DETERMINANTS

[gaunt onyx]

This may sound dumb

Now say  I have 2 eqns: ax + by = d1 ; ux + vy = d2

putting them in determinant format:
                                    | a   b |    = delta.
                                    | u   v |

|d1  b|  = delta1.
|d2  v|

now delta1= x*delta
similarly delta2 = y*delta
now for parallel lines condition: 
delta1,delta2 !=0 but delta =0

but if delta =0 delta1 and delta2 =0 na

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[gaunt onyx]

Yea thanks bro

\text{Given the system:}
\begin{aligned}
ax + by &= d_1 \
ux + vy &= d_2
\end{aligned}

\text{Define:} \quad
\Delta =
\begin{vmatrix}
a & b \
u & v
\end{vmatrix}, \quad
\Delta_1 =
\begin{vmatrix}
d_1 & b \
d_2 & v
\end{vmatrix}, \quad
\Delta_2 =
\begin{vmatrix}
a & d_1 \
u & d_2
\end{vmatrix}

\text{By Cramer's Rule:} \quad
x = \frac{\Delta_1}{\Delta}, \quad y = \frac{\Delta_2}{\Delta}

\text{Hence:} \quad
\Delta_1 = x \cdot \Delta, \quad \Delta_2 = y \cdot \Delta

\textbf{Conditions:}
\begin{itemize}
\item If (\Delta \ne 0), the system has a unique solution.
\item If (\Delta = 0) and (\Delta_1 = \Delta_2 = 0), the system has infinitely many solutions (coincident lines).
\item If (\Delta = 0) but (\Delta_1 \ne 0) or (\Delta_2 \ne 0), the system is inconsistent (parallel lines).
\end{itemize}

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[delicate bane]

Well, seems good. What's the question, though?

[gaunt onyx]

screw that man

[gaunt onyx]

@delicate bane

[delicate bane]

Why would that be implied? Take the following system:
2x + 3y = 2
4x + 6y = 3
Then Δ = 0, but Δ1, Δ2 ≠ 0.

[gaunt onyx]

but delta*x = delta 1 right

[delicate bane]

Only if a solution exists, which doesn't in this case.

[gaunt onyx]

huh

i dont get u

wait

[delicate bane]

The formulas of the form x(i) Δ = Δ(i) only apply when x(i) actually exists.

Which it doesn't in case of an inconsistent system.

[gaunt onyx]

u mean xdelta ydelta and zdelta?

[delicate bane]

Well, yeah. I'm mostly talking about the general case with variables x(1), ..., x(n).

The logic is the same, after all.

[gaunt onyx]

oh yea makes sense.. cuz parallel lines have no soln and x = delta1/delta cannot come..

hmm good one

[delicate bane]

Well, to be fair, the form x(i) = Δ(i)/Δ also doesn't work for undetermined systems, since then you get 0/0.

But x(i) Δ = Δ(i) does work.

So, using Cramer's method directly only allows you to solve determined systems.

[gaunt onyx]

thx

+close

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