The sum of the eigenvalues of the matrix $A={\left[\begin{array}{ll}1& 2\\ 3& 4\end{array}\right]}^2$ is ____________(rounded off to the nearest integer).

The correct answer is 29.

To find the sum of the eigenvalues of the matrix $A^2$ where the base matrix $A$ is given by:

$A=\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right]$

we can utilize the fundamental properties of eigenvalues and matrix traces.

First, recall that for any square matrix, the sum of its eigenvalues is equal to the trace of the matrix (the sum of the diagonal elements).

Let us compute the matrix $A^2$ by performing matrix multiplication of $A$ with itself:

$A^2=\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right]\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right]$

Now, let's calculate each entry of the resulting matrix:

The entry in the first row and first column is:
$(1\times 1)+(2\times 3)=1+6=7$

The entry in the first row and second column is:
$(1\times 2)+(2\times 4)=2+8=10$

The entry in the second row and first column is:
$(3\times 1)+(4\times 3)=3+12=15$

The entry in the second row and second column is:
$(3\times 2)+(4\times 4)=6+16=22$

So, the squared matrix is:

$A^2=\left[\begin{array}{cc}7& 10\\ 15& 22\end{array}\right]$

The sum of the eigenvalues of $A^2$ is equal to its trace, which is the sum of its diagonal elements:

$\text{Sum of eigenvalues}=\text{Trace}\left(A^2\right)=7+22=29$

Alternatively, we can find the eigenvalues of $A$ first. If the eigenvalues of $A$ are ${\lambda }_1$ and ${\lambda }_2$, then the eigenvalues of $A^2$ are ${\lambda }_1^2$ and ${\lambda }_2^2$.

For matrix $A$:
Trace of $A$: ${\lambda }_1+{\lambda }_2=1+4=5$
Determinant of $A$: ${\lambda }_1{\lambda }_2=(1\times 4)-(2\times 3)=4-6=-2$

Using the algebraic identity:

${\lambda }_1^2+{\lambda }_2^2={({\lambda }_1+{\lambda }_2)}^2-2{\lambda }_1{\lambda }_2$

We substitute the known values into the equation:

${\lambda }_1^2+{\lambda }_2^2={\left(5\right)}^2-2(-2)=25+4=29$

Both methods consistently show that the sum of the eigenvalues of $A^2$ is exactly 29.