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Linear Algebra : Reduced Row Echelon Form and Row Operations

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Example Questions

Example Question #11 : Linear Equations

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

Possible Answers:

$\begin{bmatrix}1&0&\frac{5}{9}\\0&1&\frac{1}{27}\\\end{bmatrix}$

$\begin{bmatrix}1&0&\frac{14}{23}\\0&1&-\frac{2}{23}\\\end{bmatrix}$

$\begin{bmatrix}1&0&\frac{2}{3}\\0&1&-\frac{2}{9}\\\end{bmatrix}$

$\begin{bmatrix}1&0&\frac{74}{131}\\0&1&\frac{2}{131}\\\end{bmatrix}$

Correct answer:

$\begin{bmatrix}1&0&\frac{74}{131}\\0&1&\frac{2}{131}\\\end{bmatrix}$

Explanation:

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

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Example Question #12 : Linear Equations

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

Possible Answers:

$\begin{bmatrix}1&0&-\frac{13}{23}\\0&1&\frac{26}{23}\\\end{bmatrix}$

$\begin{bmatrix}1&0&-\frac{52}{127}\\0&1&\frac{169}{127}\\\end{bmatrix}$

$\begin{bmatrix}1&0&-\frac{34}{79}\\0&1&\frac{103}{79}\\\end{bmatrix}$

$\begin{bmatrix}1&0&-\frac{8}{19}\\0&1&\frac{25}{19}\\\end{bmatrix}$

Correct answer:

$\begin{bmatrix}1&0&-\frac{34}{79}\\0&1&\frac{103}{79}\\\end{bmatrix}$

Explanation:

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

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Example Question #13 : Linear Equations

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

Possible Answers:

$\begin{bmatrix}1&0&-\frac{243}{61}\\0&1&-\frac{117}{61}\\\end{bmatrix}$

$\begin{bmatrix}1&0&-\frac{109}{88}\\0&1&-\frac{27}{176}\\\end{bmatrix}$

$\begin{bmatrix}1&0&\frac{129}{4}\\0&1&\frac{171}{8}\\\end{bmatrix}$

$\begin{bmatrix}1&0&-\frac{223}{153}\\0&1&-\frac{5}{17}\\\end{bmatrix}$

Correct answer:

$\begin{bmatrix}1&0&-\frac{109}{88}\\0&1&-\frac{27}{176}\\\end{bmatrix}$

Explanation:

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

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Example Question #14 : Linear Equations

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

Possible Answers:

$\begin{bmatrix}1&0&\frac{15}{173}&\frac{499}{346}\\0&1&-\frac{181}{173}&\frac{63}{173}\\\end{bmatrix}$

$\begin{bmatrix}1&0&\frac{81}{211}&\frac{557}{422}\\0&1&-\frac{209}{211}&\frac{72}{211}\\\end{bmatrix}$

$\begin{bmatrix}1&0&-\frac{17}{45}&\frac{49}{30}\\0&1&-\frac{17}{15}&\frac{2}{5}\\\end{bmatrix}$

$\begin{bmatrix}1&0&\frac{213}{287}&\frac{673}{574}\\0&1&-\frac{265}{287}&\frac{90}{287}\\\end{bmatrix}$

Correct answer:

$\begin{bmatrix}1&0&-\frac{17}{45}&\frac{49}{30}\\0&1&-\frac{17}{15}&\frac{2}{5}\\\end{bmatrix}$

Explanation:

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

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Example Question #11 : Reduced Row Echelon Form And Row Operations

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

Possible Answers:

$\begin{bmatrix}1&0&\frac{177}{269}\\0&1&\frac{232}{269}\\\end{bmatrix}$

$\begin{bmatrix}1&0&3\\0&1&\frac{13}{11}\\\end{bmatrix}$

$\begin{bmatrix}1&0&\frac{167}{219}\\0&1&\frac{64}{73}\\\end{bmatrix}$

$\begin{bmatrix}1&0&\frac{197}{369}\\0&1&\frac{104}{123}\\\end{bmatrix}$

Correct answer:

$\begin{bmatrix}1&0&\frac{177}{269}\\0&1&\frac{232}{269}\\\end{bmatrix}$

Explanation:

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

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Example Question #12 : Reduced Row Echelon Form And Row Operations

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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Example Question #17 : Linear Equations

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

Possible Answers:

$\begin{bmatrix}1&0&\frac{16}{35}&\frac{33}{7}\\0&1&\frac{81}{70}&\frac{26}{7}\\\end{bmatrix}$

$\begin{bmatrix}1&0&\frac{52}{275}&\frac{81}{55}\\0&1&\frac{266}{275}&\frac{78}{55}\\\end{bmatrix}$

$\begin{bmatrix}1&0&\frac{28}{29}&\frac{315}{29}\\0&1&\frac{44}{29}&\frac{234}{29}\\\end{bmatrix}$

$\begin{bmatrix}1&0&0&-\frac{35}{43}\\0&1&\frac{5}{6}&-\frac{26}{129}\\\end{bmatrix}$

Correct answer:

$\begin{bmatrix}1&0&\frac{16}{35}&\frac{33}{7}\\0&1&\frac{81}{70}&\frac{26}{7}\\\end{bmatrix}$

Explanation:

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

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Example Question #18 : Linear Equations

Give the elementary matrix that represents performing the row operation

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

in solving a three-by-three linear system.

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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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Example Question #11 : Reduced Row Echelon Form And Row Operations

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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:

1) 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.

2) 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.

3) 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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Example Question #20 : Reduced Row Echelon Form And Row Operations

Which of the following elementary matrices represents the row operation

$- \frac{2}{5}R1+R4 \rightarrow R4$

on a four-by-four system?

Possible Answers:

$\begin{bmatrix} 1& 0& 0&-\frac{2}{5}\\ 0&1 & 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1 \end{bmatrix}$

$\begin{bmatrix}-\frac{2}{5}& 0& 0& 0\\ 0&1 & 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1 \end{bmatrix}$

$\begin{bmatrix} 1& 0& 0& 0\\ 0&1 & 0& 0\\ 0& 0& 1& 0\\ -\frac{2}{5}& 0& 0& 1 \end{bmatrix}$

None of the other choices gives the correct response.

$\begin{bmatrix} 1& 0& 0& 0\\ 0&1 & 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0&-\frac{2}{5} \end{bmatrix}$

Correct answer:

$\begin{bmatrix} 1& 0& 0& 0\\ 0&1 & 0& 0\\ 0& 0& 1& 0\\ -\frac{2}{5}& 0& 0& 1 \end{bmatrix}$

Explanation:

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

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

by way of exactly one row operation. $- \frac{2}{5}R1+R4 \rightarrow R4$ represents the addition of $- \frac{2}{5}$ times each element in Row 1 to the corresponding element in Row 4, so do this in the identity matrix:

$\begin{bmatrix} 1& 0& 0& 0\\ 0&1 & 0& 0\\ 0& 0& 1& 0\\ -\frac{2}{5} \cdot 1 + 0& -\frac{2}{5} \cdot 0+ -\frac{2}{5} \cdot 0+0& 0&-\frac{2}{5} \cdot 0+ 1 \end{bmatrix}$

$=\begin{bmatrix} 1& 0& 0& 0\\ 0&1 & 0& 0\\ 0& 0& 1& 0\\ -\frac{2}{5}& 0& 0& 1 \end{bmatrix}$ ,

the correct choice.

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Linear Algebra : Reduced Row Echelon Form and Row Operations

Study concepts, example questions & explanations for Linear Algebra

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Example Question #11 : Reduced Row Echelon Form And Row Operations

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