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Linear Algebra : Linear Equations

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Example Questions

Example Question #1 : Reduced Row Echelon Form And Row Operations

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

$\begin{bmatrix} 2 &-2 &5 &1 &0 & 0\\ -2 &1 &-3 &0 &1 &0 \\ 1& 0& 1& 0&0 &1 \end{bmatrix} \sim$ 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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Example Question #1 : Linear Equations

Find the inverse using row operations

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

Possible Answers:

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

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

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

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

Correct answer:

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

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

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Example Question #2 : Linear Equations

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

Possible Answers:

$\begin{bmatrix} 1 &-1 &0 \\ 12& 14& -13\\-14 & 9& 6\end{bmatrix}$

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

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

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

Correct answer:

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

Explanation:

$\begin{bmatrix} 3 &0 &4 &1 &0 &0 \\2 &-2 &5 &0 &1 &0 \\ 4& 3& 2& 0& 0& 1\end{bmatrix} \sim$ 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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Example Question #3 : Reduced Row Echelon Form And Row Operations

Change the following matrix into reduced row echelon form.

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

Possible Answers:

$\begin{pmatrix} 1&0 &3 \\ 0&1 &4 \end{pmatrix}$

$\begin{pmatrix} 0&1 &4 \\ 1&0 &2 \end{pmatrix}$

$\begin{pmatrix} 1&0 &\frac{9}{22} \\ 0&1 &\frac{4}{11} \end{pmatrix}$

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

Correct answer:

$\begin{pmatrix} 1&0 &\frac{9}{22} \\ 0&1 &\frac{4}{11} \end{pmatrix}$

Explanation:

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

Change the following matrix into reduced row echelon form.

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

Possible Answers:

$\begin{pmatrix} 4&8 &12 \\ 0&0 &1 \end{pmatrix}$

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

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

$\begin{pmatrix} 1&0 &2 \\ 0&1 &4 \end{pmatrix}$

Correct answer:

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

Explanation:

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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Example Question #1 : Reduced Row Echelon Form And Row Operations

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

Possible Answers:

$\begin{bmatrix}1&0&\frac{34}{7}\\0&1&\frac{1}{7}\\\end{bmatrix}$

$\begin{bmatrix}1&0&-\frac{2}{7}\\0&1&-\frac{5}{7}\\\end{bmatrix}$

$\begin{bmatrix}1&0&-\frac{22}{35}\\0&1&-\frac{27}{35}\\\end{bmatrix}$

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

Correct answer:

$\begin{bmatrix}1&0&-\frac{22}{35}\\0&1&-\frac{27}{35}\\\end{bmatrix}$

Explanation:

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

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Example Question #2 : Reduced Row Echelon Form And Row Operations

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

Possible Answers:

$\begin{bmatrix}1&0&-\frac{11}{31}&-\frac{27}{31}\\0&1&-\frac{63}{403}&-\frac{25}{403}\\\end{bmatrix}$

$\begin{bmatrix}1&0&-\frac{317}{229}&-\frac{501}{229}\\0&1&-\frac{63}{229}&-\frac{49}{229}\\\end{bmatrix}$

$\begin{bmatrix}1&0&-\frac{9}{173}&-\frac{251}{519}\\0&1&-\frac{21}{173}&-\frac{3}{173}\\\end{bmatrix}$

$\begin{bmatrix}1&0&-\frac{37}{41}&-\frac{11}{7}\\0&1&-\frac{9}{41}&-\frac{1}{7}\\\end{bmatrix}$

Correct answer:

$\begin{bmatrix}1&0&-\frac{37}{41}&-\frac{11}{7}\\0&1&-\frac{9}{41}&-\frac{1}{7}\\\end{bmatrix}$

Explanation:

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

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Example Question #1 : Reduced Row Echelon Form And Row Operations

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

Possible Answers:

$\begin{bmatrix}1&0&-\frac{28}{33}\\0&1&\frac{4}{15}\\\end{bmatrix}$

$\begin{bmatrix}1&0&\frac{112}{129}\\0&1&\frac{278}{129}\\\end{bmatrix}$

$\begin{bmatrix}1&0&-7\\0&1&-\frac{13}{2}\\\end{bmatrix}$

$\begin{bmatrix}1&0&-\frac{128}{39}\\0&1&-\frac{94}{39}\\\end{bmatrix}$

Correct answer:

$\begin{bmatrix}1&0&-\frac{128}{39}\\0&1&-\frac{94}{39}\\\end{bmatrix}$

Explanation:

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

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Example Question #2 : Reduced Row Echelon Form And Row Operations

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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

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

Possible Answers:

$\begin{bmatrix}1&0&\frac{19}{35}\\0&1&-\frac{59}{49}\\\end{bmatrix}$

$\begin{bmatrix}1&0&-\frac{31}{50}\\0&1&-\frac{49}{20}\\\end{bmatrix}$

$\begin{bmatrix}1&0&-\frac{179}{13}\\0&1&-\frac{215}{13}\\\end{bmatrix}$

$\begin{bmatrix}1&0&\frac{59}{23}\\0&1&\frac{155}{161}\\\end{bmatrix}$

Correct answer:

$\begin{bmatrix}1&0&-\frac{179}{13}\\0&1&-\frac{215}{13}\\\end{bmatrix}$

Explanation:

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

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Linear Algebra : Linear Equations

Study concepts, example questions & explanations for Linear Algebra

All Linear Algebra Resources

Example Questions

Example Question #1 : Reduced Row Echelon Form And Row Operations

Example Question #1 : Linear Equations

Example Question #2 : Linear Equations

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

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

Example Question #1 : Reduced Row Echelon Form And Row Operations

Example Question #2 : Reduced Row Echelon Form And Row Operations

Example Question #1 : Reduced Row Echelon Form And Row Operations

Example Question #2 : Reduced Row Echelon Form And Row Operations

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

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