Echelon form of a Matrix, Rank, and Number of solution

Question

My book defines a matrix A to be in row Echelon form as follows:

A matrix A is said to be in row Echelon form if:

1. The first non-zero element in each row is 1
2. Every non-zero row in A preceds every zero row
3. The number of zeroes before the first non zero element in 1st, 2nd, 3rd,... rows should be in >increasing order

Example:

$$ \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 0 & 0 & 1 \\ \end{bmatrix} $$

Note: The condition that a column containing the leading 1 must have all other elements above and below it as zero is NOT mentioned as a condition and I believe that it is only necessary for Reduced Row Echelon matrices.

Now, my book defines rank of a matrix to be the number of non-zero rows in a Row-Echelon Matrix Further, it is related to number of solutions of equations as follows:

Let A be the coefficient matrix of a given system of equations, ie, AX=B, and C be the augmented matrix (Such that C = [A:B]), then

1. Consistent Equation $\rho(A)=\rho(C)$

• Unique Solution: If $\rho(A)=\rho(C)=n $, where n is the number of unknowns
• Infinite Solution: If $\rho(A)=\rho(C)= r$, such that $r < n$

2. Inconsistent Equation $\rho(A)\neq \rho(C)$, then it has no solution

Now, I have a question that says the follows:

Determine for what values of $\lambda, \mu$, the following system of equations

$$x+y+z=6$$ $$x+2y+3z=10$$ $$x+2y+ \lambda z = \mu$$

has (i) no solution, (ii) a unique solution, (iii) an infinite number of solutions

So, the solution is given, by first writing in the matrix form, then converting to the augmented form, which I am writing here....

$$ \left[ \begin{array}{ccc|c} 1&1&1&6\\ 0&1&2&4\\ 0&0& \lambda-3 & \mu-10\\ \end{array} \right] $$

Now, it is stated, that for (i) no solution, $\rho(A)\neq \rho(C)$

So, $\lambda-3 = 0 $and $\mu-10 \neq 0$

which I think is so that rank of the matrix A is 2 and rank of the matrix C is 3, which means that the ranks of the two matrices are not equal.

But, shouldn't $\lambda-3 = 1???$ This just doesnt make sense to me, as the leading element must be 1 right???? Otherwise how do we get the Elchelon form to find the rank??? Also, if $\lambda-3=0$, then atleast $\mu-10 =1$ as $\mu-10 \neq 0$, right?? But thats not given as a condition.

Should there be a leading 1 necessary to get the rank or not??

(ii) A unique solution $\rho(A)=\rho(C)=3$

$\lambda -3 \neq 0$ and $\mu \in R$

(iii) Infinite solutions, $\rho(A)=\rho(C) < 3$

$\lambda -3 = 0$ and $\mu -10=0$

This I get, as now both the ranks of A and C is 2.

But Can someone please explain the 1st two cases to me, and also the doubts that I had which I explained mid-question.

Answer 1

The basic stategy is to find the leftmost column of the matrix that has a nonzero entry (usually that will be the first column). Choose any row with a nonzero entry in that column. Use row interchange to make that row the first row of your new matrix. Multiply the first row by the reciprocal of its leading nonzero entry, producing a leading $1$ in the row. Add appropriate multiples of the first row to lower rows to get zeros under the leading $1.$ If a row has a $0$ under the leading $1$ you don't need to do anything to that row. Cover the first row of the matrix and repeat the process on the remaining part to produce a leading $1$ in the second row (it will occur in a column to the right of the first leading $1)$ and $0$'s below it. Keep going until you get a leading $1$ in as many rows as possible. The remaining rows should be all $0$'s at that point.