Równianie macierzowe

[gosia19]

\(\displaystyle{ \left[\begin{array}{cc}3&1\\-1&5\end{array}\right]X=3X + ft[\begin{array}{cc}2&3\\1&-1\end{array}\right]}\)

Jak z tego wyznaczyc X?

[yevgienij]

Ja bym to zrobil tak:

\(\displaystyle{ ((3*5)-(-1*1))X=3X + (2*(-1)-3*1)}\)

\(\displaystyle{ (15+1)X=3X-5}\)

\(\displaystyle{ 13X=-5}\)
\(\displaystyle{ X=- \frac{5}{13}}\)

[gosia19]

Ale X to macierz.

[agulka1987]

\(\displaystyle{ \left[\begin{array}{cc}3&1\\-1&5\end{array}\right]X=3X + ft[\begin{array}{cc}2&3\\1&-1\end{array}\right]}\)

Jak z tego wyznaczyc X?

za X podstaw macierz \(\displaystyle{ \begin{bmatrix} a&b\\c&d\end{bmatrix}}\)



\(\displaystyle{ \begin{bmatrix} 3&1\\-1&5\end{bmatrix} \begin{bmatrix} a&b\\c&d\end{bmatrix} = 3\begin{bmatrix} a&b\\c&d\end{bmatrix}+\begin{bmatrix} 2&3\\1&-1\end{bmatrix}}\)

\(\displaystyle{ \begin{bmatrix} 3a+c&3b+d\\-a+5c&-b+5d\end{bmatrix} = \begin{bmatrix} 3a&3b\\3c&3d\end{bmatrix}+\begin{bmatrix} 2&3\\1&-1\end{bmatrix}}\)

\(\displaystyle{ \begin{bmatrix} 3a+c&3b+d\\-a+5c&-b+5d\end{bmatrix} = \begin{bmatrix} 3a+2&3b+3\\3c+1&3d-1\end{bmatrix}}\)

\(\displaystyle{ \begin{cases} 3a+c=3a+2 c=2\\3b+d=3b+3 d=3\\-a+5c=3c+1 a=3\\ -b+5d=3d-1 b=7\end{cases}}\)

X=\(\displaystyle{ \begin{bmatrix} 3&7\\2&3\end{bmatrix}}\)

[Qń]

Prościej:

\(\displaystyle{ \left[\begin{array}{cc}3&1\\-1&5\end{array}\right]X=3\left[\begin{array}{cc}1&0\\0&1\end{array}\right]X + ft[\begin{array}{cc}2&3\\1&-1\end{array}\right] \\ \\
ft( ft[\begin{array}{cc}3&1\\-1&5\end{array}\right] - 3 ft[\begin{array}{cc}1&0\\0&1\end{array}\right] \right) X = ft[\begin{array}{cc}2&3\\1&-1\end{array}\right] \\ \\
ft[\begin{array}{cc}0&1\\-1&2\end{array}\right] X = ft[\begin{array}{cc}2&3\\1&-1\end{array}\right] \\ \\
X = ft[\begin{array}{cc}0&1\\-1&2\end{array}\right]^{-1} ft[\begin{array}{cc}2&3\\1&-1\end{array}\right] =
ft[\begin{array}{cc}2&-1\\1&0\end{array}\right] ft[\begin{array}{cc}2&3\\1&-1\end{array}\right] =
ft[\begin{array}{cc}3&7\\2&3\end{array}\right]}\)

Q.

[gosia19]

Dzięki:)

[xiikzodz]

Przenoszac wyrazy z \(\displaystyle{ X}\) na lewo i dzielac stronami (lewostronnie) przez to, co stoi przy \(\displaystyle{ X}\) otrzymujemy:

\(\displaystyle{ X=\left(\begin{pmatrix}3&1\\-1&5\end{pmatrix}-\begin{pmatrix}3&0\\0&3\end{pmatrix}\right)^{-1}\begin{pmatrix}2&3\\1&-1\end{pmatrix}}\)

Skad:

\(\displaystyle{ X=\begin{pmatrix}0&1\\-1&2\end{pmatrix}^{-1}\begin{pmatrix}2&3\\1&-1\end{pmatrix}
=\begin{pmatrix}2&-1\\1&0\end{pmatrix}\begin{pmatrix}2&3\\1&-1\end{pmatrix}=
\begin{pmatrix}3&7\\2&3\end{pmatrix}}\)