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Get the inverse matrix:
    Enter an invertible matrix, with each element separated by a comma and each row ending with a semicolon.
    Note that mathematical functions and variables are not supported.
    Current location:Linear algebra >Inverse matrix >History of inverse matrices >Answer

$$\begin{aligned}&\\ \color{black}{Calcu}&\color{black}{late\ the\ inverse\ matrix\ of\ } \ \ \begin{pmatrix} &1\ &65\ &2\ \\ &4\ &84\ &3\ \\ &34\ &7\ &89\ \end{pmatrix}\color{black}{\ .}\\ \\Solu&tion:\\ &\begin{pmatrix} &1\ &65\ &2\ \\ &4\ &84\ &3\ \\ &34\ &7\ &89\ \end{pmatrix}\\\\&\color{grey}{Using\ the\ elementary\ transformation\ of\ the\ matrix\ to\ find\ the\ inverse\ matrix:}\\&\left (\begin{array} {cccc | ccc} &1\ &65\ &2\ &1\ &0\ &0\ \\ &4\ &84\ &3\ &0\ &1\ &0\ \\ &34\ &7\ &89\ &0\ &0\ &1\ \\\end{array} \right )\\\\&\color{grey}{Transfprming\ a\ known\ matrix\ into\ an\ upper\ triangular\ matrix :}\\\\->\ \ &\left (\begin{array} {cccc | ccc} &1\ &65\ &2\ &1\ &0\ &0\ \\ &0\ &-176\ &-5\ &-4\ &1\ &0\ \\ &0\ &-2203\ &21\ &-34\ &0\ &1\ \\\end{array} \right )\\\\->\ \ &\left (\begin{array} {cccc | ccc} &1\ &65\ &2\ &1\ &0\ &0\ \\ &0\ &-176\ &-5\ &-4\ &1\ &0\ \\ &0\ &0\ &\frac{14711}{176}\ &\frac{707}{44}\ &-\frac{2203}{176}\ &1\ \\\end{array} \right )\\\\&\color{grey}{Convert\ elements\ above\ the\ diagonal\ to\ 0}\\\\->\ \ &\left (\begin{array} {cccc | ccc} &1\ &65\ &0\ &\frac{9055}{14711}\ &\frac{4406}{14711}\ &-\frac{352}{14711}\ \\ &0\ &-176\ &0\ &-\frac{44704}{14711}\ &\frac{3696}{14711}\ &\frac{880}{14711}\ \\ &0\ &0\ &\frac{14711}{176}\ &\frac{707}{44}\ &-\frac{2203}{176}\ &1\ \\\end{array} \right )\\\\->\ \ &\left (\begin{array} {cccc | ccc} &1\ &0\ &0\ &-\frac{7455}{14711}\ &\frac{5771}{14711}\ &-\frac{27}{14711}\ \\ &0\ &-176\ &0\ &-\frac{44704}{14711}\ &\frac{3696}{14711}\ &\frac{880}{14711}\ \\ &0\ &0\ &\frac{14711}{176}\ &\frac{707}{44}\ &-\frac{2203}{176}\ &1\ \\\end{array} \right )\\\\&\color{grey}{Convert\ elements\ on\ the\ main\ diagonal\ to\ 1}\\\\->\ \ &\left (\begin{array} {cccc | ccc} &1\ &0\ &0\ &-\frac{7455}{14711}\ &\frac{5771}{14711}\ &-\frac{27}{14711}\ \\ &0\ &1\ &0\ &\frac{254}{14711}\ &-\frac{21}{14711}\ &-\frac{5}{14711}\ \\ &0\ &0\ &\frac{14711}{176}\ &\frac{707}{44}\ &-\frac{2203}{176}\ &1\ \\\end{array} \right )\\\\->\ \ &\left (\begin{array} {cccc | ccc} &1\ &0\ &0\ &-\frac{7455}{14711}\ &\frac{5771}{14711}\ &-\frac{27}{14711}\ \\ &0\ &1\ &0\ &\frac{254}{14711}\ &-\frac{21}{14711}\ &-\frac{5}{14711}\ \\ &0\ &0\ &1\ &\frac{2828}{14711}\ &-\frac{2203}{14711}\ &\frac{176}{14711}\ \\\end{array} \right )\\\\&\color{grey}{The\ inverse\ matrix\ obtained\ is\ : }\\&\begin{pmatrix} &-\frac{7455}{14711}\ &\frac{5771}{14711}\ &-\frac{27}{14711}\ \\ &\frac{254}{14711}\ &-\frac{21}{14711}\ &-\frac{5}{14711}\ \\ &\frac{2828}{14711}\ &-\frac{2203}{14711}\ &\frac{176}{14711}\ \end{pmatrix}\end{aligned}$$

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Elementary transformations of matrices:


Definition:Applying the following three transformations to the rows (columns) of a matrix becomes the elementary transformation of the matrix
(1) Swap the positions of two rows (columns) in a matrix;
(2) Using non-zero constants λ Multiply a certain row (column) of a matrix;
(3) Convert a row (column) of a matrix γ Multiply to another row (column) of the matrix.



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