set of matrices

[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 \(\displaystyle \mathcal{B} = \{X \in \mathcal{A} \ | \ MX = XM\}\)

Determine which of the following elements of \(\displaystyle \mathcal{A}\) lie in \(\displaystyle \mathcal{B}\):

\(\displaystyle \begin{bmatrix}1 & 1 \\0 & 1 \end{bmatrix}, \begin{bmatrix}1 & 1 \\1 & 1 \end{bmatrix}, \begin{bmatrix}0 & 0 \\0 & 0 \end{bmatrix}, \begin{bmatrix}1 & 1 \\1 & 0 \end{bmatrix}, \begin{bmatrix}1 & 0 \\0 & 1 \end{bmatrix}, \begin{bmatrix}0 & 1 \\1 & 0 \end{bmatrix}\)

 

[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 \(\displaystyle \mathcal{B} = \{X \in \mathcal{A} \ | \ MX = XM\}\)

Determine which of the following elements of \(\displaystyle \mathcal{A}\) lie in \(\displaystyle \mathcal{B}\):

\(\displaystyle \begin{bmatrix}1 & 1 \\0 & 1 \end{bmatrix}, \begin{bmatrix}1 & 1 \\1 & 1 \end{bmatrix}, \begin{bmatrix}0 & 0 \\0 & 0 \end{bmatrix}, \begin{bmatrix}1 & 1 \\1 & 0 \end{bmatrix}, \begin{bmatrix}1 & 0 \\0 & 1 \end{bmatrix}, \begin{bmatrix}0 & 1 \\1 & 0 \end{bmatrix}\)

show us your effort/s to solve this problem.

 

[logistic_guy]

show us your effort/s to solve this problem.

Let me take the first element in \(\displaystyle \mathcal{A}\).

\(\displaystyle \begin{bmatrix}1 & 1 \\0 & 1 \end{bmatrix}\)

Since this element is just \(\displaystyle M\), then \(\displaystyle MM = MM\) and \(\displaystyle \begin{bmatrix}1 & 1 \\0 & 1 \end{bmatrix} \in \mathcal{B}\).

Let me take the second element in \(\displaystyle \mathcal{A}\).

\(\displaystyle \begin{bmatrix}1 & 1 \\1 & 1 \end{bmatrix}\)

Let us calculate and see if this element is in \(\displaystyle \mathcal{B}\).

\(\displaystyle M\begin{bmatrix}1 & 1 \\1 & 1 \end{bmatrix} = \begin{bmatrix}1 & 1 \\0 & 1 \end{bmatrix}\begin{bmatrix}1 & 1 \\1 & 1 \end{bmatrix} = \begin{bmatrix}(1)(1) + (1)(1) & (1)(1) + (1)(1) \\(0)(1) + (1)(1) & (0)(1) + (1)(1) \end{bmatrix} = \begin{bmatrix}2 & 2 \\1 & 1 \end{bmatrix}\)

Now let us switch them and calculate again.

\(\displaystyle \begin{bmatrix}1 & 1 \\1 & 1 \end{bmatrix}M = \begin{bmatrix}1 & 1 \\1 & 1 \end{bmatrix}\begin{bmatrix}1 & 1 \\0 & 1 \end{bmatrix} = \begin{bmatrix}(1)(1) + (1)(0) & (1)(1) + (1)(1) \\(1)(1) + (1)(0) & (1)(1) + (1)(1) \end{bmatrix} = \begin{bmatrix}1 & 2 \\1 & 2\end{bmatrix}\)

Since \(\displaystyle \begin{bmatrix}2 & 2 \\1 & 1 \end{bmatrix} \neq \begin{bmatrix}1 & 2 \\1 & 2\end{bmatrix}\) then \(\displaystyle \begin{bmatrix}1 & 1 \\1 & 1 \end{bmatrix} \not\in \mathcal{B}\).

I will continue in the next post.