First Principles AI

Accedmychevron_right11thchevron_rightmathchevron_rightMatrices And Determinantschevron_rightExercise 4.1

Solution

Given:

A=\begin{bmatrix}-i&2i\\1&-i\end{bmatrix},\quad B=\begin{bmatrix}-i&1\\2i&1\end{bmatrix},\quad C=\begin{bmatrix}2i&2i\\-i&i\end{bmatrix}

First compute $AB$ and $BC$

AB= \begin{bmatrix}-i&2i\\1&-i\end{bmatrix} \begin{bmatrix}-i&1\\2i&1\end{bmatrix} = \begin{bmatrix}-5&i\\2-i&1-i\end{bmatrix}

BC= \begin{bmatrix}-i&1\\2i&1\end{bmatrix} \begin{bmatrix}2i&2i\\-i&i\end{bmatrix} = \begin{bmatrix}2-i&3+i\\-5-i&-3+5i\end{bmatrix}

(i) Show that $(AB)C=A(BC)$

(AB)C= \begin{bmatrix}-5&i\\2-i&1-i\end{bmatrix} \begin{bmatrix}2i&2i\\-i&i\end{bmatrix} = \begin{bmatrix}1-10i&-1-10i\\1+3i&3+5i\end{bmatrix}

A(BC)= \begin{bmatrix}-i&2i\\1&-i\end{bmatrix} \begin{bmatrix}2-i&3+i\\-5-i&-3+5i\end{bmatrix} = \begin{bmatrix}1-10i&-1-10i\\1+3i&3+5i\end{bmatrix}

Hence:

\boxed{(AB)C=A(BC)}

(ii) Show that $A(B+C)=AB+AC$

First:

B+C= \begin{bmatrix}-i&1\\2i&1\end{bmatrix}+ \begin{bmatrix}2i&2i\\-i&i\end{bmatrix} = \begin{bmatrix}i&1+2i\\i&1+i\end{bmatrix}

Then:

A(B+C)= \begin{bmatrix}-i&2i\\1&-i\end{bmatrix} \begin{bmatrix}i&1+2i\\i&1+i\end{bmatrix} = \begin{bmatrix}-1&i\\1+i&2+i\end{bmatrix}

Also:

AC= \begin{bmatrix}-i&2i\\1&-i\end{bmatrix} \begin{bmatrix}2i&2i\\-i&i\end{bmatrix} = \begin{bmatrix}4-i&1+3i\\1-i&1+3i\end{bmatrix}

So:

AB+AC= \begin{bmatrix}-5&i\\2-i&1-i\end{bmatrix}+ \begin{bmatrix}4-i&1+3i\\1-i&1+3i\end{bmatrix} = \begin{bmatrix}-1&i\\1+i&2+i\end{bmatrix}

Therefore:

\boxed{A(B+C)=AB+AC}

Previous Question