Question

If \(A' = \begin{bmatrix} -2 & 3 \\ 1 & 2 \end{bmatrix}\) and \(B = \begin{bmatrix} -1 & 0 \\ 1 & 2 \end{bmatrix}\), find \((A+2B)'\).

Hint:

Using the properties of transpose \((X+Y)' = X' + Y'\) is often faster than first calculating A, then 2B, then A+2B, and finally taking the transpose. Using properties can reduce the number of calculations and potential errors.

Answer 1

Step 1: Understanding the Concept:
This question involves matrix operations, specifically addition, scalar multiplication, and transposition. The notation \(A'\) denotes the transpose of matrix \(A\). We need to use the properties of matrix transpose, one of which is \((X+Y)' = X' + Y'\) and another is \((kX)' = kX'\) where \(k\) is a scalar.
Step 2: Key Formula or Approach:
We can use the property of transposes: \((A+2B)' = A' + (2B)' = A' + 2B'\). 1. We are already given \(A'\).
2. Find the transpose of \(B\), which is \(B'\).
3. Calculate \(2B'\).
4. Add \(A'\) and \(2B'\) to find the result.
Step 3: Detailed Explanation:
We are given: \[ A' = \begin{bmatrix} -2 & 3 \\ 1 & 2 \end{bmatrix}, \quad B = \begin{bmatrix} -1 & 0 \\ 1 & 2 \end{bmatrix} \] We need to find \((A+2B)'\). Using the properties of transpose: \[ (A+2B)' = A' + (2B)' = A' + 2B' \] First, find \(B'\), the transpose of \(B\): \[ B' = \begin{bmatrix} -1 & 1 \\ 0 & 2 \end{bmatrix} \] Next, calculate \(2B'\): \[ 2B' = 2 \begin{bmatrix} -1 & 1 \\ 0 & 2 \end{bmatrix} = \begin{bmatrix} -2 & 2 \\ 0 & 4 \end{bmatrix} \] Finally, add \(A'\) and \(2B'\): \[ (A+2B)' = A' + 2B' = \begin{bmatrix} -2 & 3 \\ 1 & 2 \end{bmatrix} + \begin{bmatrix} -2 & 2 \\ 0 & 4 \end{bmatrix} = \begin{bmatrix} -4 & 5 \\ 1 & 6 \end{bmatrix} \] Step 4: Final Answer:
The value of \((A+2B)'\) is \(\boxed{\begin{bmatrix} -4 & 5 \\ 1 & 6 \end{bmatrix}}\).