Let \(M\) and \(N\) be two invertible square matrices over \(R\) of order 2 such that \(N\) is diagonal.…
Let \(M\) and \(N\) be two invertible square matrices over \(R\) of order 2 such that \(N\) is diagonal. Then \(M N M^{-1}\) is diagonal ____
For all \(M\)
Only when \(M\) is a scalar matrix
For all diagonal matrices \(M\)
\(M\) must be a null matrix
Solution
Let the real matrices of order 2
\(M=\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right] \text { and } N=\left[\begin{array}{cc}
n_1 & 0 \\
0 & n_2
\end{array}\right]\)
\(\because M\) and \(N\) are invertible matrices, so
\(M^{-1}=\frac{1}{a d-b c}\left[\begin{array}{cc}
d & -b \\
-c & a
\end{array}\right]\)
Therefore,
\(\begin{aligned}
M N M^{-1} & =\frac{1}{a d-b c}\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]\left[\begin{array}{cc}
n_1 & 0 \\
0 & n_2
\end{array}\right]\left[\begin{array}{cc}
d & -b \\
-c & a
\end{array}\right] \\
& =\frac{1}{a d-b c}\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]\left[\begin{array}{cc}
n_1 d & -n_1 b \\
-n_2 c & n_2 a
\end{array}\right] \\
& =\frac{1}{a d-b c}\left[\begin{array}{ll}
a n_1 d-b n_2 c & -a n_1 b+b n_2 a \\
c n_1 d-d n_2 c & -c n_1 d+d n_2 a
\end{array}\right]
\end{aligned}\)
It is given that \(M N M^{-1}\) is a diagonal matrix, then
\(a b\left(n_2-n_1\right)=0=c d\left(n_1-n_2\right)\)
and if \(M\) is a diagonal matrix, means \(b=c=0\), then the above requirement getting satisfy.
Hence, option (c) is correct.