Equations
Unfold
Math OP
Unfold
Linear algebra
Fold
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} &4\ &1\ &5\ \\ &2\ &3\ &4\ \\ &3\ &4\ &5\ \end{pmatrix}\color{black}{\ .}\\ \\Solu&tion:\\ &\begin{pmatrix} &4\ &1\ &5\ \\ &2\ &3\ &4\ \\ &3\ &4\ &5\ \end{pmatrix}\\\\&\color{grey}{Using\ the\ elementary\ transformation\ of\ the\ matrix\ to\ find\ the\ inverse\ matrix:}\\&\left (\begin{array} {cccc | ccc} &4\ &1\ &5\ &1\ &0\ &0\ \\ &2\ &3\ &4\ &0\ &1\ &0\ \\ &3\ &4\ &5\ &0\ &0\ &1\ \\\end{array} \right )\\\\&\color{grey}{Transfprming\ a\ known\ matrix\ into\ an\ upper\ triangular\ matrix :}\\\\->\ \ &\left (\begin{array} {cccc | ccc} &4\ &1\ &5\ &1\ &0\ &0\ \\ &0\ &\frac{5}{2}\ &\frac{3}{2}\ &-\frac{1}{2}\ &1\ &0\ \\ &0\ &\frac{13}{4}\ &\frac{5}{4}\ &-\frac{3}{4}\ &0\ &1\ \\\end{array} \right )\\\\->\ \ &\left (\begin{array} {cccc | ccc} &4\ &1\ &5\ &1\ &0\ &0\ \\ &0\ &\frac{5}{2}\ &\frac{3}{2}\ &-\frac{1}{2}\ &1\ &0\ \\ &0\ &0\ &-\frac{7}{10}\ &-\frac{1}{10}\ &-\frac{13}{10}\ &1\ \\\end{array} \right )\\\\&\color{grey}{Convert\ elements\ above\ the\ diagonal\ to\ 0}\\\\->\ \ &\left (\begin{array} {cccc | ccc} &4\ &1\ &0\ &\frac{2}{7}\ &-\frac{65}{7}\ &\frac{50}{7}\ \\ &0\ &\frac{5}{2}\ &0\ &-\frac{5}{7}\ &-\frac{25}{14}\ &\frac{15}{7}\ \\ &0\ &0\ &-\frac{7}{10}\ &-\frac{1}{10}\ &-\frac{13}{10}\ &1\ \\\end{array} \right )\\\\->\ \ &\left (\begin{array} {cccc | ccc} &4\ &0\ &0\ &\frac{4}{7}\ &-\frac{60}{7}\ &\frac{44}{7}\ \\ &0\ &\frac{5}{2}\ &0\ &-\frac{5}{7}\ &-\frac{25}{14}\ &\frac{15}{7}\ \\ &0\ &0\ &-\frac{7}{10}\ &-\frac{1}{10}\ &-\frac{13}{10}\ &1\ \\\end{array} \right )\\\\&\color{grey}{Convert\ elements\ on\ the\ main\ diagonal\ to\ 1}\\\\->\ \ &\left (\begin{array} {cccc | ccc} &1\ &0\ &0\ &\frac{1}{7}\ &-\frac{15}{7}\ &\frac{11}{7}\ \\ &0\ &\frac{5}{2}\ &0\ &-\frac{5}{7}\ &-\frac{25}{14}\ &\frac{15}{7}\ \\ &0\ &0\ &-\frac{7}{10}\ &-\frac{1}{10}\ &-\frac{13}{10}\ &1\ \\\end{array} \right )\\\\->\ \ &\left (\begin{array} {cccc | ccc} &1\ &0\ &0\ &\frac{1}{7}\ &-\frac{15}{7}\ &\frac{11}{7}\ \\ &0\ &1\ &0\ &-\frac{2}{7}\ &-\frac{5}{7}\ &\frac{6}{7}\ \\ &0\ &0\ &-\frac{7}{10}\ &-\frac{1}{10}\ &-\frac{13}{10}\ &1\ \\\end{array} \right )\\\\->\ \ &\left (\begin{array} {cccc | ccc} &1\ &0\ &0\ &\frac{1}{7}\ &-\frac{15}{7}\ &\frac{11}{7}\ \\ &0\ &1\ &0\ &-\frac{2}{7}\ &-\frac{5}{7}\ &\frac{6}{7}\ \\ &0\ &0\ &1\ &\frac{1}{7}\ &\frac{13}{7}\ &-\frac{10}{7}\ \\\end{array} \right )\\\\&\color{grey}{The\ inverse\ matrix\ obtained\ is\ : }\\&\begin{pmatrix} &\frac{1}{7}\ &-\frac{15}{7}\ &\frac{11}{7}\ \\ &-\frac{2}{7}\ &-\frac{5}{7}\ &\frac{6}{7}\ \\ &\frac{1}{7}\ &\frac{13}{7}\ &-\frac{10}{7}\ \end{pmatrix}\end{aligned}$$
你的问题在这里没有得到解决？请到 热门难题 里面看看吧！
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.