Question

If \(A = \begin{bmatrix} 3 & \sqrt{3} & 2 \\ 4 & 2 & 0 \end{bmatrix}\) and \(B = \begin{bmatrix} 0 & 1/4 \\ 0 & 0 \\ 1/2 & 1/8 \end{bmatrix}\), then prove that \(|C| = 1\), where \(C = (A')' B\).

Hint:

Before diving into calculations with matrices, always check if there are any properties you can use to simplify the expression. Recognizing that \((A')' = A\) is the key to making this problem much simpler.

Answer 1

Step 1: Understanding the Concept:
This problem involves matrix operations, including transpose, multiplication, and finding the determinant. The key is to first simplify the expression for matrix \(C\) and then perform the necessary calculations to find its determinant.
Step 2: Key Formula or Approach:
1. Simplify the expression for \(C\) using the matrix property \((M')' = M\).
2. Perform the matrix multiplication to find the elements of \(C\).
3. Calculate the determinant of the resulting \(2 \times 2\) matrix \(C\) using the formula \(|C| = ad - bc\).
Step 3: Detailed Explanation:
We are given the definition of \(C\) as: \[ C = (A')' B \] Using the property that the transpose of a transpose of a matrix is the matrix itself, i.e., \((A')' = A\), the expression for \(C\) simplifies to: \[ C = AB \] Now we multiply matrix \(A\) by matrix \(B\): \[ C = \begin{bmatrix} 3 & \sqrt{3} & 2 \\ 4 & 2 & 0 \end{bmatrix} \begin{bmatrix} 0 & 1/4 \\ 0 & 0 \\ 1/2 & 1/8 \end{bmatrix} \] The resulting matrix \(C\) will be \(2 \times 2\). Calculating its elements:
\[ C_{11} = (3)(0) + (\sqrt{3})(0) + (2)(1/2) = 1 \] \[ C_{12} = (3)(1/4) + (\sqrt{3})(0) + (2)(1/8) = \frac{3}{4} + \frac{1}{4} = 1 \] \[ C_{21} = (4)(0) + (2)(0) + (0)(1/2) = 0 \] \[ C_{22} = (4)(1/4) + (2)(0) + (0)(1/8) = 1 \] So, the matrix \(C\) is: \[ C = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \] Finally, calculate the determinant of \(C\): \[ |C| = (1)(1) - (1)(0) = 1 \] Step 4: Final Answer:
The determinant of \(C\) is \(1\). Hence proved.