Question
Question: Let M and N be two \(3\times 3\) matrices such that \(MN=NM\). Further, if \(M\ne {{N}^{2}}\) and \(...
Let M and N be two 3×3 matrices such that MN=NM. Further, if M=N2 and M2=N4, then find the correct options.
A. determinant of (M2+MN2) is 0.
B. there is a 3×3 matrix U such that (M2+MN2)U is the zero matrix.
C. determinant of (M2+MN2)≥1.
D. for a 3×3 matrix U, if (M2+MN2)U equals the zero matrix, then U is the zero matrix.
Solution
We consider these matrices as normal determinant form as the matrices are commutative. We try to find the determinant value of (M2+MN2). We also use a null matrix U to find (M2+MN2)U being a zero matrix.
Complete step by step answer:
We use matrix operations to find out the correct options.
It’s given that M2=N4 and M=N2.
We can say that M and N are commutative as MN=NM.
We can consider them as multiplication form M2=N4⇒(M−N2)(M+N2)=0.
As we know M=N2, we can say that
(M−N2)(M+N2)=0⇒(M+N2)=0
We have to consider them as matrix values. So, we need to use them as a determinant form.
det[(M−N2)(M+N2)]=0. We have det(M−N2)=0.
This gave us det(M+N2)=0.
We need to find det(M2+MN2).
As the matrices are commutative det(M2+MN2)=[det(M)][det(M+N2)].
Now det(M+N2)=0 which gave us
det(M2+MN2)=[det(M)][det(M+N2)]=0.
So, for any 3×3 matrix U we can find (M2+MN2)U is the zero matrix. (M2+MN2)U=O.
So, the correct answer is “Option A and B”.
Note: Considering the matrices as their determinant form and using normal binary operation is only working as the matrices are commutative. Without this condition this wouldn’t have worked.