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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\ &2\ &5\ \\ &2\ &5\ &7\ \\ &7\ &6\ &3\ \end{pmatrix}\color{black}{\ .}\\ \\Solu&tion:\\ &\begin{pmatrix} &3\ &2\ &5\ \\ &2\ &5\ &7\ \\ &7\ &6\ &3\ \end{pmatrix}\\\\&\color{grey}{Using\ the\ elementary\ transformation\ of\ the\ matrix\ to\ find\ the\ inverse\ matrix:}\\&\left (\begin{array} {cccc | ccc} &3\ &2\ &5\ &1\ &0\ &0\ \\ &2\ &5\ &7\ &0\ &1\ &0\ \\ &7\ &6\ &3\ &0\ &0\ &1\ \\\end{array} \right )\\\\&\color{grey}{Transfprming\ a\ known\ matrix\ into\ an\ upper\ triangular\ matrix :}\\\\->\ \ &\left (\begin{array} {cccc | ccc} &3\ &2\ &5\ &1\ &0\ &0\ \\ &0\ &\frac{11}{3}\ &\frac{11}{3}\ &-\frac{2}{3}\ &1\ &0\ \\ &0\ &\frac{4}{3}\ &-\frac{26}{3}\ &-\frac{7}{3}\ &0\ &1\ \\\end{array} \right )\\\\->\ \ &\left (\begin{array} {cccc | ccc} &3\ &2\ &5\ &1\ &0\ &0\ \\ &0\ &\frac{11}{3}\ &\frac{11}{3}\ &-\frac{2}{3}\ &1\ &0\ \\ &0\ &0\ &-10\ &-\frac{23}{11}\ &-\frac{4}{11}\ &1\ \\\end{array} \right )\\\\&\color{grey}{Convert\ elements\ above\ the\ diagonal\ to\ 0}\\\\->\ \ &\left (\begin{array} {cccc | ccc} &3\ &2\ &0\ &-\frac{1}{22}\ &-\frac{2}{11}\ &\frac{1}{2}\ \\ &0\ &\frac{11}{3}\ &0\ &-\frac{43}{30}\ &\frac{13}{15}\ &\frac{11}{30}\ \\ &0\ &0\ &-10\ &-\frac{23}{11}\ &-\frac{4}{11}\ &1\ \\\end{array} \right )\\\\->\ \ &\left (\begin{array} {cccc | ccc} &3\ &0\ &0\ &\frac{81}{110}\ &-\frac{36}{55}\ &\frac{3}{10}\ \\ &0\ &\frac{11}{3}\ &0\ &-\frac{43}{30}\ &\frac{13}{15}\ &\frac{11}{30}\ \\ &0\ &0\ &-10\ &-\frac{23}{11}\ &-\frac{4}{11}\ &1\ \\\end{array} \right )\\\\&\color{grey}{Convert\ elements\ on\ the\ main\ diagonal\ to\ 1}\\\\->\ \ &\left (\begin{array} {cccc | ccc} &1\ &0\ &0\ &\frac{27}{110}\ &-\frac{12}{55}\ &\frac{1}{10}\ \\ &0\ &\frac{11}{3}\ &0\ &-\frac{43}{30}\ &\frac{13}{15}\ &\frac{11}{30}\ \\ &0\ &0\ &-10\ &-\frac{23}{11}\ &-\frac{4}{11}\ &1\ \\\end{array} \right )\\\\->\ \ &\left (\begin{array} {cccc | ccc} &1\ &0\ &0\ &\frac{27}{110}\ &-\frac{12}{55}\ &\frac{1}{10}\ \\ &0\ &1\ &0\ &-\frac{43}{110}\ &\frac{13}{55}\ &\frac{1}{10}\ \\ &0\ &0\ &-10\ &-\frac{23}{11}\ &-\frac{4}{11}\ &1\ \\\end{array} \right )\\\\->\ \ &\left (\begin{array} {cccc | ccc} &1\ &0\ &0\ &\frac{27}{110}\ &-\frac{12}{55}\ &\frac{1}{10}\ \\ &0\ &1\ &0\ &-\frac{43}{110}\ &\frac{13}{55}\ &\frac{1}{10}\ \\ &0\ &0\ &1\ &\frac{23}{110}\ &\frac{2}{55}\ &-\frac{1}{10}\ \\\end{array} \right )\\\\&\color{grey}{The\ inverse\ matrix\ obtained\ is\ : }\\&\begin{pmatrix} &\frac{27}{110}\ &-\frac{12}{55}\ &\frac{1}{10}\ \\ &-\frac{43}{110}\ &\frac{13}{55}\ &\frac{1}{10}\ \\ &\frac{23}{110}\ &\frac{2}{55}\ &-\frac{1}{10}\ \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.