Factor: $\det\left[\begin{smallmatrix}\lambda-(1-a) & -a \\ -b & \lambda - (1-b)\end{smallmatrix}\right]=0$

Question

$$\text{det}\begin{bmatrix}\lambda-(1-a) & -a \\ -b & \lambda - (1-b)\end{bmatrix}=0$$

$$\Big(\lambda-(1-a)\Big)~\Big(\lambda - (1-b)\Big) -(-a)(-b) = 0$$

$$\Big(\lambda-(1-a)\Big)~\Big(\lambda - (1-b)\Big) +ab = 0$$

The textbook says the answer is:

$$(\lambda -1)(\lambda -1 +a +b)=0$$

How did they get this? Every time I try this problem... I get about two pages of multiplying terms out into polynomials and applying quadric formula... there's got to be a faster 2-3 liner way to factor this...

Answer 1

Performing the transformation $C_1\mapsto C_1+C_2$, we get $$\left| \begin{matrix} \lambda -1 & -a \\ \lambda-1 & \lambda-(1-b) \end{matrix}\right|=0 \\ \implies (\lambda-1) \left| \begin{matrix} 1& -a \\ 1 & \lambda-(1-b) \end{matrix} \right|=0 $$ Now try expanding the determinant.

Answer 2

The best way to factor something like this - in my opinion anyhow - would be to factor in terms of the roots of the polynomial. First, expand the entire thing out in terms of $\lambda$; with some simplification, you'll get

$$\lambda^2 - (2-a-b)\lambda + (1 - a - b) = 0$$

Now, let's return to your original polynomial. (Note: you made a sign error on the $ab$ term, it should be negative.) You have

$$\Big(\lambda-(1-a)\Big)~\Big(\lambda - (1-b)\Big) -ab = 0$$

Wouldn't it be nice if $\big(\lambda-(1-a)\big)\big(\lambda - (1-b)\big)=ab$, so that it could cancel with the $ab$ and get zero? For that, though, we clearly need to have no constants, so $\lambda = 1$ seems a reasonable choice ... and, indeed, we get $(-a)(-b) = ab$ as a result, which ensures $\lambda = 1$ is a root of the above.

From there, we can do division of our expanded polynomial by $\lambda - 1$ to get the factorization you desire.

Answer 3

I assume this is in the context of an eigenvalue problem. Specifically, the quantity you seek is the monic quadratic whose roots are the eigenvalues of the matrix $$A= \begin{bmatrix} 1 - a & a \\ b & 1 - b\end{bmatrix}$$ Often, you can kind of "guess" the eigenvalues by remembering that 1) the trace (sum of the diagonal entries) is equal to the sum of the two eigenvalues, counting multiplicity, and 2) the determinant is equal to the product of the eigenvalues. Here we can calculate $$\text{trace} (A) = \lambda_1 + \lambda_2 = 2 - a - b, ~~ \det A = \lambda_1 \lambda_2 = -a - b + 1$$ and from it is not too difficult to observe that choosing $\lambda_1 = -a - b + 1$ and $\lambda_2 = 1$ as the eigenvalues satisfies these relations. Hence, we can conclude that the characteristic polynomial of the matrix $A$ (which is what you originally asked for) is $$(\lambda - 1)(\lambda - 1 + a + b)$$