I have been given the matrix
$$
\begin{bmatrix}
1 & 3 & 0 & 3 \\
1 & 1 & 1 & 1 \\
0 & 4 & 2 & 8 \\
2 & 0 & 3 & 1 \\
\end{bmatrix}
$$
and told I must find the characteristic polynomial. I began by applying cofactor expansion along the top row of the matrix
$$
\begin{bmatrix}
1-\lambda & 3 & 0 & 3 \\
1 & 1-\lambda & 1 & 1 \\
0 & 4 & 2-\lambda & 8 \\
2 & 0 & 3 & 1-\lambda \\
\end{bmatrix}
$$
and attempting to multiply out my results to get the correct answer of $\lambda^4 -5\lambda^3 – 28\lambda^2 + 58\lambda – 8$. However, this takes several pages of work and I keep making calculation errors and ending up with the wrong answer.
My question is, is there an easier way to find the determinant of this specific matrix, or, once the determinant is found, to multiply out the result to find the polynomial?
The only methods I have been taught have been to either try to find or create a row with several 0's to make the cofactor expansion easier, or to get an upper or lower triangular matrix, however, those seem equally as messy here.
Best Answer
For convenience, I write $t$ for $\lambda$. Call your matrix expression $M-t I$. Divide the third row by 2, swap columns 1 and 2 and multiply the third column by 2, you get $$ \det(M-t I)= -\det\left[\begin{array}{cc|cc} 3 & 1-t & 0 & 3\\ 1-t & 1 & 2 & 1\\ \hline 2 & 0 & 2-t & 4\\ 0 & 2 & 6 & 1-t \end{array}\right] =-\det\left[\begin{array}{c|c}A&B\\ \hline C&D\end{array}\right], $$ where the leading minus sign is due to the swapping of columns.
Since $C$ and $D$ commute, we get \begin{align} \det(M-t I) &=-\det(AD-BC) \\ &=-\det\left( \pmatrix{3 & 1-t\\ 1-t & 1} \pmatrix{2-t & 4\\ 6 & 1-t} -2\pmatrix{0&3\\ 2&1} \right) \\ &=-\det\left( \pmatrix{12-9t & t^2-2t+13\\ t^2-3t+8 & 5-5t} -\pmatrix{0&6\\ 4&2} \right) \\ &=-\det\pmatrix{12-9t & t^2-2t+7\\ t^2-3t+4 & 3-5t}\\ &=t^4 - 5 t^3 - 28 t^2 + 58 t - 8. \end{align}