Reduced Row Echelon Form and Row Operations - Linear Algebra

Card 0 of 248

Question

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

Answer

$\begin{bmatrix} 2 &-2 &5 &1 &0 & 0\ -2 &1 &-3 &0 &1 &0 \ 1& 0& 1& 0&0 &1 \end{bmatrix} \sim$ 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 $\begin{bmatrix} -1 &-2 &-1 \ 1& 3& 4\1 &2 &2 \end{bmatrix}$

Compare your answer with the correct one above

Question

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

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

Answer

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.

Compare your answer with the correct one above

Question

Find the inverse using row operations

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

Answer

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

Compare your answer with the correct one above

Question

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

Answer

$\begin{bmatrix} 3 &0 &4 &1 &0 &0 \2 &-2 &5 &0 &1 &0 \ 4& 3& 2& 0& 0& 1\end{bmatrix} \sim$ 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

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

Compare your answer with the correct one above

Question

Change the following matrix into reduced row echelon form.

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

Answer

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

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

Compare your answer with the correct one above

Question

Change the following matrix into reduced row echelon form.

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

Answer

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.

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

Compare your answer with the correct one above

Question

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

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

Answer

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.

Compare your answer with the correct one above

Question

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

Answer

$\begin{align}&\text{Reduced echelon form is a matrix form in which the leading coefficient}\&\text{of each nonzero row is always to the right of the leading coefficient}\&\text{of the row above it, each leading coefficient is the only nonzero}\&\text{value in its column, and each row is scaled down to give the}\&\text{leading coefficients values of one. Beginning with our matrix,}\end{align}$

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

Compare your answer with the correct one above

Question

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

Answer

$\begin{align}&\text{Reduced echelon form is a matrix form in which the leading coefficient}\&\text{of each nonzero row is always to the right of the leading coefficient}\&\text{of the row above it, each leading coefficient is the only nonzero}\&\text{value in its column, and each row is scaled down to give the}\&\text{leading coefficients values of one. Beginning with our matrix,}\&\text{we can convert it to this form by:}\&\text{Scaling the 1st row of the matrix by }\frac{9}{5}\&\text{and subtracting it from the 2nd row:}\&\begin{bmatrix}5&5&-7\0&-28&\frac{108}{5}\\end{bmatrix}\&\text{Next we scale the 2nd row of this new matrix by }-\frac{5}{28}\&\text{and subtract it from the 1st:}\&\begin{bmatrix}5&0&-\frac{22}{7}\0&-28&\frac{108}{5}\\end{bmatrix}\&\text{Finally, we scale each row down to have one as the leading values:}\&\begin{bmatrix}1&0&-\frac{22}{35}\0&1&-\frac{27}{35}\\end{bmatrix}\end{align}$

Compare your answer with the correct one above

Question

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

Answer

$\begin{align}&\text{Reduced echelon form is a matrix form in which the leading coefficient}\&\text{of each nonzero row is always to the right of the leading coefficient}\&\text{of the row above it, each leading coefficient is the only nonzero}\&\text{value in its column, and each row is scaled down to give the}\&\text{leading coefficients values of one. Beginning with our matrix,}\&\text{we can convert it to this form by:}\&\text{Scaling the 1st row of the matrix by }\frac{17}{6}\&\text{and subtracting it from the 2nd row:}\&\begin{bmatrix}6&-9&2\0&\frac{13}{2}&-\frac{47}{3}\\end{bmatrix}\&\text{Next we scale the 2nd row of this new matrix by }-\frac{18}{13}\&\text{and subtract it from the 1st:}\&\begin{bmatrix}6&0&-\frac{256}{13}\0&\frac{13}{2}&-\frac{47}{3}\\end{bmatrix}\&\text{Finally, we scale each row down to have one as the leading values:}\&\begin{bmatrix}1&0&-\frac{128}{39}\0&1&-\frac{94}{39}\\end{bmatrix}\end{align}$

Compare your answer with the correct one above

Question

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

Answer

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

Compare your answer with the correct one above

Question

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

Answer

$\begin{align}&\text{Reduced echelon form is a matrix form in which the leading coefficient}\&\text{of each nonzero row is always to the right of the leading coefficient}\&\text{of the row above it, each leading coefficient is the only nonzero}\&\text{value in its column, and each row is scaled down to give the}\&\text{leading coefficients values of one. Beginning with our matrix,}\&\text{we can convert it to this form by:}\&\text{Scaling the 1st row of the matrix by }\frac{17}{2}\&\text{and subtracting it from the 2nd row:}\&\begin{bmatrix}-2&1&11\0&\frac{13}{2}&-\frac{215}{2}\\end{bmatrix}\&\text{Next we scale the 2nd row of this new matrix by }\frac{2}{13}\&\text{and subtract it from the 1st:}\&\begin{bmatrix}-2&0&\frac{358}{13}\0&\frac{13}{2}&-\frac{215}{2}\\end{bmatrix}\&\text{Finally, we scale each row down to have one as the leading values:}\&\begin{bmatrix}1&0&-\frac{179}{13}\0&1&-\frac{215}{13}\\end{bmatrix}\end{align}$

Compare your answer with the correct one above

Question

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

Answer

$\begin{align}&\text{Reduced echelon form is a matrix form in which the leading coefficient}\&\text{of each nonzero row is always to the right of the leading coefficient}\&\text{of the row above it, each leading coefficient is the only nonzero}\&\text{value in its column, and each row is scaled down to give the}\&\text{leading coefficients values of one. Beginning with our matrix,}\&\text{we can convert it to this form by:}\&\text{Scaling the 1st row of the matrix by }\frac{11}{4}\&\text{and subtracting it from the 2nd row:}\&\begin{bmatrix}4&-17&2\0&\frac{131}{4}&\frac{1}{2}\\end{bmatrix}\&\text{Next we scale the 2nd row of this new matrix by }-\frac{68}{131}\&\text{and subtract it from the 1st:}\&\begin{bmatrix}4&0&\frac{296}{131}\0&\frac{131}{4}&\frac{1}{2}\\end{bmatrix}\&\text{Finally, we scale each row down to have one as the leading values:}\&\begin{bmatrix}1&0&\frac{74}{131}\0&1&\frac{2}{131}\\end{bmatrix}\end{align}$

Compare your answer with the correct one above

Question

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

Answer

$\begin{align}&\text{Reduced echelon form is a matrix form in which the leading coefficient}\&\text{of each nonzero row is always to the right of the leading coefficient}\&\text{of the row above it, each leading coefficient is the only nonzero}\&\text{value in its column, and each row is scaled down to give the}\&\text{leading coefficients values of one. Beginning with our matrix,}\&\text{we can convert it to this form by:}\&\text{Scaling the 1st row of the matrix by }\frac{11}{20}\&\text{and subtracting it from the 2nd row:}\&\begin{bmatrix}-20&-2&6\0&-\frac{79}{10}&-\frac{103}{10}\\end{bmatrix}\&\text{Next we scale the 2nd row of this new matrix by }\frac{20}{79}\&\text{and subtract it from the 1st:}\&\begin{bmatrix}-20&0&\frac{680}{79}\0&-\frac{79}{10}&-\frac{103}{10}\\end{bmatrix}\&\text{Finally, we scale each row down to have one as the leading values:}\&\begin{bmatrix}1&0&-\frac{34}{79}\0&1&\frac{103}{79}\\end{bmatrix}\end{align}$

Compare your answer with the correct one above

Question

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

Answer

$\begin{align}&\text{Reduced echelon form is a matrix form in which the leading coefficient}\&\text{of each nonzero row is always to the right of the leading coefficient}\&\text{of the row above it, each leading coefficient is the only nonzero}\&\text{value in its column, and each row is scaled down to give the}\&\text{leading coefficients values of one. Beginning with our matrix,}\&\text{we can convert it to this form by:}\&\text{Scaling the 1st row of the matrix by }10\&\text{and subtracting it from the 2nd row:}\&\begin{bmatrix}-1&-18&4\0&176&-27\\end{bmatrix}\&\text{Next we scale the 2nd row of this new matrix by }-\frac{9}{88}\&\text{and subtract it from the 1st:}\&\begin{bmatrix}-1&0&\frac{109}{88}\0&176&-27\\end{bmatrix}\&\text{Finally, we scale each row down to have one as the leading values:}\&\begin{bmatrix}1&0&-\frac{109}{88}\0&1&-\frac{27}{176}\\end{bmatrix}\end{align}$

Compare your answer with the correct one above

Question

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

Answer

$\begin{align}&\text{Reduced echelon form is a matrix form in which the leading coefficient}\&\text{of each nonzero row is always to the right of the leading coefficient}\&\text{of the row above it, each leading coefficient is the only nonzero}\&\text{value in its column, and each row is scaled down to give the}\&\text{leading coefficients values of one. Beginning with our matrix,}\&\text{we can convert it to this form by:}\&\text{Scaling the 1st row of the matrix by }2\&\text{and subtracting it from the 2nd row:}\&\begin{bmatrix}6&13&-17&15\0&-45&51&-18\\end{bmatrix}\&\text{Next we scale the 2nd row of this new matrix by }-\frac{13}{45}\&\text{and subtract it from the 1st:}\&\begin{bmatrix}6&\frac{1}{562949953421312}&-\frac{34}{15}&\frac{49}{5}\0&-45&51&-18\\end{bmatrix}\&\text{Finally, we scale each row down to have one as the leading values:}\&\begin{bmatrix}1&0&-\frac{17}{45}&\frac{49}{30}\0&1&-\frac{17}{15}&\frac{2}{5}\\end{bmatrix}\end{align}$

Compare your answer with the correct one above

Question

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

Answer

$\begin{align}&\text{Reduced echelon form is a matrix form in which the leading coefficient}\&\text{of each nonzero row is always to the right of the leading coefficient}\&\text{of the row above it, each leading coefficient is the only nonzero}\&\text{value in its column, and each row is scaled down to give the}\&\text{leading coefficients values of one. Beginning with our matrix,}\&\text{we can convert it to this form by:}\&\text{Scaling the 1st row of the matrix by }\frac{11}{14}\&\text{and subtracting it from the 2nd row:}\&\begin{bmatrix}-14&13&2\0&-\frac{269}{14}&-\frac{116}{7}\\end{bmatrix}\&\text{Next we scale the 2nd row of this new matrix by }-\frac{182}{269}\&\text{and subtract it from the 1st:}\&\begin{bmatrix}-14&0&-\frac{2478}{269}\0&-\frac{269}{14}&-\frac{116}{7}\\end{bmatrix}\&\text{Finally, we scale each row down to have one as the leading values:}\&\begin{bmatrix}1&0&\frac{177}{269}\0&1&\frac{232}{269}\\end{bmatrix}\end{align}$

Compare your answer with the correct one above

Question

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

Answer

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

Compare your answer with the correct one above

Question

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

Answer

$\begin{align}&\text{Reduced echelon form is a matrix form in which the leading coefficient}\&\text{of each nonzero row is always to the right of the leading coefficient}\&\text{of the row above it, each leading coefficient is the only nonzero}\&\text{value in its column, and each row is scaled down to give the}\&\text{leading coefficients values of one. Beginning with our matrix,}\&\text{we can convert it to this form by:}\&\text{Scaling the 1st row of the matrix by }-\frac{10}{7}\&\text{and subtracting it from the 2nd row:}\&\begin{bmatrix}-7&14&13&19\0&10&\frac{81}{7}&\frac{260}{7}\\end{bmatrix}\&\text{Next we scale the 2nd row of this new matrix by }\frac{7}{5}\&\text{and subtract it from the 1st:}\&\begin{bmatrix}-7&0&-\frac{16}{5}&-33\0&10&\frac{81}{7}&\frac{260}{7}\\end{bmatrix}\&\text{Finally, we scale each row down to have one as the leading values:}\&\begin{bmatrix}1&0&\frac{16}{35}&\frac{33}{7}\0&1&\frac{81}{70}&\frac{26}{7}\\end{bmatrix}\end{align}$

Compare your answer with the correct one above

Question

Give the elementary matrix that represents performing the row operation

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

in solving a three-by-three linear system.

Answer

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.

Compare your answer with the correct one above

Tap the card to reveal the answer