Linear Algebra - Reduced Row Echelon Form and Row Operations

$\begin{align*}&\text{Convert }A=\begin{bmatrix}4&-9&-10\-11&-13&-5\\end{bmatrix}\&\text{to its reduced echelon form.}\end{align*}$

$\begin{bmatrix}1&0&-\frac{85}{151}\0&1&\frac{130}{151}\\end{bmatrix}$

$\begin{bmatrix}1&0&-\frac{47}{81}\0&1&\frac{25}{27}\\end{bmatrix}$

$\begin{bmatrix}1&0&-\frac{31}{85}\0&1&\frac{2}{17}\\end{bmatrix}$

$\begin{bmatrix}1&0&-\frac{11}{37}\0&1&-\frac{5}{37}\\end{bmatrix}$

Explanation

$\begin{align*}&\text{Reduced echelon form is a matrix form in which the leading coefficient}\&\text{of each nonzero row is always to the right of the leading coefficient}\&\text{of the row above it, each leading coefficient is the only nonzero}\&\text{value in its column, and each row is scaled down to give the}\&\text{leading coefficients values of one. Beginning with our matrix,}\&\text{we can convert it to this form by:}\&\text{Scaling the 1st row of the matrix by }-\frac{11}{4}\&\text{and subtracting it from the 2nd row:}\&\begin{bmatrix}4&-9&-10\0&-\frac{151}{4}&-\frac{65}{2}\\end{bmatrix}\&\text{Next we scale the 2nd row of this new matrix by }\frac{36}{151}\&\text{and subtract it from the 1st:}\&\begin{bmatrix}4&0&-\frac{340}{151}\0&-\frac{151}{4}&-\frac{65}{2}\\end{bmatrix}\&\text{Finally, we scale each row down to have one as the leading values:}\&\begin{bmatrix}1&0&-\frac{85}{151}\0&1&\frac{130}{151}\\end{bmatrix}\end{align*}$

Find the inverse using row operations

$\begin{bmatrix} 2 &7 &1 \-1 &5 &3 \ 2& 2& -1\end{bmatrix}$

$\begin{bmatrix} -11 &9 &16 \5 &-4 & -7\ -12& 10& 17\end{bmatrix}$

$\begin{bmatrix} -15 &9 &16 \7 &-4 & -7\ -16& 10& 17\end{bmatrix}$

$\begin{bmatrix} -11 &9 &16 \5 &-4 & -7\ -16& 10& 17\end{bmatrix}$

$\begin{bmatrix} 1 &0&1 \1&-4 & 1\ -16& 10& 1\end{bmatrix}$

Explanation

To find the inverse, use row operations:

$\begin{bmatrix} 2 &7 &1 &1 &0 &0 \-1 &5 &3 &0 &1 &0 \ 2& 2& -1& 0& 0& 1\end{bmatrix} \sim$ add the third row to the second

$\begin{bmatrix} 2 &7 &1 &1 &0 &0 \1 &7 &2 &0 &1 &1 \ 2& 2& -1& 0& 0& 1\end{bmatrix} \sim$ subtract the second row from the top

$\begin{bmatrix} 1 &0 &-1 &1 &-1 &-1 \1 &7 &2 &0 &1 &1 \ 2& 2& -1& 0& 0& 1\end{bmatrix} \sim$ subtract the first row from the second

$\begin{bmatrix} 1 &0 &-1 &1 &-1 &-1 \0&7 &3 &-1&2 &2 \ 2& 2& -1& 0& 0& 1\end{bmatrix} \sim$ subtract two times the first row from the bottom row

$\begin{bmatrix} 1 &0 &-1 &1 &-1 &-1 \0&7 &3 &-1&2 &2 \ 0& 2& 1& -2& 2&3\end{bmatrix} \sim$ subtract three times the bottom row from the second row

$\begin{bmatrix} 1 &0 &-1 &1 &-1 &-1 \0&1 &0 &5&-4 &-7 \ 0& 2& 1& -2& 2&3\end{bmatrix} \sim$ subtract 2 times the middle row from the bottom row

$\begin{bmatrix} 1 &0 &-1 &1 &-1 &-1 \0&1 &0 &5&-4 &-7 \ 0& 0& 1& -12& 10&17\end{bmatrix} \sim$ add the bottom row to the top

$\begin{bmatrix} 1 &0 &0 &-11 &9&16 \0&1 &0 &5&-4 &-7 \ 0& 0& 1& -12& 10&17\end{bmatrix}$

The inverse is $\begin{bmatrix} -11 & 9& 16\ 5& -4& -7\ -12& 10& 17\end{bmatrix}$

$\begin{align*}&\text{Find the reduced echelon form of the matrix:}\&\begin{bmatrix}-1&11&-19\-13&-17&4\\end{bmatrix}\end{align*}$

$\begin{bmatrix}1&0&\frac{279}{160}\0&1&-\frac{251}{160}\\end{bmatrix}$

$\begin{bmatrix}1&0&\frac{9}{8}\0&1&-\frac{73}{56}\\end{bmatrix}$

$\begin{bmatrix}1&0&\frac{3}{10}\0&1&-\frac{19}{20}\\end{bmatrix}$

$\begin{bmatrix}1&0&\frac{369}{152}\0&1&-\frac{283}{152}\\end{bmatrix}$

Explanation

$\begin{align*}&\text{Reduced echelon form is a matrix form in which the leading coefficient}\&\text{of each nonzero row is always to the right of the leading coefficient}\&\text{of the row above it, each leading coefficient is the only nonzero}\&\text{value in its column, and each row is scaled down to give the}\&\text{leading coefficients values of one. Beginning with our matrix,}\&\text{we can convert it to this form by:}\&\text{Scaling the 1st row of the matrix by }13\&\text{and subtracting it from the 2nd row:}\&\begin{bmatrix}-1&11&-19\0&-160&251\\end{bmatrix}\&\text{Next we scale the 2nd row of this new matrix by }-\frac{11}{160}\&\text{and subtract it from the 1st:}\&\begin{bmatrix}-1&0&-\frac{279}{160}\0&-160&251\\end{bmatrix}\&\text{Finally, we scale each row down to have one as the leading values:}\&\begin{bmatrix}1&0&\frac{279}{160}\0&1&-\frac{251}{160}\\end{bmatrix}\end{align*}$

$A = \begin{bmatrix}\bigcirc & \bigcirc & \bigcirc & \bigcirc \ 0 & 1 & 1 & 0 \ 0 & 0 & 0 & 1 \end{bmatrix}$

Note that the top row is missing. How many ways can the top row be completed in order to form a matrix in reduced row-echelon form?

Infinitely many

None

One

Two

Three

Explanation

A matrix is in reduced row-echelon form if it satisfies four conditions:

All zero rows must be below all nonzero rows.
Any leading nonzero element in a nonzero row must be 1.
Each leading 1 must be to the right of the one above it.
The other elements in the same column of any leading 1 must be zeroes.

The second row has nonzero elements, so the first row cannot be all zeroes; this would violate Condition 1. There must be at least one nonzero element in the top row. The first row must have a leading 1, by Condition 2. The first nonzero element in the second row is in the second column, as seen below; the leading 1 in the first row must be in the first column,

$A = \begin{bmatrix}\textbf{{\color{Red} 1}}& \bigcirc & \bigcirc & \bigcirc \ 0 & \textbf{{\color{Blue} 1}} & 1 & 0 \ 0 & 0 & 0 & 1 \end{bmatrix}$ .

Since leading 1's are located in the second and fourth columns, by Condition 4, the second and fourth elements in the top row must be zeroes.

$A = \begin{bmatrix} \textbf{{\color{Red} 1}} & \textbf{{\color{Red} 1}}& \bigcirc & \textbf{{\color{Red} 1}} \ 0 & \textbf{{\color{Blue} 1}} & 1 & 0 \ 0 & 0 & 0 & \textbf{{\color{Blue} 1}} \end{bmatrix}$

However, there are no restrictions on the remaining element. The first three conditions are already met, and, since the 1 in the second row and third column is not a leading 1, it does not restrict the other entries in the third column.

It follows that there are infinitely many ways to fill the first row in.

True or false:

$\begin{bmatrix}1 &0 &0&0 \0 &1 &0&0 \ 0 &0 &1 &0 \1,000,000 &0&0&1\end{bmatrix}$

is an elementary matrix.

True

False

Explanation

An elementary matrix is one that can be formed by performing a single row operation on the (here, four-by-four) identity matrix

$I = \begin{bmatrix}1 &0 &0&0 \0 &1 &0&0 \ 0 &0 &1 &0 \0 &0&0&1\end{bmatrix}$

A matrix qualifies as an elementary matrix if and only if it differs from in exactly one of three ways:

Exactly two rows have changed position with each other.
Exactly one of its elements on its main diagonal - upper left to lower right - is equal to a nonzero number other than 1.
Exactly one of its nondiagonal elements is equal to a nonzero number.

$\begin{bmatrix}1 &0 &0&0 \0 &1 &0&0 \ 0 &0 &1 &0 \1,000,000 &0&0&1\end{bmatrix}$ satisfies this condition in that it differs from in only one of these ways - one non-diagonal element (Row 4, Column 1) has been changed to a nonzero number. This is an elementary matrix.

True or false:

$\begin{bmatrix} 0.000001 &0 &0&0 \0 &1 &0&0 \ 0 &0 &1 &0 \0 &0&0&1\end{bmatrix}$

is an elementary matrix.

True

False

Explanation

An elementary matrix is one that can be formed by performing a single row operation on the (here, four-by-four) identity matrix

$I = \begin{bmatrix}1 &0 &0&0 \0 &1 &0&0 \ 0 &0 &1 &0 \0 &0&0&1\end{bmatrix}$

A matrix qualifies as an elementary matrix if and only if it differs from in exactly one of three ways:

Exactly two rows have changed position with each other.
Exactly one of its diagonal elements - the element in Row 1 and Column 1, Row 2 and Column 2, and so forth - is equal to a nonzero number other than 1.
Exactly one of its nondiagonal elements is equal to a nonzero number.

$\begin{bmatrix} 0.000001 &0 &0&0 \0 &1 &0&0 \ 0 &0 &1 &0 \0 &0&0&1\end{bmatrix}$ satisfies these criteria in that it differs from only in that one diagonal element (Row 1, Column 1) has been changed to the nonzero number 0.000001.

True or false:

$\begin{bmatrix}1 &0 &0&0 \0 &1 &0&0 \ 0 &0 &1 &0 \1 & 1& 1&1\end{bmatrix}$

is an elementary matrix.

False

True

Explanation

An elementary matrix is one that can be formed by performing a single row operation on the (here, four-by-four) identity matrix

$I = \begin{bmatrix}1 &0 &0&0 \0 &1 &0&0 \ 0 &0 &1 &0 \0 &0&0&1\end{bmatrix}$

A matrix qualifies as an elementary matrix if and only if it differs from in exactly one of three ways:

Exactly two rows have changed position with each other.
Exactly one of its diagonal elements - the element in Row 1 and Column 1, Row 2 and Column 2, and so forth - is equal to a nonzero number other than 1.
Exactly one of its nondiagonal elements is equal to a nonzero number.

$\begin{bmatrix}1 &0 &0&0 \0 &1 &0&0 \ 0 &0 &1 &0 \1 & 1& 1&1\end{bmatrix}$ violates these criteria because three of its nondiagonal elements - the first three elements in its fourth row - are equal to nonzero numbers.

Is the following matrix in reduced row echelon form? Why or why not.

$\begin{bmatrix} 5& 0& 0& 7\ 0& 1& 0& 5\ 0& 0& 1& 4 \end{bmatrix}$

No, the leading entry in the first row is not

Yes, the matrix is in reduced row echelon form

No, the matrix has nonzero elements

No, the matrix does not have an all zero row

Explanation

A matrix is in reduced row echelon form if

all nonzero rows are above any all zero rows
the left most nonzero entry in each row (the leading entry) is .
the leading entry is in a column to the right of the leading entry in the column above it
all entries in a column below the leading entry are zero

For this matrix to be in reduced row echelon form, the leading entry in the first row would need to be , not .

$A= \begin{bmatrix} 0 & 1 & 0 & 1 \ 1 & 0 & 0 & 2 \ 0 & 0 & 1 & -3 \end{bmatrix}$

True or false: is an example of a matrix in reduced row-echelon form.

False

True

Explanation

One condition for a matrix to be in reduced row-echelon form is for each leading nonzero entry to be at the right of the entry in the row above it. The leading 1 in the second row is at left of the leading 1 in the first row, as seen below:

$\begin{bmatrix} 0 & \textbf{{\color{Red} 1}} & 0 & 1 \ \textbf{{\color{Red} 1}} & 0 & 0 & 2 \ 0 & 0 & 1 & -3 \end{bmatrix}$

This is a violation of that condition. is therefore not a matrix in reduced row-echelon form.

$A= \begin{bmatrix} 1 & 0 & 1 & 0 \ 0 & 1 & -2 & 0 \ \bigcirc &\bigcirc &\bigcirc &\bigcirc \end{bmatrix}$ .

Note that the entries in the bottom row are missing.

How many possible ways can the entries in the bottom row be filled in so that is in reduced row-echelon form?

Two

One

None

Infinitely many

Three

Explanation

A matrix is in reduced row-echelon form if it meets four conditions.

Any rows comprising all zeroes must be below all nonzero rows.
All leading nonzero entries must be 1's.
All other entries in the same column as a leading 1 are zeroes.
Each leading 1 is located to the right of the one above it.

None of these four conditions have been violated, at least not yet.

First, note that there are two known leading 1's. In order for the matrix to continue to meet Condition 3, it is necessary for the first two entries in the bottom row to be zeroes.

$A= \begin{bmatrix} {\color{DarkBlue} \textbf{1}} & 0 & 1 & 0 \ 0 & {\color{DarkBlue} \textbf{1}} & -2 & 0 \ {\color{Red} \textbf{0}} &{\color{Red} \textbf{0}} &\bigcirc &\bigcirc \end{bmatrix}$

If the third entry is nonzero, then it must be 1, since it would be a leading nonzero entry. But the third entry cannot be a leading 1, since the other entries in Column 3 are nonzero - a violation of Condition 3 would be created. This entry must be a zero.

$A= \begin{bmatrix} 1& 0 & {\color{DarkBlue} \textbf{1}} & 0 \ 0 & 1 & {\color{DarkBlue} \textbf{-2}} & 0 \ {\color{Red} \textbf{0}} &{\color{Red} \textbf{0}} &{\color{Red} \textbf{0}} &\bigcirc \end{bmatrix}$

If the last entry is a 0, then no conditions are violated; the only zero row is on the bottom, and the other conditions still hold:

$A= \begin{bmatrix} 1& 0 & 1 & 0 \ 0 &1& -2 & 0 \ {\color{Red} \textbf{0}} &{\color{Red} \textbf{0}} &{\color{Red} \textbf{0}} &{\color{Red} \textbf{0}} \end{bmatrix}$

If the last entry is nonzero, then it must be 1, since it would be a leading nonzero entry.No conditions are violated, since the other entries in the column are zeroes.

$A= \begin{bmatrix} 1& 0 & 1 & \textbf{{\color{DarkBlue} 0}} \ 0 &1& -2 & \textbf{{\color{DarkBlue} 0}} \ {\color{Red} \textbf{0}} &{\color{Red} \textbf{0}} &{\color{Red} \textbf{0}} &{\color{Red} \textbf{1}} \end{bmatrix}$

Therefore, the bottom row can be filled exactly two ways.

Reduced Row Echelon Form and Row Operations

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