Answer:
Step-by-step explanation:
Find the characteristic polynomial of the matrix, using either a cofactor expansion or the special formula for 3times3 determinants. [Note: Finding the characteristic polynomial of a 3times3 matrix is not easy to do with just row operations, because the variable lambda is involved.]
[tex]\texttt {let} A = \left[\begin{array}{ccc}4&-4&0\\-4&7&0\\4&6&5\end{array}\right][/tex]
The characteristics polynomial of A is [tex]|A- \lambda I|[/tex]
[tex]\to|A-\lambda I|=\left[\begin{array}{ccc}(4-\lambda)&-4&0\\-4&(7-\lambda)&0\\4&6&(5-\lambda)\end{array}\right][/tex]
[tex]=(4-\lambda)[7-\lambda)(5-\lambda)-0]+4[-4(5-\lambda)-0]+0[-4\time 6-4(7-\lambda)]\\\\=(4-\lambda)[35-7\lambda-5\lambda+\lambda^2]+4[-20+4\lambda]+0\\\\=(4-\lambda)[35-12\lambda+\lambda^2]-80+16\lambda\\\\=140-48\lambda+4\lambda^2-35\lambda+12\lambda^2-\lambda^3-80+16\lambda\\\\=-\lambda^3+16\lambda^2-67\lambda+60[/tex]