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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} &6\ &24\ &1\ \\ &13\ &16\ &10\ \\ &20\ &17\ &15\ \end{pmatrix}\color{black}{\ .}\\ \\Solu&tion:\\ &\begin{pmatrix} &6\ &24\ &1\ \\ &13\ &16\ &10\ \\ &20\ &17\ &15\ \end{pmatrix}\\\\&\color{grey}{Using\ the\ elementary\ transformation\ of\ the\ matrix\ to\ find\ the\ inverse\ matrix:}\\&\left (\begin{array} {cccc | ccc} &6\ &24\ &1\ &1\ &0\ &0\ \\ &13\ &16\ &10\ &0\ &1\ &0\ \\ &20\ &17\ &15\ &0\ &0\ &1\ \\\end{array} \right )\\\\&\color{grey}{Transfprming\ a\ known\ matrix\ into\ an\ upper\ triangular\ matrix :}\\\\->\ \ &\left (\begin{array} {cccc | ccc} &6\ &24\ &1\ &1\ &0\ &0\ \\ &0\ &-36\ &\frac{47}{6}\ &-\frac{13}{6}\ &1\ &0\ \\ &0\ &-63\ &\frac{35}{3}\ &-\frac{10}{3}\ &0\ &1\ \\\end{array} \right )\\\\->\ \ &\left (\begin{array} {cccc | ccc} &6\ &24\ &1\ &1\ &0\ &0\ \\ &0\ &-36\ &\frac{47}{6}\ &-\frac{13}{6}\ &1\ &0\ \\ &0\ &0\ &-\frac{49}{24}\ &\frac{11}{24}\ &-\frac{7}{4}\ &1\ \\\end{array} \right )\\\\&\color{grey}{Convert\ elements\ above\ the\ diagonal\ to\ 0}\\\\->\ \ &\left (\begin{array} {cccc | ccc} &6\ &24\ &0\ &\frac{60}{49}\ &-\frac{6}{7}\ &\frac{24}{49}\ \\ &0\ &-36\ &0\ &-\frac{20}{49}\ &-\frac{40}{7}\ &\frac{188}{49}\ \\ &0\ &0\ &-\frac{49}{24}\ &\frac{11}{24}\ &-\frac{7}{4}\ &1\ \\\end{array} \right )\\\\->\ \ &\left (\begin{array} {cccc | ccc} &6\ &0\ &0\ &\frac{20}{21}\ &-\frac{14}{3}\ &\frac{64}{21}\ \\ &0\ &-36\ &0\ &-\frac{20}{49}\ &-\frac{40}{7}\ &\frac{188}{49}\ \\ &0\ &0\ &-\frac{49}{24}\ &\frac{11}{24}\ &-\frac{7}{4}\ &1\ \\\end{array} \right )\\\\&\color{grey}{Convert\ elements\ on\ the\ main\ diagonal\ to\ 1}\\\\->\ \ &\left (\begin{array} {cccc | ccc} &1\ &0\ &0\ &\frac{10}{63}\ &-\frac{7}{9}\ &\frac{32}{63}\ \\ &0\ &-36\ &0\ &-\frac{20}{49}\ &-\frac{40}{7}\ &\frac{188}{49}\ \\ &0\ &0\ &-\frac{49}{24}\ &\frac{11}{24}\ &-\frac{7}{4}\ &1\ \\\end{array} \right )\\\\->\ \ &\left (\begin{array} {cccc | ccc} &1\ &0\ &0\ &\frac{10}{63}\ &-\frac{7}{9}\ &\frac{32}{63}\ \\ &0\ &1\ &0\ &\frac{5}{441}\ &\frac{10}{63}\ &-\frac{47}{441}\ \\ &0\ &0\ &-\frac{49}{24}\ &\frac{11}{24}\ &-\frac{7}{4}\ &1\ \\\end{array} \right )\\\\->\ \ &\left (\begin{array} {cccc | ccc} &1\ &0\ &0\ &\frac{10}{63}\ &-\frac{7}{9}\ &\frac{32}{63}\ \\ &0\ &1\ &0\ &\frac{5}{441}\ &\frac{10}{63}\ &-\frac{47}{441}\ \\ &0\ &0\ &1\ &-\frac{11}{49}\ &\frac{6}{7}\ &-\frac{24}{49}\ \\\end{array} \right )\\\\&\color{grey}{The\ inverse\ matrix\ obtained\ is\ : }\\&\begin{pmatrix} &\frac{10}{63}\ &-\frac{7}{9}\ &\frac{32}{63}\ \\ &\frac{5}{441}\ &\frac{10}{63}\ &-\frac{47}{441}\ \\ &-\frac{11}{49}\ &\frac{6}{7}\ &-\frac{24}{49}\ \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.