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.
$\endgroup$ 13 Answers
$\begingroup$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}
$\endgroup$ 1 $\begingroup$Let $\chi_A(t)$ be our characteristic polynomial.
Add $\DeclareMathOperator{Row}{Row}-2\cdot\Row_2$ to $\Row_4$ to obtain $$ \chi_A(t)= \det \left[\begin{array}{rrrr} -t + 1 & 3 & 0 & 3 \\ 1 & -t + 1 & 1 & 1 \\ 0 & 4 & -t + 2 & 8 \\ 0 & 2 \, t - 2 & 1 & -t - 1 \end{array}\right] $$ Then add $-(-t+1)\cdot\Row_1$ to $\Row_2$ to obtain \begin{align*} \chi_A(t) &= \det \left[\begin{array}{rrrr} 0 & -{\left(t - 1\right)}^{2} + 3 & t - 1 & t + 2 \\ 1 & -t + 1 & 1 & 1 \\ 0 & 4 & -t + 2 & 8 \\ 0 & 2 \, t - 2 & 1 & -t - 1 \end{array}\right] \\ &= - \det \left[\begin{array}{rrrr} -{\left(t - 1\right)}^{2} + 3 & t - 1 & t + 2 \\ 4 & -t + 2 & 8 \\ 2 \, t - 2 & 1 & -t - 1 \end{array}\right] \end{align*} This gives a relatively simple way to reduce your $4\times 4$ determinant to a $3\times 3$ determinant.
$\endgroup$ $\begingroup$Exchange the first two and the last two columns (the goal is to do row reduction). The two minus signs compensate, so the determinant doesn't change.
\begin{bmatrix}
3&1-\lambda & 3 & 0 \\
1-\lambda& 1 & 1 &1\\
4&0 & 8&2-\lambda \\
0&2 & 1-\lambda&3 \\
\end{bmatrix}
Now
$$
\begin{bmatrix}
3&1-\lambda & 3 & 0 \\
1-\lambda& 1 & 1 &1\\
4&0 & 8&2-\lambda \\
0&2 & 1-\lambda&3 \\
\end{bmatrix}
=3\,\begin{bmatrix}
1&(1-\lambda)/3 & 1 & 0 \\
1-\lambda& 1 & 1 &1\\
4&0 & 8&2-\lambda \\
0&2 & 1-\lambda&3 \\
\end{bmatrix}
$$
Next subtract $1-\lambda$ times the first row from the second one, and four times the first one from the third:
$$
3\,\begin{bmatrix}
1&(1-\lambda)/3 & 1 & 0 \\
0& 1-(1-\lambda)^2/3 & \lambda &1\\
0&-4(1-\lambda)/3 & 4&2-\lambda \\
0&2 & 1-\lambda&3 \\
\end{bmatrix}
=3\,\begin{bmatrix} 1-(1-\lambda)^2/3 & \lambda &1\\
-4(1-\lambda)/3 & 4&2-\lambda \\
2 & 1-\lambda&3 \\
\end{bmatrix}
$$
Next, subtract $2-\lambda$ times the first row from the second one, and three times the first row from the third one:
$$
3\,\begin{bmatrix} 1-(1-\lambda)^2/3 & \lambda &1\\
-4(1-\lambda)/3-(2-\lambda)(1-(1-\lambda)^2/3) & 4-\lambda(2-\lambda)&0\\
-1+(1-\lambda)^2 & 1-4\lambda&0 \\
\end{bmatrix}
$$
The $2\times2$ determinant can now be calculated directly, to get
$$
3\lambda^4-15\lambda^3-84\lambda^2+174\lambda-24.
$$
