Math Problem Statement
If |
a a
2 a
3 − 1
b b
2 b
3 − 1
c c
2
c
3 − 1
| = 0 , prove that abc = 1
Solution
Let's denote the given determinant by . The determinant is:
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.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Determinants
Matrix Algebra
Formulas
Determinant of a 3x2 matrix
Theorems
-
Suitable Grade Level
Advanced High School