Question

question 34 points 2

solve the system of equations using the matrix inverse method.
\\(x + 2y - 3z = -4\\)
\\(2x + 3y + 2z = 2\\)
\\(3x - 3y - 4z = 11\\).

\\(x = 3, y = -2, z = 1\\)
\\(x = -2, y = 3, z = 1\\)

Explanation:

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<reasoning>

Set up the matrix equation

Using the Coefficient Matrix knowledge point
\[
A =

$$\begin{pmatrix} 1 & 2 & -3 \\ 2 & 3 & 2 \\ 3 & -3 & -4 \end{pmatrix}$$

, \quad X =

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

, \quad B =

$$\begin{pmatrix} -4 \\ 2 \\ 11 \end{pmatrix}$$

\]

Calculate the determinant of A

Using the Matrix Determinant knowledge point
\[

$$\begin{aligned} \det(A) &= 1 \begin{vmatrix} 3 & 2 \\ -3 & -4 \end{vmatrix} - 2 \begin{vmatrix} 2 & 2 \\ 3 & -4 \end{vmatrix} + (-3) \begin{vmatrix} 2 & 3 \\ 3 & -3 \end{vmatrix} \\ &= 1(-12 - (-6)) - 2(-8 - 6) - 3(-6 - 9) \\ &= 1(-6) - 2(-14) - 3(-15) \\ &= -6 + 28 + 45 = 67 \end{aligned}$$

\]

Find the adjugate matrix of A

Using the Adjoint Matrix and Cofactor Matrix knowledge points
\[

$$\begin{aligned} C_{11} &= -6, \quad C_{12} = 14, \quad C_{13} = -15 \\ C_{21} &= -\begin{vmatrix} 2 & -3 \\ -3 & -4 \end{vmatrix} = 17, \quad C_{22} = \begin{vmatrix} 1 & -3 \\ 3 & -4 \end{vmatrix} = 5, \quad C_{23} = -\begin{vmatrix} 1 & 2 \\ 3 & -3 \end{vmatrix} = 9 \\ C_{31} &= \begin{vmatrix} 2 & -3 \\ 3 & 2 \end{vmatrix} = 13, \quad C_{32} = -\begin{vmatrix} 1 & -3 \\ 2 & 2 \end{vmatrix} = -8, \quad C_{33} = \begin{vmatrix} 1 & 2 \\ 2 & 3 \end{vmatrix} = -1 \\ \text{adj}(A) &= \begin{pmatrix} -6 & 17 & 13 \\ 14 & 5 & -8 \\ -15 & 9 & -1 \end{pmatrix} \end{aligned}$$

\]

Compute the inverse matrix of A

Using the Matrix Inverse Formula knowledge point
\[
A^{-1} = \frac{1}{67}

$$\begin{pmatrix} -6 & 17 & 13 \\ 14 & 5 & -8 \\ -15 & 9 & -1 \end{pmatrix}$$

\]

Solve for the variable vector X

Using the Solving Matrix Equations knowledge point
\[

$$\begin{aligned} X &= A^{-1} B = \frac{1}{67} \begin{pmatrix} -6 & 17 & 13 \\ 14 & 5 & -8 \\ -15 & 9 & -1 \end{pmatrix} \begin{pmatrix} -4 \\ 2 \\ 11 \end{pmatrix} \\ x &= \frac{1}{67} [(-6)(-4) + (17)(2) + (13)(11)] = \frac{1}{67} [24 + 34 + 143] = \frac{201}{67} = 3 \\ y &= \frac{1}{67} [(14)(-4) + (5)(2) + (-8)(11)] = \frac{1}{67} [-56 + 10 - 88] = \frac{-134}{67} = -2 \\ z &= \frac{1}{67} [(-15)(-4) + (9)(2) + (-1)(11)] = \frac{1}{67} [60 + 18 - 11] = \frac{67}{67} = 1 \end{aligned}$$

\]
</reasoning>

<answer>
<mcq-correct>(A) \(x = 3, y = -2, z = 1\)</mcq-correct>
<mcq-option>(B) \(x = -2, y = 3, z = 1\)</mcq-option>
</answer>

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Answer:

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<reasoning>

Set up the matrix equation

Using the Coefficient Matrix knowledge point
\[
A =

$$\begin{pmatrix} 1 & 2 & -3 \\ 2 & 3 & 2 \\ 3 & -3 & -4 \end{pmatrix}$$

, \quad X =

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

, \quad B =

$$\begin{pmatrix} -4 \\ 2 \\ 11 \end{pmatrix}$$

\]

Calculate the determinant of A

Using the Matrix Determinant knowledge point
\[

$$\begin{aligned} \det(A) &= 1 \begin{vmatrix} 3 & 2 \\ -3 & -4 \end{vmatrix} - 2 \begin{vmatrix} 2 & 2 \\ 3 & -4 \end{vmatrix} + (-3) \begin{vmatrix} 2 & 3 \\ 3 & -3 \end{vmatrix} \\ &= 1(-12 - (-6)) - 2(-8 - 6) - 3(-6 - 9) \\ &= 1(-6) - 2(-14) - 3(-15) \\ &= -6 + 28 + 45 = 67 \end{aligned}$$

\]

Find the adjugate matrix of A

Using the Adjoint Matrix and Cofactor Matrix knowledge points
\[

$$\begin{aligned} C_{11} &= -6, \quad C_{12} = 14, \quad C_{13} = -15 \\ C_{21} &= -\begin{vmatrix} 2 & -3 \\ -3 & -4 \end{vmatrix} = 17, \quad C_{22} = \begin{vmatrix} 1 & -3 \\ 3 & -4 \end{vmatrix} = 5, \quad C_{23} = -\begin{vmatrix} 1 & 2 \\ 3 & -3 \end{vmatrix} = 9 \\ C_{31} &= \begin{vmatrix} 2 & -3 \\ 3 & 2 \end{vmatrix} = 13, \quad C_{32} = -\begin{vmatrix} 1 & -3 \\ 2 & 2 \end{vmatrix} = -8, \quad C_{33} = \begin{vmatrix} 1 & 2 \\ 2 & 3 \end{vmatrix} = -1 \\ \text{adj}(A) &= \begin{pmatrix} -6 & 17 & 13 \\ 14 & 5 & -8 \\ -15 & 9 & -1 \end{pmatrix} \end{aligned}$$

\]

Compute the inverse matrix of A

Using the Matrix Inverse Formula knowledge point
\[
A^{-1} = \frac{1}{67}

$$\begin{pmatrix} -6 & 17 & 13 \\ 14 & 5 & -8 \\ -15 & 9 & -1 \end{pmatrix}$$

\]

Solve for the variable vector X

Using the Solving Matrix Equations knowledge point
\[

$$\begin{aligned} X &= A^{-1} B = \frac{1}{67} \begin{pmatrix} -6 & 17 & 13 \\ 14 & 5 & -8 \\ -15 & 9 & -1 \end{pmatrix} \begin{pmatrix} -4 \\ 2 \\ 11 \end{pmatrix} \\ x &= \frac{1}{67} [(-6)(-4) + (17)(2) + (13)(11)] = \frac{1}{67} [24 + 34 + 143] = \frac{201}{67} = 3 \\ y &= \frac{1}{67} [(14)(-4) + (5)(2) + (-8)(11)] = \frac{1}{67} [-56 + 10 - 88] = \frac{-134}{67} = -2 \\ z &= \frac{1}{67} [(-15)(-4) + (9)(2) + (-1)(11)] = \frac{1}{67} [60 + 18 - 11] = \frac{67}{67} = 1 \end{aligned}$$

\]
</reasoning>

<answer>
<mcq-correct>(A) \(x = 3, y = -2, z = 1\)</mcq-correct>
<mcq-option>(B) \(x = -2, y = 3, z = 1\)</mcq-option>
</answer>

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