Loser66
  • Loser66
\[a_{11}x_1+a_{12}x_2=b_1\\a_{21}x_1+a_{22}x_2=b_2\] Define \[\triangle = a_{11}a{22}-a_{12}a_{21}\\\triangle_1=a_{22}b_1-a_{12}b_2\\\triangle_2=a_{11}b_2-a_{12}b_1\] a) Show that the given system has a unique solution if and only if \(\triangle \neq 0\)and that unique solution in this case is \(x_1=\frac{\triangle_1}{\triangle}\) and \(x_2\frac{\triangle_2}{\triangle}\) b) If \(\triangle =0\), and \(a_{11}\neq 0\), determine the condition on \(\triangle_2\) that would guarantee that the system has i) no solution, ii) an infinite number of solution c) Interpret your results in term of intersections of straight lines Please help
Mathematics
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katieb
  • katieb
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ybarrap
  • ybarrap
Hey @Loser, is cramer's rule. \(\triangle\ne 0\ )means that there are no combinations of x's that make the b's zero other than when the x's are all zero (i.e. the trivial solution) This is the same thing as saying that the the columns that make up the a's are independent and therefore the system of equations has a unique solution. Conversely, if the matrix of a's is independent with a unique solution, the ranks of the column space of A (column vectors filled with the a's) is full and the null space is empty. Meaning that the deteminant is non zero.
Loser66
  • Loser66
is it not \(\triangle \) det of A?
ybarrap
  • ybarrap
yes

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ybarrap
  • ybarrap
\(\triangle \) is determinant.
Loser66
  • Loser66
I don't see the link
ybarrap
  • ybarrap
if you make one column a multiple of another column, it does not fill the space, it stays in 1 dimension. If the columns are independent, then any vector in 2 dimensions can be represented as a combination of these two vectors. Therefore, determinant equal to 0 indicates that the columns (or rows) are linear combinations of the other and not independent. Take for example a vector (a,b) and another vector (2a,2b), both these are dependent. Make these your a's above and take the determinant. The determinant will be zero.
Loser66
  • Loser66
I got it @ybarrap part a) how about part b?
ybarrap
  • ybarrap
No solution means determinant is zero. So, for determinant equation, set it equal to zero and solve for \(a_{11}\), then plug into \(\triangle_{2}\).
ybarrap
  • ybarrap
For infinite solution, we have one null space solution, other than the trial case. Set \(a_{12}=1\) and solve for \(a_{11}=1\) in terms of \(\triangle _2\) and plug into your determinant equation (after setting it equal to zero). You will than have your condition for \(\triangle _2\) to make solutions infinite.
ybarrap
  • ybarrap
*trivial not trial case. Trivial case is where the x's are zero.
ybarrap
  • ybarrap
The reason we set \(a_{12}=1\) is because we have one free variable, due to the fact that we have a nonempty null space.
Loser66
  • Loser66
I got you @ybarrap thanks a lot.

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