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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\ &5\ &4\ &7\ \\ &6\ &3\ &9\ &8\ \\ &4\ &4\ &7\ &5\ \\ &5\ &1\ &8\ &4\ \end{pmatrix}\color{black}{\ .}\\ \\Solu&tion:\\ &\begin{pmatrix} &1\ &5\ &4\ &7\ \\ &6\ &3\ &9\ &8\ \\ &4\ &4\ &7\ &5\ \\ &5\ &1\ &8\ &4\ \end{pmatrix}\\\\&\color{grey}{Using\ the\ elementary\ transformation\ of\ the\ matrix\ to\ find\ the\ inverse\ matrix:}\\&\left (\begin{array} {ccccc | cccc} &1\ &5\ &4\ &7\ &1\ &0\ &0\ &0\ \\ &6\ &3\ &9\ &8\ &0\ &1\ &0\ &0\ \\ &4\ &4\ &7\ &5\ &0\ &0\ &1\ &0\ \\ &5\ &1\ &8\ &4\ &0\ &0\ &0\ &1\ \\\end{array} \right )\\\\&\color{grey}{Transfprming\ a\ known\ matrix\ into\ an\ upper\ triangular\ matrix :}\\\\->\ \ &\left (\begin{array} {ccccc | cccc} &1\ &5\ &4\ &7\ &1\ &0\ &0\ &0\ \\ &0\ &-27\ &-15\ &-34\ &-6\ &1\ &0\ &0\ \\ &0\ &-16\ &-9\ &-23\ &-4\ &0\ &1\ &0\ \\ &0\ &-24\ &-12\ &-31\ &-5\ &0\ &0\ &1\ \\\end{array} \right )\\\\->\ \ &\left (\begin{array} {ccccc | cccc} &1\ &5\ &4\ &7\ &1\ &0\ &0\ &0\ \\ &0\ &-27\ &-15\ &-34\ &-6\ &1\ &0\ &0\ \\ &0\ &0\ &-\frac{1}{9}\ &-\frac{77}{27}\ &-\frac{4}{9}\ &-\frac{16}{27}\ &1\ &0\ \\ &0\ &0\ &\frac{4}{3}\ &-\frac{7}{9}\ &\frac{1}{3}\ &-\frac{8}{9}\ &0\ &1\ \\\end{array} \right )\\\\->\ \ &\left (\begin{array} {ccccc | cccc} &1\ &5\ &4\ &7\ &1\ &0\ &0\ &0\ \\ &0\ &-27\ &-15\ &-34\ &-6\ &1\ &0\ &0\ \\ &0\ &0\ &-\frac{1}{9}\ &-\frac{77}{27}\ &-\frac{4}{9}\ &-\frac{16}{27}\ &1\ &0\ \\ &0\ &0\ &0\ &-35\ &-5\ &-8\ &12\ &1\ \\\end{array} \right )\\\\&\color{grey}{Convert\ elements\ above\ the\ diagonal\ to\ 0}\\\\->\ \ &\left (\begin{array} {ccccc | cccc} &1\ &5\ &4\ &0\ &0\ &-\frac{8}{5}\ &\frac{12}{5}\ &\frac{1}{5}\ \\ &0\ &-27\ &-15\ &0\ &-\frac{8}{7}\ &\frac{307}{35}\ &-\frac{408}{35}\ &-\frac{34}{35}\ \\ &0\ &0\ &-\frac{1}{9}\ &0\ &-\frac{1}{27}\ &\frac{8}{135}\ &\frac{1}{45}\ &-\frac{11}{135}\ \\ &0\ &0\ &0\ &-35\ &-5\ &-8\ &12\ &1\ \\\end{array} \right )\\\\->\ \ &\left (\begin{array} {ccccc | cccc} &1\ &5\ &0\ &0\ &-\frac{4}{3}\ &\frac{8}{15}\ &\frac{16}{5}\ &-\frac{41}{15}\ \\ &0\ &-27\ &0\ &0\ &\frac{27}{7}\ &\frac{27}{35}\ &-\frac{513}{35}\ &\frac{351}{35}\ \\ &0\ &0\ &-\frac{1}{9}\ &0\ &-\frac{1}{27}\ &\frac{8}{135}\ &\frac{1}{45}\ &-\frac{11}{135}\ \\ &0\ &0\ &0\ &-35\ &-5\ &-8\ &12\ &1\ \\\end{array} \right )\\\\->\ \ &\left (\begin{array} {ccccc | cccc} &1\ &0\ &0\ &0\ &-\frac{13}{21}\ &\frac{71}{105}\ &\frac{17}{35}\ &-\frac{92}{105}\ \\ &0\ &-27\ &0\ &0\ &\frac{27}{7}\ &\frac{27}{35}\ &-\frac{513}{35}\ &\frac{351}{35}\ \\ &0\ &0\ &-\frac{1}{9}\ &0\ &-\frac{1}{27}\ &\frac{8}{135}\ &\frac{1}{45}\ &-\frac{11}{135}\ \\ &0\ &0\ &0\ &-35\ &-5\ &-8\ &12\ &1\ \\\end{array} \right )\\\\&\color{grey}{Convert\ elements\ on\ the\ main\ diagonal\ to\ 1}\\\\->\ \ &\left (\begin{array} {ccccc | cccc} &1\ &0\ &0\ &0\ &-\frac{13}{21}\ &\frac{71}{105}\ &\frac{17}{35}\ &-\frac{92}{105}\ \\ &0\ &1\ &0\ &0\ &-\frac{1}{7}\ &-\frac{1}{35}\ &\frac{19}{35}\ &-\frac{13}{35}\ \\ &0\ &0\ &-\frac{1}{9}\ &0\ &-\frac{1}{27}\ &\frac{8}{135}\ &\frac{1}{45}\ &-\frac{11}{135}\ \\ &0\ &0\ &0\ &-35\ &-5\ &-8\ &12\ &1\ \\\end{array} \right )\\\\->\ \ &\left (\begin{array} {ccccc | cccc} &1\ &0\ &0\ &0\ &-\frac{13}{21}\ &\frac{71}{105}\ &\frac{17}{35}\ &-\frac{92}{105}\ \\ &0\ &1\ &0\ &0\ &-\frac{1}{7}\ &-\frac{1}{35}\ &\frac{19}{35}\ &-\frac{13}{35}\ \\ &0\ &0\ &1\ &0\ &\frac{1}{3}\ &-\frac{8}{15}\ &-\frac{1}{5}\ &\frac{11}{15}\ \\ &0\ &0\ &0\ &-35\ &-5\ &-8\ &12\ &1\ \\\end{array} \right )\\\\->\ \ &\left (\begin{array} {ccccc | cccc} &1\ &0\ &0\ &0\ &-\frac{13}{21}\ &\frac{71}{105}\ &\frac{17}{35}\ &-\frac{92}{105}\ \\ &0\ &1\ &0\ &0\ &-\frac{1}{7}\ &-\frac{1}{35}\ &\frac{19}{35}\ &-\frac{13}{35}\ \\ &0\ &0\ &1\ &0\ &\frac{1}{3}\ &-\frac{8}{15}\ &-\frac{1}{5}\ &\frac{11}{15}\ \\ &0\ &0\ &0\ &1\ &\frac{1}{7}\ &\frac{8}{35}\ &-\frac{12}{35}\ &-\frac{1}{35}\ \\\end{array} \right )\\\\&\color{grey}{The\ inverse\ matrix\ obtained\ is\ : }\\&\begin{pmatrix} &-\frac{13}{21}\ &\frac{71}{105}\ &\frac{17}{35}\ &-\frac{92}{105}\ \\ &-\frac{1}{7}\ &-\frac{1}{35}\ &\frac{19}{35}\ &-\frac{13}{35}\ \\ &\frac{1}{3}\ &-\frac{8}{15}\ &-\frac{1}{5}\ &\frac{11}{15}\ \\ &\frac{1}{7}\ &\frac{8}{35}\ &-\frac{12}{35}\ &-\frac{1}{35}\ \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.