Definicija
uredi
Za dani matriki A {\displaystyle A\,} B {\displaystyle B\,}
A = [ 1 3 2 2 0 1 5 2 2 ] , B = [ 4 3 1 ] . {\displaystyle A={\begin{bmatrix}1&3&2\\2&0&1\\5&2&2\end{bmatrix}},\quad B={\begin{bmatrix}4\\3\\1\end{bmatrix}}.} se razširjena matrika piše kot
( A | B ) = [ 1 3 2 4 2 0 1 3 5 2 2 1 ] . {\displaystyle (A|B)=\left[{\begin{array}{ccc|c}1&3&2&4\\2&0&1&3\\5&2&2&1\end{array}}\right].}
Takšen način pisanja je zelo uporaben pri reševanju sistema linearnih enačb in pri določanju obratnih matrik .
Določanje obratne matrike
uredi
Imamo kvadratno matriko z razsežnostjo 2 × 2 {\displaystyle 2\times 2\,}
C = [ 1 3 − 5 0 ] . {\displaystyle C={\begin{bmatrix}1&3\\-5&0\end{bmatrix}}.} Za določanje obratne matrike moramo narediti razširjeno matriko C | I {\displaystyle C|I\,} I {\displaystyle I\,} enotska matrika z razsežnostjo 2 × 2 {\displaystyle 2\times 2\,} C − 1 {\displaystyle C^{-1}\,} C {\displaystyle C\,} ( C | I ) = [ 1 3 1 0 − 5 0 0 1 ] {\displaystyle (C|I)=\left[{\begin{array}{cc|cc}1&3&1&0\\-5&0&0&1\end{array}}\right]}
( I | C − 1 ) = [ 1 0 0 − 1 5 0 1 1 3 1 15 ] {\displaystyle (I|C^{-1})=\left[{\begin{array}{cc|cc}1&0&0&-{\frac {1}{5}}\\0&1&{\frac {1}{3}}&{\frac {1}{15}}\end{array}}\right]}
Reševanje sistema linearnih enačb
uredi
Dane imamo tri linearne enačbe
x 1 + 2 x 2 + 3 x 3 = 0 3 x 1 + 4 x 2 + 7 x 3 = 2 6 x 1 + 5 x 2 + 9 x 3 = 11 {\displaystyle {\begin{aligned}x_{1}+2x_{2}+3x_{3}&=0\\3x_{1}+4x_{2}+7x_{3}&=2\\6x_{1}+5x_{2}+9x_{3}&=11\end{aligned}}} To pa odgovarja naslednjim matrikam
A = [ 1 2 3 3 4 7 6 5 9 ] , B = [ 0 2 11 ] . {\displaystyle A={\begin{bmatrix}1&2&3\\3&4&7\\6&5&9\end{bmatrix}},\quad B={\begin{bmatrix}0\\2\\11\end{bmatrix}}.} ( A | B ) = [ 1 2 3 0 3 4 7 2 6 5 9 11 ] , {\displaystyle (A|B)=\left[{\begin{array}{ccc|c}1&2&3&0\\3&4&7&2\\6&5&9&11\end{array}}\right],} ali
( A | B ) = [ 1 0 0 4 0 1 0 1 0 0 1 − 2 ] . {\displaystyle (A|B)=\left[{\begin{array}{ccc|c}1&0&0&4\\0&1&0&1\\0&0&1&-2\\\end{array}}\right].} To pomeni, da so rešitve zgornjega sistema linearnih enačb
x 1 = 4 {\displaystyle x_{1}=4\,} x 2 = 1 {\displaystyle x_{2}=1\,} x 3 = − 2 {\displaystyle x_{3}=-2\,} Zunanje povezave
uredi
![{\displaystyle (A|B)=\left[{\begin{array}{ccc|c}1&3&2&4\\2&0&1&3\\5&2&2&1\end{array}}\right].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9f14272674b4b17491481e7ae9bcfcbc168ccb58)
![{\displaystyle (C|I)=\left[{\begin{array}{cc|cc}1&3&1&0\\-5&0&0&1\end{array}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8f5ad0f062318891216bc045d12b5fa019c00a8d)
![{\displaystyle (I|C^{-1})=\left[{\begin{array}{cc|cc}1&0&0&-{\frac {1}{5}}\\0&1&{\frac {1}{3}}&{\frac {1}{15}}\end{array}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/59be87e78725b558b9c1baf01edc097e9cb7ed7f)
![{\displaystyle (A|B)=\left[{\begin{array}{ccc|c}1&2&3&0\\3&4&7&2\\6&5&9&11\end{array}}\right],}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7d824a756a96bc77970c21efb74a2f82de1bbe9a)
![{\displaystyle (A|B)=\left[{\begin{array}{ccc|c}1&0&0&4\\0&1&0&1\\0&0&1&-2\\\end{array}}\right].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b3c8beb24663ffb74a5d6fcb7d111c218e51f3cb)
