
Let M and N be two $3\times 3$ matrices such that $MN=NM$. Further, if $M\ne {{N}^{2}}$ and ${{M}^{2}}={{N}^{4}}$, then find the correct options.
A. determinant of $\left( {{M}^{2}}+M{{N}^{2}} \right)$ is 0.
B. there is a $3\times 3$ matrix U such that $\left( {{M}^{2}}+M{{N}^{2}} \right)U$ is the zero matrix.
C. determinant of $\left( {{M}^{2}}+M{{N}^{2}} \right)\ge 1$.
D. for a $3\times 3$ matrix U, if $\left( {{M}^{2}}+M{{N}^{2}} \right)U$ equals the zero matrix, then U is the zero matrix.
Answer
562.2k+ views
Hint: We consider these matrices as normal determinant form as the matrices are commutative. We try to find the determinant value of $\left( {{M}^{2}}+M{{N}^{2}} \right)$. We also use a null matrix U to find $\left( {{M}^{2}}+M{{N}^{2}} \right)U$ being a zero matrix.
Complete step by step answer:
We use matrix operations to find out the correct options.
It’s given that ${{M}^{2}}={{N}^{4}}$ and $M\ne {{N}^{2}}$.
We can say that M and N are commutative as $MN=NM$.
We can consider them as multiplication form ${{M}^{2}}={{N}^{4}}\Rightarrow \left( M-{{N}^{2}} \right)\left( M+{{N}^{2}} \right)=0$.
As we know $M\ne {{N}^{2}}$, we can say that
$\begin{align}
& \left( M-{{N}^{2}} \right)\left( M+{{N}^{2}} \right)=0 \\
& \Rightarrow \left( M+{{N}^{2}} \right)=0 \\
\end{align}$
We have to consider them as matrix values. So, we need to use them as a determinant form.
$\det \left[ \left( M-{{N}^{2}} \right)\left( M+{{N}^{2}} \right) \right]=0$. We have $\det \left( M-{{N}^{2}} \right)\ne 0$.
This gave us $\det \left( M+{{N}^{2}} \right)=0$.
We need to find $\det \left( {{M}^{2}}+M{{N}^{2}} \right)$.
As the matrices are commutative $\det \left( {{M}^{2}}+M{{N}^{2}} \right)=\left[ \det \left( M \right) \right]\left[ \det \left( M+{{N}^{2}} \right) \right]$.
Now $\det \left( M+{{N}^{2}} \right)=0$ which gave us
$\det \left( {{M}^{2}}+M{{N}^{2}} \right)=\left[ \det \left( M \right) \right]\left[ \det \left( M+{{N}^{2}} \right) \right]=0$.
So, for any $3\times 3$ matrix U we can find $\left( {{M}^{2}}+M{{N}^{2}} \right)U$ is the zero matrix. $\left( {{M}^{2}}+M{{N}^{2}} \right)U=O$.
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.
Complete step by step answer:
We use matrix operations to find out the correct options.
It’s given that ${{M}^{2}}={{N}^{4}}$ and $M\ne {{N}^{2}}$.
We can say that M and N are commutative as $MN=NM$.
We can consider them as multiplication form ${{M}^{2}}={{N}^{4}}\Rightarrow \left( M-{{N}^{2}} \right)\left( M+{{N}^{2}} \right)=0$.
As we know $M\ne {{N}^{2}}$, we can say that
$\begin{align}
& \left( M-{{N}^{2}} \right)\left( M+{{N}^{2}} \right)=0 \\
& \Rightarrow \left( M+{{N}^{2}} \right)=0 \\
\end{align}$
We have to consider them as matrix values. So, we need to use them as a determinant form.
$\det \left[ \left( M-{{N}^{2}} \right)\left( M+{{N}^{2}} \right) \right]=0$. We have $\det \left( M-{{N}^{2}} \right)\ne 0$.
This gave us $\det \left( M+{{N}^{2}} \right)=0$.
We need to find $\det \left( {{M}^{2}}+M{{N}^{2}} \right)$.
As the matrices are commutative $\det \left( {{M}^{2}}+M{{N}^{2}} \right)=\left[ \det \left( M \right) \right]\left[ \det \left( M+{{N}^{2}} \right) \right]$.
Now $\det \left( M+{{N}^{2}} \right)=0$ which gave us
$\det \left( {{M}^{2}}+M{{N}^{2}} \right)=\left[ \det \left( M \right) \right]\left[ \det \left( M+{{N}^{2}} \right) \right]=0$.
So, for any $3\times 3$ matrix U we can find $\left( {{M}^{2}}+M{{N}^{2}} \right)U$ is the zero matrix. $\left( {{M}^{2}}+M{{N}^{2}} \right)U=O$.
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.
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