Question

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

Hint:

Before starting a complex matrix calculation, always check for properties that can simplify the expression, such as \((A')' = A\) or \((AB)' = B'A'\). This can save a significant amount of time and reduce the chances of calculation errors.

Answer 1

Step 1: Understanding the Concept:
This problem involves matrix operations, specifically the transpose of a matrix and matrix multiplication. The goal is to compute the left-hand side (LHS) of the equation and show that it is equal to the right-hand side (RHS).
Step 2: Key Formula or Approach:
We will use two key properties of matrices:
1. Double Transpose Property: The transpose of the transpose of a matrix is the original matrix itself. That is, \((A')' = A\).
2. Matrix Multiplication: To find the element in the i-th row and j-th column of the product matrix \(AB\), we take the dot product of the i-th row of \(A\) and the j-th column of \(B\).
Step 3: Detailed Explanation:
First, we simplify the left-hand side (LHS) of the equation to be proved: \[ \text{LHS} = (A')' B \] Using the double transpose property, we know that \((A')' = A\). So, the equation becomes: \[ AB = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \] Now we perform the matrix multiplication \(A \times B\).
Matrix \(A\) has dimensions \(2 \times 3\), and matrix \(B\) has dimensions \(3 \times 2\). The product \(AB\) will be a \(2 \times 2\) matrix.
\[ AB = \begin{bmatrix} 3 & \sqrt{3} & 2 \\ 4 & 2 & 0 \end{bmatrix} \begin{bmatrix} 0 & 1/4 \\ 0 & 0 \\ 1/2 & 1/8 \end{bmatrix} \] Let's calculate each element of the resulting matrix:
Element (1,1): (Row 1 of A) \(\cdot\) (Column 1 of B) \[ (3)(0) + (\sqrt{3})(0) + (2)(1/2) = 0 + 0 + 1 = 1 \] Element (1,2): (Row 1 of A) \(\cdot\) (Column 2 of B) \[ (3)(1/4) + (\sqrt{3})(0) + (2)(1/8) = \frac{3}{4} + 0 + \frac{2}{8} = \frac{3}{4} + \frac{1}{4} = 1 \] Element (2,1): (Row 2 of A) \(\cdot\) (Column 1 of B) \[ (4)(0) + (2)(0) + (0)(1/2) = 0 + 0 + 0 = 0 \] Element (2,2): (Row 2 of A) \(\cdot\) (Column 2 of B) \[ (4)(1/4) + (2)(0) + (0)(1/8) = 1 + 0 + 0 = 1 \] Assembling the resulting matrix: \[ AB = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \] This is equal to the right-hand side (RHS) of the equation.
Step 4: Final Answer:
Since LHS = RHS, the statement \((A')' B = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}\) is proved.