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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 3×33\times 3 matrices such that MN=NMMN=NM. Further, if MN2M\ne {{N}^{2}} and M2=N4{{M}^{2}}={{N}^{4}}, then find the correct options.
A. determinant of (M2+MN2)\left( {{M}^{2}}+M{{N}^{2}} \right) is 0.
B. there is a 3×33\times 3 matrix U such that (M2+MN2)U\left( {{M}^{2}}+M{{N}^{2}} \right)U is the zero matrix.
C. determinant of (M2+MN2)1\left( {{M}^{2}}+M{{N}^{2}} \right)\ge 1.
D. for a 3×33\times 3 matrix U, if (M2+MN2)U\left( {{M}^{2}}+M{{N}^{2}} \right)U equals the zero matrix, then U is the zero matrix.

Explanation

Solution

We consider these matrices as normal determinant form as the matrices are commutative. We try to find the determinant value of (M2+MN2)\left( {{M}^{2}}+M{{N}^{2}} \right). We also use a null matrix U to find (M2+MN2)U\left( {{M}^{2}}+M{{N}^{2}} \right)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{{M}^{2}}={{N}^{4}} and MN2M\ne {{N}^{2}}.
We can say that M and N are commutative as MN=NMMN=NM.
We can consider them as multiplication form M2=N4(MN2)(M+N2)=0{{M}^{2}}={{N}^{4}}\Rightarrow \left( M-{{N}^{2}} \right)\left( M+{{N}^{2}} \right)=0.
As we know MN2M\ne {{N}^{2}}, we can say that
(MN2)(M+N2)=0 (M+N2)=0 \begin{aligned} & \left( M-{{N}^{2}} \right)\left( M+{{N}^{2}} \right)=0 \\\ & \Rightarrow \left( M+{{N}^{2}} \right)=0 \\\ \end{aligned}
We have to consider them as matrix values. So, we need to use them as a determinant form.
det[(MN2)(M+N2)]=0\det \left[ \left( M-{{N}^{2}} \right)\left( M+{{N}^{2}} \right) \right]=0. We have det(MN2)0\det \left( M-{{N}^{2}} \right)\ne 0.
This gave us det(M+N2)=0\det \left( M+{{N}^{2}} \right)=0.
We need to find det(M2+MN2)\det \left( {{M}^{2}}+M{{N}^{2}} \right).
As the matrices are commutative det(M2+MN2)=[det(M)][det(M+N2)]\det \left( {{M}^{2}}+M{{N}^{2}} \right)=\left[ \det \left( M \right) \right]\left[ \det \left( M+{{N}^{2}} \right) \right].
Now det(M+N2)=0\det \left( M+{{N}^{2}} \right)=0 which gave us
det(M2+MN2)=[det(M)][det(M+N2)]=0\det \left( {{M}^{2}}+M{{N}^{2}} \right)=\left[ \det \left( M \right) \right]\left[ \det \left( M+{{N}^{2}} \right) \right]=0.
So, for any 3×33\times 3 matrix U we can find (M2+MN2)U\left( {{M}^{2}}+M{{N}^{2}} \right)U is the zero matrix. (M2+MN2)U=O\left( {{M}^{2}}+M{{N}^{2}} \right)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.