ISI2015-MMA-61

Question

Let $$ A = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 2 & 2  \\ 1 & 2 & 3 \end{pmatrix} \text{ and } B=\begin{pmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 1 & 1 \end{pmatrix}.$$ Then

• A. there exists a matrix $C$ such that $A=BC=CB$
• B. there is no matrix $C$ such that $A=BC$
• C. there exists a matrix $C$ such that $A=BC$, but $A \neq CB$
• D. there is no matrix $C$ such that $A=CB$

Comments

Is the answer (C)?

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The ans is C

Answer 1

Determinant of B is |B|=1.

Let $I$ be identity matrix.

1- Consider $A =CB$

$A$*$B^{-1}$=$C*B*B^{-1}$

$A*B^{-1}=C*I$          ($\because$ product of a matrix and it's inverse is identity matrix)

$\therefore C= A*B^{-1}$

since determinant of B is non zero,$B^{-1}$ exists.

Therefore, $A*B^{-1}$ exists.So we can say that there is a matrix C such that $A= CB$

2-

$A=BC$

$B^{-1}*A=B^{-1}*B*C$

$B^{-1}*A=I*C$      ($\because$ product of a matrix and it's inverse is identity matrix)

$C=B^{-1}*A$

since determinant of B is non zero,$B^{-1}$ exists.

Therefore, $B^{-1}*A$ exists.So we can say that there is a matrix C such that $A= BC$.

Therefore option A is true.

Comments

$AB^{-1} \neq B^{-1}A$ , So C in both case will be different. thus option C is correct