proof related to matrices - 2

[logistic_guy]

Let $\displaystyle \mathcal{A}$ be the set of $\displaystyle 2 \times 2$ matrices with real number entries. Recall that matrix multiplication is defined by

$\displaystyle \begin{bmatrix}a & b \\c & d \end{bmatrix} \begin{bmatrix}p & q \\r & s \end{bmatrix} = \begin{bmatrix}ap + br & aq + bs \\cp + dr & cq + ds \end{bmatrix}$

Let $\displaystyle M = \begin{bmatrix}1 & 1 \\0 & 1 \end{bmatrix}$

and let $\displaystyle \mathcal{B} = \{X \in \mathcal{A} \ | \ MX = XM\}$

Prove that if $\displaystyle P, Q \in \mathcal{B}$, then $\displaystyle P \cdot Q \in \mathcal{B}$ (where $\displaystyle \cdot$ denotes the usual product of two matrices).

 

[khansaheb]

Let $\displaystyle \mathcal{A}$ be the set of $\displaystyle 2 \times 2$ matrices with real number entries. Recall that matrix multiplication is defined by

$\displaystyle \begin{bmatrix}a & b \\c & d \end{bmatrix} \begin{bmatrix}p & q \\r & s \end{bmatrix} = \begin{bmatrix}ap + br & aq + bs \\cp + dr & cq + ds \end{bmatrix}$

Let $\displaystyle M = \begin{bmatrix}1 & 1 \\0 & 1 \end{bmatrix}$

and let $\displaystyle \mathcal{B} = \{X \in \mathcal{A} \ | \ MX = XM\}$

Prove that if $\displaystyle P, Q \in \mathcal{B}$, then $\displaystyle P \cdot Q \in \mathcal{B}$ (where $\displaystyle \cdot$ denotes the usual product of two matrices).

show us your effort/s to solve this problem.

 

[logistic_guy]

I will assume $\displaystyle P, Q \in \mathcal{B}$. And I will use the associative property of matrix multiplication.

$\displaystyle M(P \cdot Q) = (M \cdot P)Q = MPQ$

Since $\displaystyle P \in \mathcal{B}$, I can switch the positions of $\displaystyle M$ and $\displaystyle P$.

$\displaystyle MPQ = PMQ$

Since $\displaystyle Q \in \mathcal{B}$, I can switch the positions of $\displaystyle M$ and $\displaystyle Q$.

$\displaystyle PMQ = PQM = (P \cdot Q)M$

Since $\displaystyle M(P \cdot Q) = (P \cdot Q)M$, then $\displaystyle P \cdot Q \in \mathcal{B}$.