Determinants 1

Find this content useful?
Submitted by roamingfree on Sun, 09/26/2021 - 22:01
skills
Manipulate matrices, calculate determinants,eignvalues
Question

Given that the matrix M is singular, determine the possible value of $a$. $$ M = \left(\begin{array}{ccc}a & 1 & 2\\ 1 & a & 3 \\ 1 & 1 & 2\end{array}\right) $$

In each case, determine the eigenvalues of $M$.

answer --- press button to toggle display
$a=1,\frac{3}{2}$
solution --- press button to toggle display

$$\det(M) = a((a)(2) - (1)(3)) - 1((1)(2) - (3)(1)) + 2((1)(1) - (a)(1))$$

This simplifies to $$\det(M) = 2a^2 - 5a + 3$$

Since M is singular, then $\det(M) = 0$ and hence $a = 1,\frac{3}{2}$

The eigenvalues are the solutions of the equations $\det(M-\lambda) = 0$.

 

Let $a=1$, then then eigenvalues are the solution of $$ \det(M-\lambda) = \det\left(\begin{array}{ccc}1-\lambda & 1 & 2\\ 1 & 1-\lambda & 3 \\ 1 & 1 & 2-\lambda\end{array}\right) $$ 

Simplification yields a characteristic polynomial of $-\lambda^3 + 4\lambda^2 + \lambda = 0$, which in turn has solutions of $\lambda = 0,\; \lambda = 2\pm \sqrt{5}$

Let $a=\frac{3}{2}$, then the eigenvalues are solutions of $$ \det(M-\lambda) = \det\left(\begin{array}{ccc}\frac{3}{2}-\lambda & 1 & 2\\ 1 & \frac{3}{2}-\lambda & 3 \\ 1 & 1 & 2-\lambda\end{array}\right) $$

Simplification yields a characteristic polynomial of $-\lambda^3 + 5\lambda^2 - \frac{9}{4}\lambda = 0$, which yields eigenvalues of $\lambda = 0,\;\frac{9}{2},\;\frac{1}{2}$