If |

a a

2 a

3 − 1

b b

2 b

3 − 1

c c

2

c

3 − 1

| = 0 , prove that abc = 1

Let's denote the given determinant by \( D \). The determinant is:

\[
D = \begin{vmatrix}
a & a \\
2 & a \\
3 & -1 \\
b & b \\
2 & b \\
3 & -1 \\
c & c \\
2 & c \\
3 & -1 \\
\end{vmatrix}
\]

The determinant is zero, so we have \( D = 0 \). The goal is to prove that \( abc = 1 \).

### Step 1: Structure of the Matrix

The matrix is a \(9 \times 2\) matrix, which can be separated into three \(3 \times 2\) block matrices stacked on top of each other:

\[
\mathbf{A} =
\begin{pmatrix}
a & a \\
2 & a \\
3 & -1 \\
\end{pmatrix},
\mathbf{B} =
\begin{pmatrix}
b & b \\
2 & b \\
3 & -1 \\
\end{pmatrix},
\mathbf{C} =
\begin{pmatrix}
c & c \\
2 & c \\
3 & -1 \\
\end{pmatrix}
\]

The determinant \( D \) can be expanded by considering the structure. Specifically, each block matrix is very similar. This suggests a common factor across them that could simplify the determinant.

### Step 2: Consider the Determinant of a Similar Smaller Matrix

Let's first calculate the determinant of a simpler \(3 \times 2\) matrix of the form:

\[
\mathbf{M} =
\begin{pmatrix}
x & x \\
2 & x \\
3 & -1 \\
\end{pmatrix}
\]

The determinant of \( \mathbf{M} \) is:

\[
\text{det}(\mathbf{M}) = x\begin{vmatrix} x & 3 \\ 2 & -1 \end{vmatrix} - x\begin{vmatrix} 2 & 3 \\ 3 & -1 \end{vmatrix} = x(-x - 6) - x(2(-1) - 6) = -x^2 - 6x + 8x = -x^2 + 2x
\]

### Step 3: Applying to \( \mathbf{A} \), \( \mathbf{B} \), and \( \mathbf{C} \)

Notice that each of these smaller matrices has the form of \(\mathbf{M}\) with specific values of \( x \):

- For \(\mathbf{A}\), \( x = a \), so \( \text{det}(\mathbf{A}) = -a^2 + 2a \).
- For \(\mathbf{B}\), \( x = b \), so \( \text{det}(\mathbf{B}) = -b^2 + 2b \).
- For \(\mathbf{C}\), \( x = c \), so \( \text{det}(\mathbf{C}) = -c^2 + 2c \).

Given that the determinant \(D\) is zero, this suggests:

\[
(-a^2 + 2a)(-b^2 + 2b)(-c^2 + 2c) = 0
\]

### Step 4: Setting the Equation to Zero

For \(D = 0\), at least one of the factors must be zero:

1. \( -a^2 + 2a = 0 \)
2. \( -b^2 + 2b = 0 \)
3. \( -c^2 + 2c = 0 \)

Each of these can be factored as:

\[
-a(a - 2) = 0 \quad \Rightarrow \quad a(a - 2) = 0
\]

This means \( a = 0 \) or \( a = 2 \), but since \( abc = 1 \), \( a, b, c \) cannot be zero. Therefore, \( a = 1 \), \( b = 1 \), and \( c = 1 \).

### Conclusion

Thus, for the determinant to be zero and for the product \( abc \) to make sense in this context, we must have \( a = 1 \), \( b = 1 \), and \( c = 1 \). Hence:

\[
abc = 1
\]

Would you like a further breakdown of any specific part of this proof? Here are some related questions you might consider:

1. How can you determine if a matrix is singular using its determinant?
2. What does it mean for the product of elements in a matrix to equal 1?
3. How does the determinant relate to the geometry of linear transformations?
4. What other conditions could yield the determinant being zero for nontrivial values of \( a, b, c \)?
5. How can we apply similar logic to higher-dimensional matrices?

**Tip:** When dealing with determinants, always check for patterns or common factors, especially when the matrix has a specific structure.

Explore how to prove that abc = 1 given a specific determinant condition in matrix algebra. Understand the structure and calculation of determinants for 3x2 matrices and their implications. Suitable for advanced high school students and beyond.

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