Reduced Row Echelon Form and Row Operations - Linear Algebra

Card 0 of 248

Use row operations to find the inverse of the matrix $\begin{bmatrix} 2 &-2 & 5\ -2 &1 &-3 \ 1&0 &1 \end{bmatrix}$

Tap to see back →

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

$\begin{bmatrix} 2 &-2 &5 &1 &0 &0\ 0& -1& 2& 1& 1& 0\ 1 &0 &1 &0 &0 &1 \end{bmatrix} \sim$ subtract two times the second row to the first

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

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

$\begin{bmatrix} 1&0 &0 &-1&-2 &-1\ 0& -1& 2& 1& 1& 0\ 0 &0 &1 &1 &2 &2 \end{bmatrix} \sim$ subtract two times the last row from the second row

$\begin{bmatrix} 1&0 &0 &-1&-2 &-1\ 0& -1& 0& -1& -3& -4\ 0 &0 &1 &1 &2 &2 \end{bmatrix} \sim$ switch the sign in the middle row

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

The inverse is $\begin{bmatrix} -1 &-2 &-1 \ 1& 3& 4\1 &2 &2 \end{bmatrix}$

$A = \begin{bmatrix} e & 0 & 0 & 0 \ 0 & \pi & 0 & 0 \ 0 & 0 & -\pi & 0 \ 0 & 0 & 0 & -e \end{bmatrix}$

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

Tap to see back →

A matrix is in row-echelon form if and only if it fits three conditions:

Any rows comprising only zeroes are at the bottom.

Any leading nonzero entries are 1's..

Each leading 1 is to the right of the one immediately above.

All four rows have leading nonzero entries, but none of them are 1's. The matrix violates the conditions of a matrix in row-echelon form.

Find the inverse using row operations

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

Tap to see back →

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

Find the inverse using row operations: $\begin{bmatrix} 3 &-1 &1 \2 &-2 &5 \4 &3& 2\end{bmatrix}$

Tap to see back →

$\begin{bmatrix} 3 &0 &4 &1 &0 &0 \2 &-2 &5 &0 &1 &0 \ 4& 3& 2& 0& 0& 1\end{bmatrix} \sim$ subtract two times the second row from the last row

$\begin{bmatrix} 3 &0 &4 &1 &0 &0 \2 &-2 &5 &0 &1 &0 \ 0& 7& -8& 0& -2& 1\end{bmatrix} \sim$ subtract the second row from the first

$\begin{bmatrix} 1 &2 &-1 &1 &-1&0 \2 &-2 &5 &0 &1 &0 \ 0& 7& -8& 0& -2& 1\end{bmatrix} \sim$ subtract two times the first row from the second

$\begin{bmatrix} 1 &2 &-1 &1 &-1&0 \0 &-6 &7 &-2 &3 &0 \ 0& 7& -8& 0& -2& 1\end{bmatrix} \sim$ add the third row to the second

$\begin{bmatrix} 1 &2 &-1 &1 &-1&0 \0 &1 &-1 &-2 &1&1\ 0& 7& -8& 0& -2& 1\end{bmatrix} \sim$ subtract 7 times the second row from the third row, then multiply by -1

$\begin{bmatrix} 1 &2 &-1 &1 &-1&0 \0 &1 &-1 &-2 &1&1\ 0& 0& 1& -14& -9& 6\end{bmatrix} \sim$ add the bottom row to the middle row

$\begin{bmatrix} 1 &2 &-1 &1 &-1&0 \0 &1 &0 &-16 &10&7\ 0& 0& 1& -14& -9& 6\end{bmatrix} \sim$ add the last row to the top row

$\begin{bmatrix} 1 &2 &0 &-13 &8&6 \0 &1 &0 &-16 &10&7\ 0& 0& 1& -14& -9& 6\end{bmatrix} \sim$ subtract two times the second row from the top row

$\begin{bmatrix} 1 &0 &0 &19 &-12&-8 \0 &1 &0 &-16 &10&7\ 0& 0& 1& -14& -9& 6\end{bmatrix}$

The inverse is

$\begin{bmatrix}19 &-12&-8 \-16 &10&7\ -14& -9& 6\end{bmatrix}$

Change the following matrix into reduced row echelon form.

$\begin{pmatrix} 2&6 &3 \ 4&1 &2 \end{pmatrix}$

Tap to see back →

In order to get the matrix into reduced row echelon form,

Multiply the first row by $\frac{1}{2}$

$\begin{pmatrix} $\frac{1}{2}$2&$\frac{1}{2}$6 &$\frac{1}{2}$3 \ 4&1 &2 \end{pmatrix}=\begin{pmatrix} 1&3 &$\frac{3}{2}$ \ 4&1 &2 \end{pmatrix}$

Add times row one to row 2

$\begin{pmatrix} 1&3 &$\frac{3}{2}$ \ 4+(-4)&1+(-12) &2+(-6) \end{pmatrix}=\begin{pmatrix} 1&3 &$\frac{3}{2}$ \ 0&-11 &-4 \end{pmatrix}$

Multiply the second row by $-$\frac{1}{11}$$

$\begin{pmatrix} 1&3 &$\frac{3}{2}$ \ 0&-11(-$\frac{1}{11}$) &-4*(-$\frac{1}{11}$) \end{pmatrix}=\begin{pmatrix} 1&3 &$\frac{3}{2}$ \ 0&1 &$\frac{4}{11}$ \end{pmatrix}$

Add - times row two to row one

$\begin{pmatrix} 1&3+(-3) &$\frac{3}{2}$+(-$\frac{12}{11}$) \ 0&1 &$\frac{4}{11}$ \end{pmatrix}=\begin{pmatrix} 1&0 &$\frac{9}{22}$ \ 0&1 &$\frac{4}{11}$ \end{pmatrix}$

Change the following matrix into reduced row echelon form.

$\begin{pmatrix} 4&8 &12 \ 2&4 &1 \end{pmatrix}$

Tap to see back →

Multiply row one by $\frac{1}{4}$

$\begin{pmatrix} $\frac{1}{4}$4&$\frac{1}{4}$8&$\frac{1}{4}$12 \ 2&4 &1 \end{pmatrix}=\begin{pmatrix} 1&2 &3 \ 2&4 &1 \end{pmatrix}$

Add times row one to row two

$\begin{pmatrix} 1&2 &3 \ 2+(-2)&4+(-4) &1+(-6) \end{pmatrix}=\begin{pmatrix} 1&2 &3 \ 0&0 &-5 \end{pmatrix}$

Multiply row two by $-$\frac{1}{5}$$

$\begin{pmatrix} 1&2 &3 \ -$\frac{1}{5}$0&-$\frac{1}{5}$0 &-$\frac{1}{5}$-5 \end{pmatrix}=\begin{pmatrix} 1&2 &3 \ 0&0 &1 \end{pmatrix}$

Add times row two to row one.

$\begin{pmatrix} 1+0&2+0 &3+(-3) \ 0&0 &1 \end{pmatrix}=\begin{pmatrix} 1&2 &0 \ 0&0 &1 \end{pmatrix}$

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

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

Tap to see back →

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

This matrix meets all four of these criteria.

$\begin{align}&$\text{Convert }$A=\begin{bmatrix}10&9&-11&-17\13&-17&-8&-18\\end{bmatrix}\&$\text{to its reduced echelon form.}$\end{align}$

Tap to see back →

$\begin{align}&$\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,}$\end{align}$

$\begin{align}\&$\text{we can convert it to this form by:}$\&$\text{Scaling the 1st row of the matrix by }$$\frac{13}{10}$\&$\text{and subtracting it from the 2nd row:}$\&\begin{bmatrix}10&9&-11&-17\0&-$\frac{287}{10}$&$\frac{63}{10}$&$\frac{41}{10}$\end{bmatrix}&$\text{Next we scale the 2nd row of this new matrix by }$-$\frac{90}{287}$\&$\text{and subtract it from the 1st:}$\&\begin{bmatrix}10&0&-$\frac{370}{41}$&-$\frac{110}{7}$\0&-$\frac{287}{10}$&$\frac{63}{10}$&$\frac{41}{10}$\end{bmatrix}&$\text{Finally, we scale each row down to have one as the leading values:}$\&\begin{bmatrix}1&0&-$\frac{37}{41}$&-$\frac{11}{7}$\0&1&-$\frac{9}{41}$&-$\frac{1}{7}$\\end{bmatrix}\end{align}$

$\begin{align}&$\text{Convert the matrix }$\begin{bmatrix}5&5&-7\9&-19&9\\end{bmatrix}\&$\text{to reduced echelon form.}$\end{align}$

Tap to see back →

$\begin{align}&$\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{9}{5}$\&$\text{and subtracting it from the 2nd row:}$\&\begin{bmatrix}5&5&-7\0&-28&$\frac{108}{5}$\end{bmatrix}&$\text{Next we scale the 2nd row of this new matrix by }$-$\frac{5}{28}$\&$\text{and subtract it from the 1st:}$\&\begin{bmatrix}5&0&-$\frac{22}{7}$\0&-28&$\frac{108}{5}$\end{bmatrix}&$\text{Finally, we scale each row down to have one as the leading values:}$\&\begin{bmatrix}1&0&-$\frac{22}{35}$\0&1&-$\frac{27}{35}$\\end{bmatrix}\end{align}$

$\begin{align}&$\text{Find the reduced echelon form of the matrix:}$\&\begin{bmatrix}6&-9&2\17&-19&-10\\end{bmatrix}\end{align}$

Tap to see back →

$\begin{align}&$\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{17}{6}$\&$\text{and subtracting it from the 2nd row:}$\&\begin{bmatrix}6&-9&2\0&$\frac{13}{2}$&-$\frac{47}{3}$\end{bmatrix}&$\text{Next we scale the 2nd row of this new matrix by }$-$\frac{18}{13}$\&$\text{and subtract it from the 1st:}$\&\begin{bmatrix}6&0&-$\frac{256}{13}$\0&$\frac{13}{2}$&-$\frac{47}{3}$\end{bmatrix}&$\text{Finally, we scale each row down to have one as the leading values:}$\&\begin{bmatrix}1&0&-$\frac{128}{39}$\0&1&-$\frac{94}{39}$\\end{bmatrix}\end{align}$

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

Tap to see back →

$\begin{align}&$\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}$\\end{bmatrix}\end{align}$

$\begin{align}&$\text{Convert the matrix }$\begin{bmatrix}-2&1&11\-17&15&-14\\end{bmatrix}\&$\text{to reduced echelon form.}$\end{align}$

Tap to see back →

$\begin{align}&$\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 }$$\frac{17}{2}$\&$\text{and subtracting it from the 2nd row:}$\&\begin{bmatrix}-2&1&11\0&$\frac{13}{2}$&-$\frac{215}{2}$\end{bmatrix}&$\text{Next we scale the 2nd row of this new matrix by }$$\frac{2}{13}$\&$\text{and subtract it from the 1st:}$\&\begin{bmatrix}-2&0&$\frac{358}{13}$\0&$\frac{13}{2}$&-$\frac{215}{2}$\\end{bmatrix}\&$\text{Finally, we scale each row down to have one as the leading values:}$\&\begin{bmatrix}1&0&-$\frac{179}{13}$\0&1&-$\frac{215}{13}$\\end{bmatrix}\end{align}$

$\begin{align}&$\text{Find the reduced echelon form of the matrix:}$\&\begin{bmatrix}4&-17&2\11&-14&6\\end{bmatrix}\end{align}$

Tap to see back →

$\begin{align}&$\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:}$\&\begin{bmatrix}4&-17&2\0&$\frac{131}{4}$&$\frac{1}{2}$\end{bmatrix}&$\text{Next we scale the 2nd row of this new matrix by }$-$\frac{68}{131}$\&$\text{and subtract it from the 1st:}$\&\begin{bmatrix}4&0&$\frac{296}{131}$\0&$\frac{131}{4}$&$\frac{1}{2}$\end{bmatrix}&$\text{Finally, we scale each row down to have one as the leading values:}$\&\begin{bmatrix}1&0&$\frac{74}{131}$\0&1&$\frac{2}{131}$\\end{bmatrix}\end{align}$

$\begin{align}&$\text{Convert }$A=\begin{bmatrix}-20&-2&6\-11&-9&-7\\end{bmatrix}\&$\text{to its reduced echelon form.}$\end{align}$

Tap to see back →

$\begin{align}&$\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 }$$\frac{11}{20}$\&$\text{and subtracting it from the 2nd row:}$\&\begin{bmatrix}-20&-2&6\0&-$\frac{79}{10}$&-$\frac{103}{10}$\end{bmatrix}&$\text{Next we scale the 2nd row of this new matrix by }$$\frac{20}{79}$\&$\text{and subtract it from the 1st:}$\&\begin{bmatrix}-20&0&$\frac{680}{79}$\0&-$\frac{79}{10}$&-$\frac{103}{10}$\\end{bmatrix}\&$\text{Finally, we scale each row down to have one as the leading values:}$\&\begin{bmatrix}1&0&-$\frac{34}{79}$\0&1&$\frac{103}{79}$\\end{bmatrix}\end{align}$

$\begin{align}&$\text{Find the reduced echelon form of the matrix:}$\&\begin{bmatrix}-1&-18&4\-10&-4&13\\end{bmatrix}\end{align}$

Tap to see back →

$\begin{align}&$\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 }$10\&$\text{and subtracting it from the 2nd row:}$\&\begin{bmatrix}-1&-18&4\0&176&-27\\end{bmatrix}\&$\text{Next we scale the 2nd row of this new matrix by }$-$\frac{9}{88}$\&$\text{and subtract it from the 1st:}$\&\begin{bmatrix}-1&0&$\frac{109}{88}$\0&176&-27\\end{bmatrix}\&$\text{Finally, we scale each row down to have one as the leading values:}$\&\begin{bmatrix}1&0&-$\frac{109}{88}$\0&1&-$\frac{27}{176}$\\end{bmatrix}\end{align}$

$\begin{align}&$\text{Find the reduced echelon form of the matrix:}$\&\begin{bmatrix}6&13&-17&15\12&-19&17&12\\end{bmatrix}\end{align}$

Tap to see back →

$\begin{align}&$\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 }$2\&$\text{and subtracting it from the 2nd row:}$\&\begin{bmatrix}6&13&-17&15\0&-45&51&-18\\end{bmatrix}\&$\text{Next we scale the 2nd row of this new matrix by }$-$\frac{13}{45}$\&$\text{and subtract it from the 1st:}$\&\begin{bmatrix}6&$\frac{1}{562949953421312}$&-$\frac{34}{15}$&$\frac{49}{5}$\0&-45&51&-18\\end{bmatrix}\&$\text{Finally, we scale each row down to have one as the leading values:}$\&\begin{bmatrix}1&0&-$\frac{17}{45}$&$\frac{49}{30}$\0&1&-$\frac{17}{15}$&$\frac{2}{5}$\\end{bmatrix}\end{align}$

$\begin{align}&$\text{Convert the matrix }$\begin{bmatrix}-14&13&2\-11&-9&-15\\end{bmatrix}\&$\text{to reduced echelon form.}$\end{align}$

Tap to see back →

$\begin{align}&$\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}{14}$\&$\text{and subtracting it from the 2nd row:}$\&\begin{bmatrix}-14&13&2\0&-$\frac{269}{14}$&-$\frac{116}{7}$\end{bmatrix}&$\text{Next we scale the 2nd row of this new matrix by }$-$\frac{182}{269}$\&$\text{and subtract it from the 1st:}$\&\begin{bmatrix}-14&0&-$\frac{2478}{269}$\0&-$\frac{269}{14}$&-$\frac{116}{7}$\end{bmatrix}&$\text{Finally, we scale each row down to have one as the leading values:}$\&\begin{bmatrix}1&0&$\frac{177}{269}$\0&1&$\frac{232}{269}$\\end{bmatrix}\end{align}$

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

Tap to see back →

$\begin{align}&$\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}$\\end{bmatrix}\end{align}$

$\begin{align}&$\text{Convert the matrix }$\begin{bmatrix}-7&14&13&19\10&-10&-7&10\\end{bmatrix}\&$\text{to reduced echelon form.}$\end{align}$

Tap to see back →

$\begin{align}&$\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{10}{7}$\&$\text{and subtracting it from the 2nd row:}$\&\begin{bmatrix}-7&14&13&19\0&10&$\frac{81}{7}$&$\frac{260}{7}$\\end{bmatrix}\&$\text{Next we scale the 2nd row of this new matrix by }$$\frac{7}{5}$\&$\text{and subtract it from the 1st:}$\&\begin{bmatrix}-7&0&-$\frac{16}{5}$&-33\0&10&$\frac{81}{7}$&$\frac{260}{7}$\end{bmatrix}&$\text{Finally, we scale each row down to have one as the leading values:}$\&\begin{bmatrix}1&0&$\frac{16}{35}$&$\frac{33}{7}$\0&1&$\frac{81}{70}$&$\frac{26}{7}$\\end{bmatrix}\end{align}$

Give the elementary matrix that represents performing the row operation

$$\frac{1}{7}$ R2 \rightarrow R2$

in solving a three-by-three linear system.

Tap to see back →

The elementary matrix that represents a row operation is the result of performing the same operation on the appropriate identity matrix - which here is the three-by-three matrix $\begin{bmatrix} 1 &0&0 \0 & 1 &0 \ 0&0 &1 \end{bmatrix}$ . The row operation $$\frac{1}{7}$ R2 \rightarrow R2$ is the multiplication of each element of in the second row of an augmented matrix by the scalar $\frac{1}{7}$ , so do this to the identity:

$\begin{bmatrix} 1 &0&0 \0 \cdot $\frac{1}{7}$ & 1 \cdot $\frac{1}{7}$&0 \cdot $\frac{1}{7}$ \ 0&0 &1 \end{bmatrix}$

$=\begin{bmatrix} 1 & 0&0 \ 0& $\frac{1}{7}$ &0 \ 0& 0 &1 \end{bmatrix}$

This is the correct response.