Existence of solution of system of equations

Question

I have some doubt with the existence of solution of system of 3 linear equations.

Representing $$\begin{cases} a_1x+b_1y+c_1z=d_1 \\ a_2x+b_2y+c_2z=d_2 \\ a_3x+b_3y+c_3z=d_3 \end{cases} $$

by the matrix $AX = B$. where $A$ is $3\times 3$ coefficient matrix, $X$ is $3\times 1$ variable matrix and $B$ is $3\times 1$ matrix having $d_1,d_2,d_3$.

Solution is given by $X = A^{-1}B = \frac{adj(A)\times B}{det(A)}$

Now what happens when $A$ is singular matrix $(det(A) = 0)$ and $adj(A) B = O$ (Null matrix)?

All questions where this occurs which I have solved to date have infinitely many solutions?

However my textbook states there maybe no solution or infinite solutions. (And there is a similar result in Cramer's rule as well)

Is my textbook correct? And if it is, can anybody give an example of no solution case?

Answer 1

Yes , your textbook is correct.

$$A=\left[\begin{matrix}1&2&3\\ 2&4&6\\3&6&9\end{matrix}\right];B=\left[\begin{matrix}1\\2\\\color{red}4\end{matrix}\right]$$

This system has $det(A)=0$ as well as $adj(A)\times B=0 $ ..But no solutions!

BUT..

$$A=\left[\begin{matrix}1&2&3\\ 2&4&6\\3&6&9\end{matrix}\right];B=\left[\begin{matrix}1\\2\\\color{blue}3\end{matrix}\right]$$

This system has $det(A)=0$ as well as $adj(A)\times B=0 $ ..But infinitely many solutions!

Answer 2

When $A$ is singular, Cramer's rule (which requires that $A$ must have nonzero determinant) cannot be applied.

For an example of "no solution" case, consider: $$ 0x+0y+0z=2016. $$ For an example of "infinite solutions" case, consider: $$ 0x+0y+0z=0. $$

Answer 3

With two equations and two unknowns you can find systems with no solution $$\left\{\begin{align}x+y&=1 \\ x+y& =2\end{align}\right.$$ and with infinitely many solutions

$$\left\{\begin{align}x+y&=1 \\ 2x+2y& =2\end{align}\right.$$ (You can write similar examples for three equations.) In both cases, assuming you have the same number of equations and unknowns, the coefficients matrix must be singular.

The solution $X=A^{-1}B$ only works if $A$ is non-singular.

Have a look at Rouché-Capelli theorem for more information about that https://en.wikipedia.org/wiki/Rouch%C3%A9%E2%80%93Capelli_theorem.