Let M be the vector space of 2×2 real-valued matrices. M=[abcd] and define M#=[dbca] Characterize the matrices M such that M#=M−1

Step 1 Let M be the vector space of 2×2 real-valued matrices. M=[abcd] and define M#=[dbca] To characterize matrix M such that M#=M−1 Therefore for findiny M−1 we want det(M)≠0 Therefore det(M)=ad−bc≠0 If det(M)≠0 then M−1=1ad−bc[d−b−ca] So, M#=M−1 ⇒[dbca]=1(ad−bc)[d−b−ca] ⇒[(ad−bc)d(ad−bc)b(ad−bc)c(ad−bc)a]=[d−b−ca] Case (1) By comparing if either a=0 pr d=0 of both then ab−bc=−bc≠0(as ad−bc≠0) So, b and c both non zero ⇒−b=b(ad−bc)≠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≠0 then a=(ad−bc)a⇒ad−bc=1 So, b=−b and c=−c ⇒2b=0 and 2c=0 ⇒b=0⇒c=0 So, M=[a00d] with ad=1

Resolved: "Let M be the vector space of 2×2..."

Jaya Legge

Answered question

2020-11-30

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

Answer & Explanation

FieniChoonin

Skilled2020-12-01Added 102 answers

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

Jeffrey Jordon

Expert2022-01-24Added 2605 answers

Answer is given below (on video)

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