Question

Let M be the vector space of 2 times 2 real-valued matrices.M=begin{bmatrix}a & b c & d end{bmatrix}and define M^{#}=begin{bmatrix}d & b c & a end{bmatrix}

Matrices
ANSWERED
asked 2020-11-30

Let M be the vector space of \(2 \times 2\) real-valued matrices.
\(M=\begin{bmatrix}a & b \\c & d \end{bmatrix}\)
and define \(M^{\#}=\begin{bmatrix}d & b \\c & a \end{bmatrix}\) Characterize the matrices M such that \(M^{\#}=M^{-1}\)

Expert Answers (1)

2020-12-01

Step 1
Let M be the vector space of \(2 \times 2\) real-valued matrices.
\(M=\begin{bmatrix}a & b \\c & d \end{bmatrix}\)
and define \(M^{\#}=\begin{bmatrix}d & b \\c & a \end{bmatrix}\)
To characterize matrix M such that \(M^{\#}=M^{-1}\)
Therefore for findiny \(M^{-1}\) we want \(det(M) \neq 0\)
Therefore \(det(M)=ad-bc \neq 0\)
If \(det(M) \neq 0\)
then \(M^{-1}=\frac{1}{ad-bc} \begin{bmatrix}d & -b \\ -c & a \end{bmatrix}\)
So, \(M^{\#}=M^{-1}\)
\(\Rightarrow \begin{bmatrix}d & b \\c & a \end{bmatrix}= \frac{1}{(ad-bc)} \begin{bmatrix}d & -b \\ -c & a \end{bmatrix}\)
\(\Rightarrow \begin{bmatrix}(ad-bc)d & (ad-bc)b \\ (ad-bc)c & (ad-bc)a \end{bmatrix} = \begin{bmatrix}d & -b \\ -c & a \end{bmatrix}\)
Case (1)
By comparing if either a=0 pr d=0 of both then \(ab-bc=-bc \neq 0 (\text{as } ad-bc \neq 0 )\)
So, b and c both non zero
\(\Rightarrow -b=b(ad-bc) \neq ab-bc=-1\)
Step 2
\(M^{\#}=M^{-1}\) and if either a=0 or d=0 of both then ad-bc=-1
Case ( 2 )
If \(a,d \neq 0\) then
\(a=(ad-bc)a \Rightarrow ad-bc=1\)
So, \(b=-b \text{ and } c=-c\)
\(\Rightarrow 2b=0\ and\ 2c=0\)
\(\Rightarrow b=0 \Rightarrow c=0\)
So, \(M=\begin{bmatrix}a & 0 \\ 0 & d \end{bmatrix}\) with ad=1

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