Reduced Row Echelon Form and Hexadecimal Numbers

Algebra Level 4

$A = \left[ \begin{array}{c c c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \\ \end{array}\right] \quad, \quad B= \left[ \begin{array}{ c c c} 1 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 2 \\ \end{array}\right] \quad,\quad C= \left[ \begin{array}{c c c c} 1 & 0 & 2 & 5\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ \end{array}\right], \\ D = \left[ \begin{array}{c c c} 1 & 0 & 0\\ 0 & 0 & 1 \\ 0 & 1 & 0 \\ \end{array}\right] \quad,\quad E = \left[ \begin{array}{c c} 1 & 0\\ 0 & 0 \\ 0 & 0 \\ 0 & 1 \\ \end{array}\right]$

Which of the matrices above are in their reduced row echelon form ?

Concatenate the letters of the matrices that are in their reduced row echelon form in alphabetical order and convert your answer from hexadecimal to decimal.

As an explicit example, if you think matrices $A, B$ and $C$ are in their reduced row echelon form, then your answer is $ABC_{16} = 10 \times 16^{2} + 11 \times 16 + 12 = \boxed{2748}$ .

You may find the rules for reduced row echelon form on this page to be helpful.

The answer is 12.

1 solution

Pranshu Gaba
Dec 10, 2015

We will look at each of the five matrices and check whether they are in their reduced row echelon form.
We will refer to the rules on this page - Solving Linear Systems Using Matrices

Matrix A is not in its reduced row echelon form because it does not follow rule 2. Since the '1' in the first row is a pivot, the first column must be $\left[ \begin{array} {c} 1 \\ 0 \\ 0 \end{array} \right]$ for matrix $A$ to be in its reduced row echelon form.

We can subtract row 1 from row 3 to obtain the reduced row echelon form of matrix $A$ .

Matrix B is not in its reduced row echelon form because it does not follow rule 1. The leftmost non-zero element in each row must be $1$ . In rows 2 and 3 of matrix $B$ the leftmost non-zero elements are 5 and 2 respectively, so rule 1 is not followed.

Rows 2 and 3 can be divided by 5 and 2 respectively to obtain the reduced row echelon form of the matrix $B$ .

Matrix C follows all the rules so it is in its reduced row echelon form.

Matrix D is not in its reduced row echelon form because it does not follow rule 3. The pivot in column 2 is to the left of the pivot in column 3, and yet row 2 is above row 3.

We can switch rows 2 and 3 to obtain the reduced row echelon form of matrix $D$ .

Matrix E is not in its reduced row echelon form because it does not follow rule 4. Rows that consist of only zeroes should be at the bottom of the matrix. However we see that, in matrix $E$ that there are rows that consist of only zeroes in the middle of the matrix.

If we switch rows 2 and 4, then all the rows that have only zeroes will be at the bottom of the matrix and we will obtain the reduced row echelon form of matrix $E$ .

We see that only matrix $C$ is in its reduced row echelon form, therefore the answer is $C_{16} = \boxed{12}$ . $_\square$

Moderator note:

Thanks for identifying several ways where matrices look like they are in rref, but actually are not!

Reduced Row Echelon Form and Hexadecimal Numbers

You may find the rules for reduced row echelon form on this page to be helpful.

The answer is 12.

1 solution

Moderator note:

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