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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} &3\ &5\ &7\ \\ &2\ &4\ &8\ \\ &6\ &3\ &7\ \end{pmatrix}\color{black}{\ .}\\ \\Solu&tion:\\ &\begin{pmatrix} &3\ &5\ &7\ \\ &2\ &4\ &8\ \\ &6\ &3\ &7\ \end{pmatrix}\\\\&\color{grey}{Using\ the\ elementary\ transformation\ of\ the\ matrix\ to\ find\ the\ inverse\ matrix:}\\&\left (\begin{array} {cccc | ccc} &3\ &5\ &7\ &1\ &0\ &0\ \\ &2\ &4\ &8\ &0\ &1\ &0\ \\ &6\ &3\ &7\ &0\ &0\ &1\ \\\end{array} \right )\\\\&\color{grey}{Transfprming\ a\ known\ matrix\ into\ an\ upper\ triangular\ matrix :}\\\\->\ \ &\left (\begin{array} {cccc | ccc} &3\ &5\ &7\ &1\ &0\ &0\ \\ &0\ &\frac{2}{3}\ &\frac{10}{3}\ &-\frac{2}{3}\ &1\ &0\ \\ &0\ &-7\ &-7\ &-2\ &0\ &1\ \\\end{array} \right )\\\\->\ \ &\left (\begin{array} {cccc | ccc} &3\ &5\ &7\ &1\ &0\ &0\ \\ &0\ &\frac{2}{3}\ &\frac{10}{3}\ &-\frac{2}{3}\ &1\ &0\ \\ &0\ &0\ &28\ &-9\ &\frac{21}{2}\ &1\ \\\end{array} \right )\\\\&\color{grey}{Convert\ elements\ above\ the\ diagonal\ to\ 0}\\\\->\ \ &\left (\begin{array} {cccc | ccc} &3\ &5\ &0\ &\frac{13}{4}\ &-\frac{21}{8}\ &-\frac{1}{4}\ \\ &0\ &\frac{2}{3}\ &0\ &\frac{17}{42}\ &-\frac{1}{4}\ &-\frac{5}{42}\ \\ &0\ &0\ &28\ &-9\ &\frac{21}{2}\ &1\ \\\end{array} \right )\\\\->\ \ &\left (\begin{array} {cccc | ccc} &3\ &0\ &0\ &\frac{3}{14}\ &-\frac{3}{4}\ &\frac{9}{14}\ \\ &0\ &\frac{2}{3}\ &0\ &\frac{17}{42}\ &-\frac{1}{4}\ &-\frac{5}{42}\ \\ &0\ &0\ &28\ &-9\ &\frac{21}{2}\ &1\ \\\end{array} \right )\\\\&\color{grey}{Convert\ elements\ on\ the\ main\ diagonal\ to\ 1}\\\\->\ \ &\left (\begin{array} {cccc | ccc} &1\ &0\ &0\ &\frac{1}{14}\ &-\frac{1}{4}\ &\frac{3}{14}\ \\ &0\ &\frac{2}{3}\ &0\ &\frac{17}{42}\ &-\frac{1}{4}\ &-\frac{5}{42}\ \\ &0\ &0\ &28\ &-9\ &\frac{21}{2}\ &1\ \\\end{array} \right )\\\\->\ \ &\left (\begin{array} {cccc | ccc} &1\ &0\ &0\ &\frac{1}{14}\ &-\frac{1}{4}\ &\frac{3}{14}\ \\ &0\ &1\ &0\ &\frac{17}{28}\ &-\frac{3}{8}\ &-\frac{5}{28}\ \\ &0\ &0\ &28\ &-9\ &\frac{21}{2}\ &1\ \\\end{array} \right )\\\\->\ \ &\left (\begin{array} {cccc | ccc} &1\ &0\ &0\ &\frac{1}{14}\ &-\frac{1}{4}\ &\frac{3}{14}\ \\ &0\ &1\ &0\ &\frac{17}{28}\ &-\frac{3}{8}\ &-\frac{5}{28}\ \\ &0\ &0\ &1\ &-\frac{9}{28}\ &\frac{3}{8}\ &\frac{1}{28}\ \\\end{array} \right )\\\\&\color{grey}{The\ inverse\ matrix\ obtained\ is\ : }\\&\begin{pmatrix} &\frac{1}{14}\ &-\frac{1}{4}\ &\frac{3}{14}\ \\ &\frac{17}{28}\ &-\frac{3}{8}\ &-\frac{5}{28}\ \\ &-\frac{9}{28}\ &\frac{3}{8}\ &\frac{1}{28}\ \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.