Question

the sum of three numbers is 6. if the first number is added to twice the third number, the result is 7. the sum of the second, the third and thrice the first number is 12. find the numbers using the matrix inverse method.

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

Explanation:

<pre_analysis>
{
"quality": "clear",
"question_count": 1,
"skills_matched": [
"step_cot",
"multiple_choice_formatter",
"explain_then_answer",
"tutor",
"mastery_adaptive_brevity"
],
"needs_drawing": false,
"concepts_used": [
"Matrix Construction",
"Matrix Determinant",
"Matrix Inverse Formula",
"Solving Matrix Equations"
],
"new_concepts": [],
"current_concepts": [
"Matrix Construction",
"Matrix Determinant",
"Matrix Inverse Formula",
"Solving Matrix Equations"
]
}
</pre_analysis>

<reasoning>

Formulate the system of equations

Using the Matrix Construction knowledge point
Let the three numbers be \(x\), \(y\), and \(z\).
\[

$$\begin{cases} x + y + z = 6 \\ x + 2z = 7 \\ 3x + y + z = 12 \end{cases}$$

\]
This system can be written in matrix form \(A X = B\):
\[

$$\begin{bmatrix} 1 & 1 & 1 \\ 1 & 0 & 2 \\ 3 & 1 & 1 \end{bmatrix}$$

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

=

$$\begin{bmatrix} 6 \\ 7 \\ 12 \end{bmatrix}$$

\]

Calculate the determinant of the coefficient matrix

Using the Matrix Determinant knowledge point
\[

$$\begin{aligned} \det(A) &= 1(0 \cdot 1 - 2 \cdot 1) - 1(1 \cdot 1 - 2 \cdot 3) + 1(1 \cdot 1 - 0 \cdot 3) \\ &= 1(-2) - 1(-5) + 1(1) \\ &= -2 + 5 + 1 \\ &= 4 \end{aligned}$$

\]

Find the inverse of the coefficient matrix

Using the Matrix Inverse Formula knowledge point
\[
A^{-1} = \frac{1}{4}

$$\begin{bmatrix} -2 & 0 & 2 \\ 5 & -2 & -1 \\ 1 & 2 & -1 \end{bmatrix}$$

\]

Solve for the variables

Using the Solving Matrix Equations knowledge point
\[

$$\begin{aligned} X &= A^{-1} B \\ \begin{bmatrix} x \\ y \\ z \end{bmatrix} &= \frac{1}{4} \begin{bmatrix} -2 & 0 & 2 \\ 5 & -2 & -1 \\ 1 & 2 & -1 \end{bmatrix} \begin{bmatrix} 6 \\ 7 \\ 12 \end{bmatrix} \\ &= \frac{1}{4} \begin{bmatrix} -12 + 0 + 24 \\ 30 - 14 - 12 \\ 6 + 14 - 12 \end{bmatrix} \\ &= \frac{1}{4} \begin{bmatrix} 12 \\ 4 \\ 8 \end{bmatrix} \\ &= \begin{bmatrix} 3 \\ 1 \\ 2 \end{bmatrix} \end{aligned}$$

\]
</reasoning>

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

<post_analysis>
{
"subject": "Mathematics",
"question_type": "Multiple Choice",
"knowledge_point": [
"Mathematics",
"Algebra",
"Solving Matrix Equations"
]
}
</post_analysis>

Answer:

<pre_analysis>
{
"quality": "clear",
"question_count": 1,
"skills_matched": [
"step_cot",
"multiple_choice_formatter",
"explain_then_answer",
"tutor",
"mastery_adaptive_brevity"
],
"needs_drawing": false,
"concepts_used": [
"Matrix Construction",
"Matrix Determinant",
"Matrix Inverse Formula",
"Solving Matrix Equations"
],
"new_concepts": [],
"current_concepts": [
"Matrix Construction",
"Matrix Determinant",
"Matrix Inverse Formula",
"Solving Matrix Equations"
]
}
</pre_analysis>

<reasoning>

Formulate the system of equations

Using the Matrix Construction knowledge point
Let the three numbers be \(x\), \(y\), and \(z\).
\[

$$\begin{cases} x + y + z = 6 \\ x + 2z = 7 \\ 3x + y + z = 12 \end{cases}$$

\]
This system can be written in matrix form \(A X = B\):
\[

$$\begin{bmatrix} 1 & 1 & 1 \\ 1 & 0 & 2 \\ 3 & 1 & 1 \end{bmatrix}$$

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

=

$$\begin{bmatrix} 6 \\ 7 \\ 12 \end{bmatrix}$$

\]

Calculate the determinant of the coefficient matrix

Using the Matrix Determinant knowledge point
\[

$$\begin{aligned} \det(A) &= 1(0 \cdot 1 - 2 \cdot 1) - 1(1 \cdot 1 - 2 \cdot 3) + 1(1 \cdot 1 - 0 \cdot 3) \\ &= 1(-2) - 1(-5) + 1(1) \\ &= -2 + 5 + 1 \\ &= 4 \end{aligned}$$

\]

Find the inverse of the coefficient matrix

Using the Matrix Inverse Formula knowledge point
\[
A^{-1} = \frac{1}{4}

$$\begin{bmatrix} -2 & 0 & 2 \\ 5 & -2 & -1 \\ 1 & 2 & -1 \end{bmatrix}$$

\]

Solve for the variables

Using the Solving Matrix Equations knowledge point
\[

$$\begin{aligned} X &= A^{-1} B \\ \begin{bmatrix} x \\ y \\ z \end{bmatrix} &= \frac{1}{4} \begin{bmatrix} -2 & 0 & 2 \\ 5 & -2 & -1 \\ 1 & 2 & -1 \end{bmatrix} \begin{bmatrix} 6 \\ 7 \\ 12 \end{bmatrix} \\ &= \frac{1}{4} \begin{bmatrix} -12 + 0 + 24 \\ 30 - 14 - 12 \\ 6 + 14 - 12 \end{bmatrix} \\ &= \frac{1}{4} \begin{bmatrix} 12 \\ 4 \\ 8 \end{bmatrix} \\ &= \begin{bmatrix} 3 \\ 1 \\ 2 \end{bmatrix} \end{aligned}$$

\]
</reasoning>

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

<post_analysis>
{
"subject": "Mathematics",
"question_type": "Multiple Choice",
"knowledge_point": [
"Mathematics",
"Algebra",
"Solving Matrix Equations"
]
}
</post_analysis>