Krushnaraj Pandya said:
Homework Statement
Consider three equations in three variables, To check their consistency usually the adjoint of the matrix (from coefficients of x,y and z are found) and the inverse is found for the solution. I have heard of cramer's rule but it has a similar complexity.
These same kind of questions are repeated every year in the JEE which makes it very important for students like me to figure out something fast since there's no time to be calculating inverses and adjoints and risk mistakes as well.
As a sample- Here's a question from 2017 JEE Mains
Find the set of distinct values of b for which the following equations has no solution.
x+y+z=1; x+ay+z=1; ax+by+z=0
Homework Equations
Any pertaining to algebra
The Attempt at a Solution
I thought of some ways which might be faster but I don't know if they are foolproof-
1)Solving them pairwise- (easier for a solution- but I don't know how consistency can be checked)
2) Using 3d geometry, these are equations of planes. I'm not sure how exactly we can proceed using that though
3) As for the sample question, setting determinant=0 gives a=1 which means two of the equations are identical- x+y+z can't be 0 and 1 at the same time (I suppose but I'm not sure) so b is 1. Is that the right way to go about it?
This question is in the wrong forum: it should be in "precalculus" mathematics. However, I will answer it anyway.
The standard way is to use Gaussian elimination. So, use the first equation to solve for x in terms of y and z. Substitute that expression for x into the second and third equations. Now the last two equations contain y and z only, so are easier to examine. If need be, you can even use the new second equation to solve for one of y and z in terms of the other. Then substitute that expression into the third equation, and you will be left with a single equation in a single variable.
Solving ##(1): \; x+y+z=1## for ##x## gives ##(1a): \;x = 1 - y - z.## Substituting (1a) into ##(2): \; x + ay + z = 1## gives ##1-y-z+ay+z=1,## which simplifies to ##(2a): \; (a-1) y =0.## Substituting (1a) into ##(3):\; ax + by + z = 0## gives ##a -ay - az +by + z = 0,## which simplifies to ##(3a): \; (a-b) y +(a-1) z = a.##
So, our new equations are ## \; x+y+z = 1, \: (a-1) y = 0, \: (a-b)y +(a-1)z = a.## Now we can start to examine cases.
(a) If ##a \neq 1## the second equation forces ##y = 0.## Then, the third equation has solution ##z = a/(a-1).## We get ##x## from the first equation.
(b) If ##a = 1## the second equation allows ##y = \text{anything}.## You can then look further into consequences of the third equation (when you have a unique solution for ##z## or no solution at all for ##z##).
Note that instead of solving and substituting, we can obtain the same results by adding multiples of one equation onto the others, to eliminate a variable. All of that can be sped up and mechanised by looking at the detached-coefficient forms. The original equations are
$$\begin{array}{ccc|c}
x & y & z & \text{RHS} \\ \hline
1&1&1&1\\
1&a&1&1\\
a&b&1&0
\end{array}$$
To eliminate ##x## from the last two rows we use row-operations: (New Row 2) = (old row 2) - (Row 1) and (New Row 3) = (Old Row 3) - a×(Row 1). That gives
$$
\begin{array}{ccc|c}
x & y & z & \text{RHS} \\ \hline
1&1&1&1\\
0& a-1 & 0 & 0 \\
0 & b-a & 1-a & -a
\end{array}$$
We can abbreviate the row operations as ##R2 \leftarrow R2 - R1## and ##R3 \leftarrow R3 - a \times R1.##
All such row operations can be turned into matrix multiplications on the left, and that is essentially the way modern linear solving methods deal with such systems. Almost nobody uses determinants on large systems, unless they happen to have special structure. (In fact, the way most computer packages evaluate determinants is to first perform Gaussian elimination steps to obtain a so-called "LU decomposition", then compute the determinants of ##L## and ##U##, both of which are very easy.)
