Question

question 30 points 2

find the determinant of the 3 by 3 matrix shown below:
\\\

$$\begin{vmatrix} 1 & 2 & 5 \\\\ 2 & 1 & 4 \\\\ 3 & 5 & 1 \\end{vmatrix}$$

\\

30

36

Explanation:

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<reasoning>

Identify the matrix elements

Using the Matrix Representation knowledge point, we write the given \(3 \times 3\) matrix \(A\) as:
\[
A =

$$\begin{pmatrix} 1 & 2 & 5 \\ 2 & 1 & 4 \\ 3 & 5 & 1 \end{pmatrix}$$

\]
We need to find its determinant, denoted as:
\[
\det(A) =

$$\begin{vmatrix} 1 & 2 & 5 \\ 2 & 1 & 4 \\ 3 & 5 & 1 \end{vmatrix}$$

\]

Apply cofactor expansion

We expand along the first row to calculate the determinant:
\[
\det(A) = a_{11}C_{11} + a_{12}C_{12} + a_{13}C_{13}
\]
This gives:
\[
\det(A) = 1 \cdot

$$\begin{vmatrix} 1 & 4 \\ 5 & 1 \end{vmatrix}$$

• 2 \cdot

$$\begin{vmatrix} 2 & 4 \\ 3 & 1 \end{vmatrix}$$

+ 5 \cdot

$$\begin{vmatrix} 2 & 1 \\ 3 & 5 \end{vmatrix}$$

\]

Calculate \(2 \times 2\) determinants

We evaluate each of the \(2 \times 2\) sub-determinants:
\[

$$\begin{vmatrix} 1 & 4 \\ 5 & 1 \end{vmatrix}$$

= (1)(1) - (4)(5) = 1 - 20 = -19
\]
\[

$$\begin{vmatrix} 2 & 4 \\ 3 & 1 \end{vmatrix}$$

= (2)(1) - (4)(3) = 2 - 12 = -10
\]
\[

$$\begin{vmatrix} 2 & 1 \\ 3 & 5 \end{vmatrix}$$

= (2)(5) - (1)(3) = 10 - 3 = 7
\]

Combine the terms

We substitute these values back into our expansion formula:
\[
\det(A) = 1(-19) - 2(-10) + 5(7)
\]
\[
\det(A) = -19 + 20 + 35
\]
\[
\det(A) = 1 + 35 = 36
\]
</reasoning>

<answer>
<mcq-option>30</mcq-option>
<mcq-correct>36</mcq-correct>
</answer>

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Answer:

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<reasoning>

Identify the matrix elements

Using the Matrix Representation knowledge point, we write the given \(3 \times 3\) matrix \(A\) as:
\[
A =

$$\begin{pmatrix} 1 & 2 & 5 \\ 2 & 1 & 4 \\ 3 & 5 & 1 \end{pmatrix}$$

\]
We need to find its determinant, denoted as:
\[
\det(A) =

$$\begin{vmatrix} 1 & 2 & 5 \\ 2 & 1 & 4 \\ 3 & 5 & 1 \end{vmatrix}$$

\]

Apply cofactor expansion

We expand along the first row to calculate the determinant:
\[
\det(A) = a_{11}C_{11} + a_{12}C_{12} + a_{13}C_{13}
\]
This gives:
\[
\det(A) = 1 \cdot

$$\begin{vmatrix} 1 & 4 \\ 5 & 1 \end{vmatrix}$$

• 2 \cdot

$$\begin{vmatrix} 2 & 4 \\ 3 & 1 \end{vmatrix}$$

+ 5 \cdot

$$\begin{vmatrix} 2 & 1 \\ 3 & 5 \end{vmatrix}$$

\]

Calculate \(2 \times 2\) determinants

We evaluate each of the \(2 \times 2\) sub-determinants:
\[

$$\begin{vmatrix} 1 & 4 \\ 5 & 1 \end{vmatrix}$$

= (1)(1) - (4)(5) = 1 - 20 = -19
\]
\[

$$\begin{vmatrix} 2 & 4 \\ 3 & 1 \end{vmatrix}$$

= (2)(1) - (4)(3) = 2 - 12 = -10
\]
\[

$$\begin{vmatrix} 2 & 1 \\ 3 & 5 \end{vmatrix}$$

= (2)(5) - (1)(3) = 10 - 3 = 7
\]

Combine the terms

We substitute these values back into our expansion formula:
\[
\det(A) = 1(-19) - 2(-10) + 5(7)
\]
\[
\det(A) = -19 + 20 + 35
\]
\[
\det(A) = 1 + 35 = 36
\]
</reasoning>

<answer>
<mcq-option>30</mcq-option>
<mcq-correct>36</mcq-correct>
</answer>

<post_analysis>
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"subject": "Mathematics",
"question_type": "Multiple Choice",
"knowledge_point": [
"Mathematics",
"Algebra",
"Matrix Determinant"
]
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