Reduced Row Echelon Form and Row Operations - Linear Algebra

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Question

Find the inverse using row operations

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

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

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Question

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

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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

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

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Question

Change the following matrix into reduced row echelon form.

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

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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

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

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Question

Change the following matrix into reduced row echelon form.

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

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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.

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

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Question

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

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

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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.

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Question

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

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Answer

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

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

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Question

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

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

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Question

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

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Answer

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

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Question

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

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Answer

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

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Question

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

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Answer

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

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Question

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

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Answer

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

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Question

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

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Answer

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

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Question

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

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Answer

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

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Question

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

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Answer

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

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Question

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

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Answer

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

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Question

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

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Answer

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

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Question

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

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Answer

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

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Question

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

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Answer

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

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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.

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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.

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Question

Which of the following is an example of an elementary matrix?

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Answer

An elementary matrix is one that can be formed from the (here, three-by-three) identity matrix

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

by way of exactly one row operation. An elementary matrix can have one of the following characteristics:

Exactly two rows are switched. The only choice that repositions rows is

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

but this choice rearranges all three rows, so this is incorrect.

All of the "1" elements in the diagonal remain unchanged, but exactly one "0" is changed to a nonzero number. The choice that leaves all "1" elements unchanged is

$\begin{bmatrix} 1& 2& -6\ 0& 1& 0\ 0& 0& 1 \end{bmatrix}$

but this matrix changes two of the zeroes to nonzero elements.

One of the "1" elements in the diagonal is changed to another nonzero element. The other three choices change these elements. But, of them:

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

changes two of the "1" elements to other nonzero numbers, and

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

also changes a "0" to a nonzero number.

$\begin{bmatrix} 4 & 0& 0\ 0& 1& 0\ 0& 0& 1 \end{bmatrix}$ ,

however, makes one such change and no others, so it is an elementary matrix, and it is the correct choice.

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