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

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Kathleen
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in Linear Algebra
Oct 10, 2014

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13 votes

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 $B$.

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Best answer

$AB = I$, 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} $

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

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

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

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2 votes

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

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