MHB 311.2.2.6 use inverse matrix to solve system of equations

karush
Gold Member
MHB
Messages
3,240
Reaction score
5
$\tiny{311.2.2.6}$
Use the inverse to solve the system
$\begin{array}{rrrrr}
7x_1&+3x_2&=-9\\
-2x_1&+x_2&=10
\end{array}$

$A^{-1}b=x$
$\begin{bmatrix}
\frac{1}{13}&-\frac{3}{13} \\ \\ \frac{2}{13}& \frac{7}{13} \end{bmatrix}
\begin{bmatrix}
-9\\ \\ 10
\end{bmatrix}
=
\begin{bmatrix}
-3\\ \\ 4
\end{bmatrix}$

the thing I could not get here without a calculator is $A^{-1}$
 
Physics news on Phys.org
Given [math]A = \left ( \begin{matrix} a & b \\ c & d \end{matrix} \right )[/math]

When all else fails:
[math]A^{-1} = \dfrac{1}{ |A| } \left ( \begin{matrix} d & -b \\ -c & a \end{matrix} \right )[/math]

I have this one memorized.

-Dan
 
Last edited by a moderator:
ok i think i had a and switched
 
The world of 2\times 2 complex matrices is very colorful. They form a Banach-algebra, they act on spinors, they contain the quaternions, SU(2), su(2), SL(2,\mathbb C), sl(2,\mathbb C). Furthermore, with the determinant as Euclidean or pseudo-Euclidean norm, isu(2) is a 3-dimensional Euclidean space, \mathbb RI\oplus isu(2) is a Minkowski space with signature (1,3), i\mathbb RI\oplus su(2) is a Minkowski space with signature (3,1), SU(2) is the double cover of SO(3), sl(2,\mathbb C) is the...

Similar threads

Replies
5
Views
2K
Replies
2
Views
1K
Replies
7
Views
1K
Replies
2
Views
1K
Replies
2
Views
3K
Replies
2
Views
2K