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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} &1\ &4\ &2\ &8\ &5\ &7\ \\ &2\ &8\ &5\ &7\ &1\ &4\ \\ &4\ &2\ &8\ &5\ &7\ &1\ \\ &5\ &7\ &1\ &4\ &2\ &8\ \\ &7\ &1\ &4\ &2\ &8\ &5\ \\ &8\ &5\ &7\ &1\ &4\ &2\ \end{pmatrix}\color{black}{\ .}\\ \\Solu&tion:\\ &\begin{pmatrix} &1\ &4\ &2\ &8\ &5\ &7\ \\ &2\ &8\ &5\ &7\ &1\ &4\ \\ &4\ &2\ &8\ &5\ &7\ &1\ \\ &5\ &7\ &1\ &4\ &2\ &8\ \\ &7\ &1\ &4\ &2\ &8\ &5\ \\ &8\ &5\ &7\ &1\ &4\ &2\ \end{pmatrix}\\\\&\color{grey}{Using\ the\ elementary\ transformation\ of\ the\ matrix\ to\ find\ the\ inverse\ matrix:}\\&\left (\begin{array} {ccccccc | cccccc} &1\ &4\ &2\ &8\ &5\ &7\ &1\ &0\ &0\ &0\ &0\ &0\ \\ &2\ &8\ &5\ &7\ &1\ &4\ &0\ &1\ &0\ &0\ &0\ &0\ \\ &4\ &2\ &8\ &5\ &7\ &1\ &0\ &0\ &1\ &0\ &0\ &0\ \\ &5\ &7\ &1\ &4\ &2\ &8\ &0\ &0\ &0\ &1\ &0\ &0\ \\ &7\ &1\ &4\ &2\ &8\ &5\ &0\ &0\ &0\ &0\ &1\ &0\ \\ &8\ &5\ &7\ &1\ &4\ &2\ &0\ &0\ &0\ &0\ &0\ &1\ \\\end{array} \right )\\\\&\color{grey}{Transfprming\ a\ known\ matrix\ into\ an\ upper\ triangular\ matrix :}\\\\->\ \ &\left (\begin{array} {ccccccc | cccccc} &1\ &4\ &2\ &8\ &5\ &7\ &1\ &0\ &0\ &0\ &0\ &0\ \\ &0\ &0\ &1\ &-9\ &-9\ &-10\ &-2\ &1\ &0\ &0\ &0\ &0\ \\ &0\ &-14\ &0\ &-27\ &-13\ &-27\ &-4\ &0\ &1\ &0\ &0\ &0\ \\ &0\ &-13\ &-9\ &-36\ &-23\ &-27\ &-5\ &0\ &0\ &1\ &0\ &0\ \\ &0\ &-27\ &-10\ &-54\ &-27\ &-44\ &-7\ &0\ &0\ &0\ &1\ &0\ \\ &0\ &-27\ &-9\ &-63\ &-36\ &-54\ &-8\ &0\ &0\ &0\ &0\ &1\ \\\end{array} \right )\\\\->\ \ &\left (\begin{array} {ccccccc | cccccc} &1\ &4\ &2\ &8\ &5\ &7\ &1\ &0\ &0\ &0\ &0\ &0\ \\ &0\ &-14\ &0\ &-27\ &-13\ &-27\ &-4\ &0\ &1\ &0\ &0\ &0\ \\ &0\ &0\ &1\ &-9\ &-9\ &-10\ &-2\ &1\ &0\ &0\ &0\ &0\ \\ &0\ &0\ &-9\ &-\frac{153}{14}\ &-\frac{153}{14}\ &-\frac{27}{14}\ &-\frac{9}{7}\ &0\ &-\frac{13}{14}\ &1\ &0\ &0\ \\ &0\ &0\ &-10\ &-\frac{27}{14}\ &-\frac{27}{14}\ &\frac{113}{14}\ &\frac{5}{7}\ &0\ &-\frac{27}{14}\ &0\ &1\ &0\ \\ &0\ &0\ &-9\ &-\frac{153}{14}\ &-\frac{153}{14}\ &-\frac{27}{14}\ &-\frac{2}{7}\ &0\ &-\frac{27}{14}\ &0\ &0\ &1\ \\\end{array} \right )\\\\->\ \ &\left (\begin{array} {ccccccc | cccccc} &1\ &4\ &2\ &8\ &5\ &7\ &1\ &0\ &0\ &0\ &0\ &0\ \\ &0\ &-14\ &0\ &-27\ &-13\ &-27\ &-4\ &0\ &1\ &0\ &0\ &0\ \\ &0\ &0\ &1\ &-9\ &-9\ &-10\ &-2\ &1\ &0\ &0\ &0\ &0\ \\ &0\ &0\ &0\ &-\frac{1287}{14}\ &-\frac{1287}{14}\ &-\frac{1287}{14}\ &-\frac{135}{7}\ &9\ &-\frac{13}{14}\ &1\ &0\ &0\ \\ &0\ &0\ &0\ &-\frac{1287}{14}\ &-\frac{1287}{14}\ &-\frac{1287}{14}\ &-\frac{135}{7}\ &10\ &-\frac{27}{14}\ &0\ &1\ &0\ \\ &0\ &0\ &0\ &-\frac{1287}{14}\ &-\frac{1287}{14}\ &-\frac{1287}{14}\ &-\frac{128}{7}\ &9\ &-\frac{27}{14}\ &0\ &0\ &1\ \\\end{array} \right )\\\\->\ \ &\left (\begin{array} {ccccccc | cccccc} &1\ &4\ &2\ &8\ &5\ &7\ &1\ &0\ &0\ &0\ &0\ &0\ \\ &0\ &-14\ &0\ &-27\ &-13\ &-27\ &-4\ &0\ &1\ &0\ &0\ &0\ \\ &0\ &0\ &1\ &-9\ &-9\ &-10\ &-2\ &1\ &0\ &0\ &0\ &0\ \\ &0\ &0\ &0\ &-\frac{1287}{14}\ &-\frac{1287}{14}\ &-\frac{1287}{14}\ &-\frac{135}{7}\ &9\ &-\frac{13}{14}\ &1\ &0\ &0\ \\ &0\ &0\ &0\ &0\ &0\ &0\ &0\ &1\ &-1\ &-1\ &1\ &0\ \\ &0\ &0\ &0\ &0\ &0\ &0\ &1\ &0\ &-1\ &-1\ &0\ &1\ \\\end{array} \right )\\\ \ &\color{red}{This\ matrix\ is\ an\ irreversible\ matrix.}\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.