Subtraction of Matrices

Given,

$$A=\begin{pmatrix}1&2\\-1&3\end{pmatrix}$$

First find $$A^2=A\cdot A$$

$$A^2=\begin{pmatrix}1&2\\-1&3\end{pmatrix}\begin{pmatrix}1&2\\-1&3\end{pmatrix}$$

First row first column : $$1(1)+2(-1)=1-2=-1$$

First row second column : $$1(2)+2(3)=2+6=8$$

Second row first column : $$(-1)(1)+3(-1)=-1-3=-4$$

Second row second column : $$(-1)(2)+3(3)=-2+9=7$$

Therefore, $$A^2=\begin{pmatrix}-1&8\\-4&7\end{pmatrix}$$

Now find $$4A$$

$$4A=4\begin{pmatrix}1&2\\-1&3\end{pmatrix}=\begin{pmatrix}4&8\\-4&12\end{pmatrix}$$

Now calculate $$A^2-4A$$

$$=\begin{pmatrix}-1&8\\-4&7\end{pmatrix}-\begin{pmatrix}4&8\\-4&12\end{pmatrix}$$

Subtract corresponding entries :

$$=\begin{pmatrix}-1-4&8-8\\-4-(-4)&7-12\end{pmatrix}$$

$$=\begin{pmatrix}-5&0\\0&-5\end{pmatrix}$$

Trace of a matrix is sum of diagonal elements.

Therefore,

$$\text{Trace}(A^2-4A)=-5+(-5)=-10$$

Final Answer :

$$-10$$