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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.
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} &7\ &8\ &30\ \\ &3\ &9\ &2\ \\ &8\ &5\ &3\ \end{pmatrix}\color{black}{\ .}\\ \\Solu&tion:\\ &\begin{pmatrix} &7\ &8\ &30\ \\ &3\ &9\ &2\ \\ &8\ &5\ &3\ \end{pmatrix}\\\\&\color{grey}{Using\ the\ elementary\ transformation\ of\ the\ matrix\ to\ find\ the\ inverse\ matrix:}\\&\left (\begin{array} {cccc | ccc} &7\ &8\ &30\ &1\ &0\ &0\ \\ &3\ &9\ &2\ &0\ &1\ &0\ \\ &8\ &5\ &3\ &0\ &0\ &1\ \\\end{array} \right )\\\\&\color{grey}{Transfprming\ a\ known\ matrix\ into\ an\ upper\ triangular\ matrix :}\\\\->\ \ &\left (\begin{array} {cccc | ccc} &7\ &8\ &30\ &1\ &0\ &0\ \\ &0\ &\frac{39}{7}\ &-\frac{76}{7}\ &-\frac{3}{7}\ &1\ &0\ \\ &0\ &-\frac{29}{7}\ &-\frac{219}{7}\ &-\frac{8}{7}\ &0\ &1\ \\\end{array} \right )\\\\->\ \ &\left (\begin{array} {cccc | ccc} &7\ &8\ &30\ &1\ &0\ &0\ \\ &0\ &\frac{39}{7}\ &-\frac{76}{7}\ &-\frac{3}{7}\ &1\ &0\ \\ &0\ &0\ &-\frac{1535}{39}\ &-\frac{19}{13}\ &\frac{29}{39}\ &1\ \\\end{array} \right )\\\\&\color{grey}{Convert\ elements\ above\ the\ diagonal\ to\ 0}\\\\->\ \ &\left (\begin{array} {cccc | ccc} &7\ &8\ &0\ &-\frac{35}{307}\ &\frac{174}{307}\ &\frac{234}{307}\ \\ &0\ &\frac{39}{7}\ &0\ &-\frac{39}{1535}\ &\frac{8541}{10745}\ &-\frac{2964}{10745}\ \\ &0\ &0\ &-\frac{1535}{39}\ &-\frac{19}{13}\ &\frac{29}{39}\ &1\ \\\end{array} \right )\\\\->\ \ &\left (\begin{array} {cccc | ccc} &7\ &0\ &0\ &-\frac{36533}{471245}\ &-\frac{882}{1535}\ &\frac{1778}{1535}\ \\ &0\ &\frac{39}{7}\ &0\ &-\frac{39}{1535}\ &\frac{8541}{10745}\ &-\frac{2964}{10745}\ \\ &0\ &0\ &-\frac{1535}{39}\ &-\frac{19}{13}\ &\frac{29}{39}\ &1\ \\\end{array} \right )\\\\&\color{grey}{Convert\ elements\ on\ the\ main\ diagonal\ to\ 1}\\\\->\ \ &\left (\begin{array} {cccc | ccc} &1\ &0\ &0\ &-\frac{5219}{471245}\ &-\frac{126}{1535}\ &\frac{254}{1535}\ \\ &0\ &\frac{39}{7}\ &0\ &-\frac{39}{1535}\ &\frac{8541}{10745}\ &-\frac{2964}{10745}\ \\ &0\ &0\ &-\frac{1535}{39}\ &-\frac{19}{13}\ &\frac{29}{39}\ &1\ \\\end{array} \right )\\\\->\ \ &\left (\begin{array} {cccc | ccc} &1\ &0\ &0\ &-\frac{5219}{471245}\ &-\frac{126}{1535}\ &\frac{254}{1535}\ \\ &0\ &1\ &0\ &-\frac{7}{1535}\ &\frac{219}{1535}\ &-\frac{76}{1535}\ \\ &0\ &0\ &-\frac{1535}{39}\ &-\frac{19}{13}\ &\frac{29}{39}\ &1\ \\\end{array} \right )\\\\->\ \ &\left (\begin{array} {cccc | ccc} &1\ &0\ &0\ &-\frac{5219}{471245}\ &-\frac{126}{1535}\ &\frac{254}{1535}\ \\ &0\ &1\ &0\ &-\frac{7}{1535}\ &\frac{219}{1535}\ &-\frac{76}{1535}\ \\ &0\ &0\ &1\ &\frac{57}{1535}\ &-\frac{29}{1535}\ &-\frac{39}{1535}\ \\\end{array} \right )\\\\&\color{grey}{The\ inverse\ matrix\ obtained\ is\ : }\\&\begin{pmatrix} &-\frac{5219}{471245}\ &-\frac{126}{1535}\ &\frac{254}{1535}\ \\ &-\frac{7}{1535}\ &\frac{219}{1535}\ &-\frac{76}{1535}\ \\ &\frac{57}{1535}\ &-\frac{29}{1535}\ &-\frac{39}{1535}\ \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.