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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\ &1\ &1\ &1\ \\ &1\ &2\ &-1\ &4\ \\ &2\ &-3\ &-1\ &-5\ \\ &3\ &1\ &2\ &11\ \end{pmatrix}\color{black}{\ .}\\ \\Solu&tion:\\ &\begin{pmatrix} &1\ &1\ &1\ &1\ \\ &1\ &2\ &-1\ &4\ \\ &2\ &-3\ &-1\ &-5\ \\ &3\ &1\ &2\ &11\ \end{pmatrix}\\\\&\color{grey}{Using\ the\ elementary\ transformation\ of\ the\ matrix\ to\ find\ the\ inverse\ matrix:}\\&\left (\begin{array} {ccccc | cccc} &1\ &1\ &1\ &1\ &1\ &0\ &0\ &0\ \\ &1\ &2\ &-1\ &4\ &0\ &1\ &0\ &0\ \\ &2\ &-3\ &-1\ &-5\ &0\ &0\ &1\ &0\ \\ &3\ &1\ &2\ &11\ &0\ &0\ &0\ &1\ \\\end{array} \right )\\\\&\color{grey}{Transfprming\ a\ known\ matrix\ into\ an\ upper\ triangular\ matrix :}\\\\->\ \ &\left (\begin{array} {ccccc | cccc} &1\ &1\ &1\ &1\ &1\ &0\ &0\ &0\ \\ &0\ &1\ &-2\ &3\ &-1\ &1\ &0\ &0\ \\ &0\ &-5\ &-3\ &-7\ &-2\ &0\ &1\ &0\ \\ &0\ &-2\ &-1\ &8\ &-3\ &0\ &0\ &1\ \\\end{array} \right )\\\\->\ \ &\left (\begin{array} {ccccc | cccc} &1\ &1\ &1\ &1\ &1\ &0\ &0\ &0\ \\ &0\ &1\ &-2\ &3\ &-1\ &1\ &0\ &0\ \\ &0\ &0\ &-13\ &8\ &-7\ &5\ &1\ &0\ \\ &0\ &0\ &-5\ &14\ &-5\ &2\ &0\ &1\ \\\end{array} \right )\\\\->\ \ &\left (\begin{array} {ccccc | cccc} &1\ &1\ &1\ &1\ &1\ &0\ &0\ &0\ \\ &0\ &1\ &-2\ &3\ &-1\ &1\ &0\ &0\ \\ &0\ &0\ &-13\ &8\ &-7\ &5\ &1\ &0\ \\ &0\ &0\ &0\ &\frac{142}{13}\ &-\frac{30}{13}\ &\frac{1}{13}\ &-\frac{5}{13}\ &1\ \\\end{array} \right )\\\\&\color{grey}{Convert\ elements\ above\ the\ diagonal\ to\ 0}\\\\->\ \ &\left (\begin{array} {ccccc | cccc} &1\ &1\ &1\ &0\ &\frac{86}{71}\ &-\frac{1}{142}\ &\frac{5}{142}\ &-\frac{13}{142}\ \\ &0\ &1\ &-2\ &0\ &-\frac{26}{71}\ &\frac{139}{142}\ &\frac{15}{142}\ &-\frac{39}{142}\ \\ &0\ &0\ &-13\ &0\ &-\frac{377}{71}\ &\frac{351}{71}\ &\frac{91}{71}\ &-\frac{52}{71}\ \\ &0\ &0\ &0\ &\frac{142}{13}\ &-\frac{30}{13}\ &\frac{1}{13}\ &-\frac{5}{13}\ &1\ \\\end{array} \right )\\\\->\ \ &\left (\begin{array} {ccccc | cccc} &1\ &1\ &0\ &0\ &\frac{57}{71}\ &\frac{3763}{10082}\ &\frac{1349}{10082}\ &-\frac{1491}{10082}\ \\ &0\ &1\ &0\ &0\ &\frac{32}{71}\ &\frac{2201}{10082}\ &-\frac{13}{142}\ &-\frac{1633}{10082}\ \\ &0\ &0\ &-13\ &0\ &-\frac{377}{71}\ &\frac{351}{71}\ &\frac{91}{71}\ &-\frac{52}{71}\ \\ &0\ &0\ &0\ &\frac{142}{13}\ &-\frac{30}{13}\ &\frac{1}{13}\ &-\frac{5}{13}\ &1\ \\\end{array} \right )\\\\->\ \ &\left (\begin{array} {ccccc | cccc} &1\ &0\ &0\ &0\ &\frac{25}{71}\ &\frac{11}{71}\ &\frac{1136}{5041}\ &\frac{1}{71}\ \\ &0\ &1\ &0\ &0\ &\frac{32}{71}\ &\frac{2201}{10082}\ &-\frac{13}{142}\ &-\frac{1633}{10082}\ \\ &0\ &0\ &-13\ &0\ &-\frac{377}{71}\ &\frac{351}{71}\ &\frac{91}{71}\ &-\frac{52}{71}\ \\ &0\ &0\ &0\ &\frac{142}{13}\ &-\frac{30}{13}\ &\frac{1}{13}\ &-\frac{5}{13}\ &1\ \\\end{array} \right )\\\\&\color{grey}{Convert\ elements\ on\ the\ main\ diagonal\ to\ 1}\\\\->\ \ &\left (\begin{array} {ccccc | cccc} &1\ &0\ &0\ &0\ &\frac{25}{71}\ &\frac{11}{71}\ &\frac{1136}{5041}\ &\frac{1}{71}\ \\ &0\ &1\ &0\ &0\ &\frac{32}{71}\ &\frac{2201}{10082}\ &-\frac{13}{142}\ &-\frac{1633}{10082}\ \\ &0\ &0\ &1\ &0\ &\frac{29}{71}\ &-\frac{27}{71}\ &-\frac{7}{71}\ &\frac{4}{71}\ \\ &0\ &0\ &0\ &\frac{142}{13}\ &-\frac{30}{13}\ &\frac{1}{13}\ &-\frac{5}{13}\ &1\ \\\end{array} \right )\\\\->\ \ &\left (\begin{array} {ccccc | cccc} &1\ &0\ &0\ &0\ &\frac{25}{71}\ &\frac{11}{71}\ &\frac{1136}{5041}\ &\frac{1}{71}\ \\ &0\ &1\ &0\ &0\ &\frac{32}{71}\ &\frac{2201}{10082}\ &-\frac{13}{142}\ &-\frac{1633}{10082}\ \\ &0\ &0\ &1\ &0\ &\frac{29}{71}\ &-\frac{27}{71}\ &-\frac{7}{71}\ &\frac{4}{71}\ \\ &0\ &0\ &0\ &1\ &-\frac{15}{71}\ &\frac{1}{142}\ &-\frac{5}{142}\ &\frac{13}{142}\ \\\end{array} \right )\\\\&\color{grey}{The\ inverse\ matrix\ obtained\ is\ : }\\&\begin{pmatrix} &\frac{25}{71}\ &\frac{11}{71}\ &\frac{1136}{5041}\ &\frac{1}{71}\ \\ &\frac{32}{71}\ &\frac{2201}{10082}\ &-\frac{13}{142}\ &-\frac{1633}{10082}\ \\ &\frac{29}{71}\ &-\frac{27}{71}\ &-\frac{7}{71}\ &\frac{4}{71}\ \\ &-\frac{15}{71}\ &\frac{1}{142}\ &-\frac{5}{142}\ &\frac{13}{142}\ \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.