Scalar Multiplication

Solution :

Given,

$$A=\begin{bmatrix}1 & 2 \\3 & 4\end{bmatrix}$$

First calculate

$$A^2$$

$$A^2=\begin{bmatrix}1 & 2 \\3 & 4\end{bmatrix}\begin{bmatrix}1 & 2 \\3 & 4\end{bmatrix}$$

$$=\begin{bmatrix}7 & 10 \\15 & 22\end{bmatrix}$$

Now,
$$5A=\begin{bmatrix}5 & 10 \\15 & 20\end{bmatrix}$$

Therefore,

$$A^2-5A=\begin{bmatrix}7 & 10 \\15 & 22\end{bmatrix} - \begin{bmatrix}5 & 10 \\15 & 20\end{bmatrix}$$

$$=\begin{bmatrix}2 & 0 \\0 & 2\end{bmatrix}$$

Given,

$$A^2-5A+kI=0$$

Therefore,

$$2I+kI=0$$

$$(2+k)I=0$$

$$2+k=0$$

$$k=-2$$

Final Answer :

$$-2$$

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 :

$$-10$$