrix Determinant knowledge point
\[

$$\begin{aligned} \det(A) &= 1((-1)(-3) - (1)(1)) - 1((2)(-3) - (1)(2)) + 1((2)(1) - (-1)(2)) \\ &= 1(3 - 1) - 1(-6 - 2) + 1(2 + 2) \\ &= 2 + 8 + 4 = 14 \end{aligned}$$

\]

Find the inverse matrix of A

Using the Matrix Inverse Formula knowledge point
\[
C =

$$\begin{bmatrix} 2 & 8 & 4 \\ 4 & -5 & 1 \\ 2 & 1 & -3 \end{bmatrix}$$

\implies A^{-1} = \frac{1}{14}

$$\begin{bmatrix} 2 & 4 & 2 \\ 8 & -5 & 1 \\ 4 & 1 & -3 \end{bmatrix}$$

\]

Solve for the variable vector X

Using the Solving Matrix Equations knowledge point
\[

$$\begin{aligned} X &= A^{-1} B = \frac{1}{14} \begin{bmatrix} 2 & 4 & 2 \\ 8 & -5 & 1 \\ 4 & 1 & -3 \end{bmatrix} \begin{bmatrix} 3 \\ -1 \\ -9 \end{bmatrix} \\ &= \frac{1}{14} \begin{bmatrix} 2(3) + 4(-1) + 2(-9) \\ 8(3) - 5(-1) + 1(-9) \\ 4(3) + 1(-1) - 3(-9) \end{bmatrix} = \frac{1}{14} \begin{bmatrix} -16 \\ 20 \\ 38 \end{bmatrix} = \begin{bmatrix} -8/7 \\ 10/7 \\ 19/7 \end{bmatrix} \end{aligned}$$

\]
</reasoning>

<answer>
<mcq-option>(A) \(x = 1/5, y = 3/5, z = 8/7\)</mcq-option>
<mcq-correct>(B) \(x= -8/7, y = 10/7, z = 19/7\)</mcq-correct>
</answer>

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Answer:

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<reasoning>

Write the system in matrix form

Using the Coefficient Matrix knowledge point
\[
A =

$$\begin{bmatrix} 1 & 1 & 1 \\ 2 & -1 & 1 \\ 2 & 1 & -3 \end{bmatrix}$$

, \quad X =

$$\begin{bmatrix} x \\ y \\ z \end{bmatrix}$$

, \quad B =

$$\begin{bmatrix} 3 \\ -1 \\ -9 \end{bmatrix}$$

\]

Calculate the determinant of A

Using the Matrix Determinant knowledge point
\[

$$\begin{aligned} \det(A) &= 1((-1)(-3) - (1)(1)) - 1((2)(-3) - (1)(2)) + 1((2)(1) - (-1)(2)) \\ &= 1(3 - 1) - 1(-6 - 2) + 1(2 + 2) \\ &= 2 + 8 + 4 = 14 \end{aligned}$$

\]

Find the inverse matrix of A

Using the Matrix Inverse Formula knowledge point
\[
C =

$$\begin{bmatrix} 2 & 8 & 4 \\ 4 & -5 & 1 \\ 2 & 1 & -3 \end{bmatrix}$$

\implies A^{-1} = \frac{1}{14}

$$\begin{bmatrix} 2 & 4 & 2 \\ 8 & -5 & 1 \\ 4 & 1 & -3 \end{bmatrix}$$

\]

Solve for the variable vector X

Using the Solving Matrix Equations knowledge point
\[

$$\begin{aligned} X &= A^{-1} B = \frac{1}{14} \begin{bmatrix} 2 & 4 & 2 \\ 8 & -5 & 1 \\ 4 & 1 & -3 \end{bmatrix} \begin{bmatrix} 3 \\ -1 \\ -9 \end{bmatrix} \\ &= \frac{1}{14} \begin{bmatrix} 2(3) + 4(-1) + 2(-9) \\ 8(3) - 5(-1) + 1(-9) \\ 4(3) + 1(-1) - 3(-9) \end{bmatrix} = \frac{1}{14} \begin{bmatrix} -16 \\ 20 \\ 38 \end{bmatrix} = \begin{bmatrix} -8/7 \\ 10/7 \\ 19/7 \end{bmatrix} \end{aligned}$$

\]
</reasoning>

<answer>
<mcq-option>(A) \(x = 1/5, y = 3/5, z = 8/7\)</mcq-option>
<mcq-correct>(B) \(x= -8/7, y = 10/7, z = 19/7\)</mcq-correct>
</answer>

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