<s>[INST] Calculate the determinant of $A-5I$ Question
Let $ A = 
\begin{bmatrix}
1 & 2 & 3 & 4 & 5 \\
6 & 7 & 8 & 9 & 10 \\
11 & 12 & 13 & 14 & 15 \\
16 & 17 & 18 & 19 & 20 \\
21& 22 & 23 & 24 & 25
\end{bmatrix}
$.
Calculate the determinant of $A-5I$.
My approach
the nullity of $A$ is $3$, so the algebraic multiplicity of $\lambda = 0$ is $3$, i.e. $\lambda_1 = \lambda_2 = \lambda_3 = 0.$
Now trace($A$) = $\lambda_4 + \lambda_5 = 1+6+13+19+25 = 65$
Then det($A-5I$) = $(\lambda_1-5)(\lambda_2-5)(\lambda_3-5)(\lambda_4-5)(\lambda_5-5)=(-5)^3(\lambda_4\lambda_5 - 5 \times 65 + 25)$
We need to calculate the value of $\lambda_4 \lambda_5$, which includes sum of lots of determinant of $2 \times 2$ matrices.
Is there any fast way to calculate the determinant?
 [/INST] Row-reducing the matrix $A-5I$ is quicker.
$$
\begin{gather}
\begin{pmatrix}
-4 & 2 & 3 & 4 & 5 \\
6 & 2 & 8 & 9 & 10 \\
11 & 12 & 8 & 14 & 15 \\
16 & 17 & 18 & 14 & 20 \\
21 & 22 & 23 & 24 & 20
\end{pmatrix} \longrightarrow
\begin{pmatrix}
-4 & 2 & 3 & 4 & 5 \\
14 & -2 & 2 & 1 & 0 \\
23 & 6 & -1 & 2 & 0 \\
32 & 9 & 6 & -2 & 0 \\
37 & 14 & 11 & 8 & 0
\end{pmatrix} \\
\longrightarrow \begin{pmatrix}
-4 & 2 & 3 & 4 & 5 \\
14 & -2 & 2 & 1 & 0 \\
-5 & 10 & -5 & 0 & 0 \\
60 & 5 & 10 & 0 & 0 \\
-75 & 30 & -5 & 0 & 0
\end{pmatrix} \longrightarrow 
\begin{pmatrix}
-4 & 2 & 3 & 4 & 5 \\
14 & -2 & 2 & 1 & 0 \\
-5 & 10 & -5 & 0 & 0 \\
50 & 25 & 0 & 0 & 0 \\
-70 & 20 & 0 & 0 & 0
\end{pmatrix} \\
\longrightarrow \begin{pmatrix}
-4 & 2 & 3 & 4 & 5 \\
14 & -2 & 2 & 1 & 0 \\
-5 & 10 & -5 & 0 & 0 \\
50 & 25 & 0 & 0 & 0 \\
-110 & 0 & 0 & 0 & 0
\end{pmatrix}
\end{gather}
$$
This takes not more than $30$ multiplications in all. The determinant is clearly $$5\cdot (-1) \cdot (-5) \cdot (-25) \cdot (-110) = 68750.$$
I chose to row-reduce the matrix in this fashion to make use of the fact that every entry in the last column is a multiple of $5$. It made the calculations easier when doing it by hand.
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