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Let $A = \begin{bmatrix} a_{11} && a_{12} \\ a_{21} && a_{22} \end{bmatrix} \text { and } B = \begin{bmatrix} b_{11} && b_{12} \\ b_{21} && b_{22} \end{bmatrix}$ be two matrices such that $AB=I$. Let $C = A \begin{bmatrix} 1 && 0 \\ 1 && 1 \end{bmatrix}$ and $CD =I$. Express the elements of $D$ in terms of the elements of $B$.

### 1 comment

caluclating we get the inversse of X  = (1\(det X)) * adjoint of X, and then multiply by B to get the desired matrix

$AB = I$, B is equal to the inverse of $A$ and vice versa.

So, $B= A^{-1}$

Now $CD =I$, $C$ is equal to the inverse of $D$ and vice versa.

So, $D =C^{-1}$ ​​​​​​

​$=\left(A.\begin{bmatrix} 1& 0 \\ 1&1 \end{bmatrix}\right)^{-1}$

Remark: $(AB)^{-1} = B^{-1}A^{-1}$

​$=\begin{bmatrix} 1& 0 \\ 1&1 \end{bmatrix}^{-1}.A^{-1}$

​$=\begin{bmatrix} 1& 0 \\ {-1}&1 \end{bmatrix}.B$

​$=\begin{bmatrix} b_{11}& b_{12} \\ b_{21}-b_{11}&b_{22}-b_{12} \end{bmatrix}$

which formula in last line?
^ Multiplication of matrices $I and B$  :)
$(1 - \lambda)^{2} = 0\\ 2\lambda - \lambda ^{2} = 1\\ 2 - \lambda = \lambda^{-1}\\ X^{-1} = 2I - X$
If $A/B = A * inverse(B)$

I did the same and:

$inverse(C) = inverse(A) * inverse($\begin{bmatrix} 1 & 0\\ 1 & 1 \end{bmatrix}$)$

What am I doing wrong?

AB=CD

let x=[ 1  0 ]

[ 1  1]

AB=AXD

D=inv(X) *B

ANS IS          [b11                b12  ]

[ b21-b11   b22-b12  ]

by
Suppose D = $\begin{bmatrix} d_{11} & d_{12} \\ d_{21} & d_{22} \end{bmatrix}$

Now, CD = AB  where C= A$\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}$

so,  A$\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}$ = AB  [here A will be cancel out because on bothside A is a matrix]

$\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}$ . $\begin{bmatrix} d_{11} & d_{12} \\ d_{21} & d_{22} \end{bmatrix}$ = A$\begin{bmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{bmatrix}$

$\begin{bmatrix} d_{11} & d_{12} \\ d_{11} + d_{21} & d_{12} + d_{22} \end{bmatrix}$ = $\begin{bmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{bmatrix}$

two matrix equally same means they are element wise equal.by this we can easily calculate, D = $\begin{bmatrix} b_{11} & b_{12} \\ b_{21} - b_{11} & b_{22} - b_{12} \end{bmatrix}$
by

### 1 comment

in 3rd line why have you not written D after substituting value of C ?
Let |1 0| is X.

|1 1|

C = AX

AB = I, A = 1/B

C = (1/B)*X

CD = (1/B)*XD = 1*I XD = B. // left multiplication by B

D = (1/X)*B. // left multiplication by 1/X

(1/X)*B = |1 -1| * |b11 b12|

|0 1|     |b21 b22|

= |(b11-b21) b12-b22|

| b21 b22 |

= D