Question

A = $\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}$, C = $\begin{bmatrix} 7 & 8 & 9 \\ 10 & 11 & 12 \\ 13 & 14 & 15 \end{bmatrix}$, B = $\begin{bmatrix} 1 & 2 & 3 \\ 7 & 8 & 9 \\ 4 & 5 & 6 \end{bmatrix}$. P is a matrix of order 3 x 3 such that AP = C, then matrix BP is:

• A. $\begin{bmatrix} 7 & 8 & 9 \\ 10 & 11 & 12 \\ 13 & 14 & 15 \end{bmatrix}$
• B. $\begin{bmatrix} 7 & 9 & 8 \\ 10 & 12 & 11 \\ 13 & 15 & 14 \end{bmatrix}$
• C. $\begin{bmatrix} 7 & 8 & 9 \\ 13 & 14 & 15 \\ 10 & 11 & 12 \end{bmatrix}$
• D. $\begin{bmatrix} 8 & 7 & 9 \\ 11 & 10 & 12 \\ 14 & 13 & 15 \end{bmatrix}$

Answer 1

Answer: (C)

$\begin{bmatrix} 7 & 8 & 9 \\ 13 & 14 & 15 \\ 10 & 11 & 12 \end{bmatrix}$

Answer 2

The key to solving this problem lies in recognizing the row operations performed on matrix A to obtain matrix B, and then applying the same row operations to matrix C.

1. Observe the relationship between A and B:

   Matrix B is obtained from matrix A by swapping the second and third rows.

2. Apply the same row operation to C:

   Since AP = C, pre-multiplying by B implies performing the same row swaps on C. Therefore, swap the second and third rows of matrix C to find BP.

3. Resulting Matrix BP:

   $BP = \begin{bmatrix} 7 & 8 & 9 \\ 13 & 14 & 15 \\ 10 & 11 & 12 \end{bmatrix}$