For the matrix A=[3211], find the numbers a and b such that A2+aA+bI=0
Given A=[3211]
Also, A2+aA+bI=O ....(1)
Now, A2=[3211][3211]
⇒A2=[9+26+23+12+1]
⇒A2=[11843]
Substituting the values in (1), we get
[11843]+a[3211]+b[1001]=[0000]
⇒[11843]+[3a2aaa]+[b00b]=[0000]
⇒[11+3a+b8+2a4+a3+a+b]=[0000]
On comparing corresponding elements of both matrix, we get
a+4=0
⇒a=−4
Also, 3+a+b=0
⇒3−4+b=0
⇒b=1
Hence, the values of a and b are -4,1 respectively.