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Linear Algebra : Linear Algebra

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Example Questions

Example Question #65 : Linear Equations

$\begin{align*}&\text{Among the choices given, find the matrix that is not }A=\begin{bmatrix}-13&11&17&-10\\-12&-15&-5&-2\\\end{bmatrix}\\&\text{in an equivalent configuration.}\end{align*}$

Possible Answers:

$\begin{bmatrix}\frac{13}{4}&-\frac{11}{4}&-\frac{17}{4}&\frac{5}{2}\\3&\frac{15}{4}&\frac{5}{4}&\frac{1}{2}\\\end{bmatrix}$

$\begin{bmatrix}-1&26&22&-8\\-12&-15&-5&-2\\\end{bmatrix}$

$\begin{bmatrix}-14&10&16&-11\\-13&-16&-6&-3\\\end{bmatrix}$

$\begin{bmatrix}\frac{13}{3}&-\frac{11}{3}&-\frac{17}{3}&\frac{10}{3}\\-2&-\frac{5}{2}&-\frac{5}{6}&-\frac{1}{3}\\\end{bmatrix}$

Correct answer:

$\begin{bmatrix}-14&10&16&-11\\-13&-16&-6&-3\\\end{bmatrix}$

Explanation:

$\begin{align*}&\text{Two matrices are equivalent if one can be turned into another}\\&\text{via row operations: scaling a row by multiplication of some}\\&\text{constant, addition/subtraction of one row with another, or a}\\&\text{combination of these processes..}\\&\text{The matrix: }\begin{bmatrix}\frac{13}{4}&-\frac{11}{4}&-\frac{17}{4}&\frac{5}{2}\\3&\frac{15}{4}&\frac{5}{4}&\frac{1}{2}\\\end{bmatrix}\text{ is our entire initial matrix scaled by }-\frac{1}{4}.\\&\text{The matrix }\begin{bmatrix}\frac{13}{3}&-\frac{11}{3}&-\frac{17}{3}&\frac{10}{3}\\-2&-\frac{5}{2}&-\frac{5}{6}&-\frac{1}{3}\\\end{bmatrix}\text{ can be formed by scaling the first and second rows of}\\&\text{the orignal by }-\frac{1}{3}\text{ and }\frac{1}{6}.\\&\text{Finally, subtracting one row from another creates }\begin{bmatrix}-1&26&22&-8\\-12&-15&-5&-2\\\end{bmatrix}.\\&\text{No row operations, however, will create }\begin{bmatrix}-14&10&16&-11\\-13&-16&-6&-3\\\end{bmatrix}.\end{align*}$

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Example Question #71 : Linear Equations

$\begin{align*}&\text{Which of the following matrices is not equivalent to}\\&\begin{bmatrix}-12&-1&8\\10&3&3\\\end{bmatrix}?\end{align*}$

Possible Answers:

$\begin{bmatrix}-4&-\frac{1}{3}&\frac{8}{3}\\50&15&15\\\end{bmatrix}$

$\begin{bmatrix}-12&-1&8\\22&4&-5\\\end{bmatrix}$

$\begin{bmatrix}-14&-3&6\\8&1&1\\\end{bmatrix}$

$\begin{bmatrix}-6&-\frac{1}{2}&4\\5&\frac{3}{2}&\frac{3}{2}\\\end{bmatrix}$

Correct answer:

$\begin{bmatrix}-14&-3&6\\8&1&1\\\end{bmatrix}$

Explanation:

$\begin{align*}&\text{Two matrices are equivalent if one can be turned into another}\\&\text{via row operations: scaling a row by multiplication of some}\\&\text{constant, addition/subtraction of one row with another, or a}\\&\text{combination of these processes..}\\&\text{The matrix: }\begin{bmatrix}-6&-\frac{1}{2}&4\\5&\frac{3}{2}&\frac{3}{2}\\\end{bmatrix}\text{ is our entire initial matrix scaled by }\frac{1}{2}.\\&\text{The matrix }\begin{bmatrix}-4&-\frac{1}{3}&\frac{8}{3}\\50&15&15\\\end{bmatrix}\text{ can be formed by scaling the first and second rows of}\\&\text{the orignal by }\frac{1}{3}\text{ and }5.\\&\text{Finally, subtracting one row from another creates }\begin{bmatrix}-12&-1&8\\22&4&-5\\\end{bmatrix}.\\&\text{No row operations, however, will create }\begin{bmatrix}-14&-3&6\\8&1&1\\\end{bmatrix}.\end{align*}$

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Example Question #3 : Row Operations And Equivalent Systems

$\begin{align*}&\text{Which of the following matrices is not equivalent to}\\&\begin{bmatrix}-12&9&19\\3&-4&12\\\end{bmatrix}?\end{align*}$

Possible Answers:

$\begin{bmatrix}\frac{12}{5}&-\frac{9}{5}&-\frac{19}{5}\\-\frac{3}{5}&\frac{4}{5}&-\frac{12}{5}\\\end{bmatrix}$

$\begin{bmatrix}-15&13&7\\3&-4&12\\\end{bmatrix}$

$\begin{bmatrix}-10&11&21\\5&-2&14\\\end{bmatrix}$

$\begin{bmatrix}6&-\frac{9}{2}&-\frac{19}{2}\\-\frac{1}{2}&\frac{2}{3}&-2\\\end{bmatrix}$

Correct answer:

$\begin{bmatrix}-10&11&21\\5&-2&14\\\end{bmatrix}$

Explanation:

$\begin{align*}&\text{Two matrices are equivalent if one can be turned into another}\\&\text{via row operations: scaling a row by multiplication of some}\\&\text{constant, addition/subtraction of one row with another, or a}\\&\text{combination of these processes..}\\&\text{The matrix: }\begin{bmatrix}\frac{12}{5}&-\frac{9}{5}&-\frac{19}{5}\\-\frac{3}{5}&\frac{4}{5}&-\frac{12}{5}\\\end{bmatrix}\text{ is our entire initial matrix scaled by }-\frac{1}{5}.\\&\text{The matrix }\begin{bmatrix}6&-\frac{9}{2}&-\frac{19}{2}\\-\frac{1}{2}&\frac{2}{3}&-2\\\end{bmatrix}\text{ can be formed by scaling the first and second rows of}\\&\text{the orignal by }-\frac{1}{2}\text{ and }-\frac{1}{6}.\\&\text{Finally, subtracting one row from another creates }\begin{bmatrix}-15&13&7\\3&-4&12\\\end{bmatrix}.\\&\text{No row operations, however, will create }\begin{bmatrix}-10&11&21\\5&-2&14\\\end{bmatrix}.\end{align*}$

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Example Question #71 : Linear Equations

$\begin{align*}&\text{Identify the matrix which is not a equivalent to:}\\&\begin{bmatrix}19&6&-10\\-1&-12&7\\\end{bmatrix}\end{align*}$

Possible Answers:

$\begin{bmatrix}\frac{19}{2}&3&-5\\\frac{1}{6}&2&-\frac{7}{6}\\\end{bmatrix}$

$\begin{bmatrix}-38&-12&20\\2&24&-14\\\end{bmatrix}$

$\begin{bmatrix}20&18&-17\\-1&-12&7\\\end{bmatrix}$

$\begin{bmatrix}22&9&-7\\2&-9&10\\\end{bmatrix}$

Correct answer:

$\begin{bmatrix}22&9&-7\\2&-9&10\\\end{bmatrix}$

Explanation:

$\begin{align*}&\text{Two matrices are equivalent if one can be turned into another}\\&\text{via row operations: scaling a row by multiplication of some}\\&\text{constant, addition/subtraction of one row with another, or a}\\&\text{combination of these processes.}\\&\text{The matrix: }\begin{bmatrix}-38&-12&20\\2&24&-14\\\end{bmatrix}\text{ is our entire initial matrix scaled by }-2.\\&\text{The matrix }\begin{bmatrix}\frac{19}{2}&3&-5\\\frac{1}{6}&2&-\frac{7}{6}\\\end{bmatrix}\text{ can be formed by scaling the first and second rows of}\\&\text{the orignal by }\frac{1}{2}\text{ and }-\frac{1}{6}.\\&\text{Finally, subtracting one row from another creates }\begin{bmatrix}20&18&-17\\-1&-12&7\\\end{bmatrix}.\\&\text{No row operations, however, will create }\begin{bmatrix}22&9&-7\\2&-9&10\\\end{bmatrix}.\end{align*}$

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Example Question #632 : Linear Algebra

$\begin{align*}&\text{Among the choices given, find the matrix that is not }A=\begin{bmatrix}17&4&-2&7\\-5&14&19&-15\\\end{bmatrix}\\&\text{in an equivalent configuration.}\end{align*}$

Possible Answers:

$\begin{bmatrix}-68&-16&8&-28\\20&-56&-76&60\\\end{bmatrix}$

$\begin{bmatrix}34&8&-4&14\\-30&84&114&-90\\\end{bmatrix}$

$\begin{bmatrix}18&5&-1&8\\-4&15&20&-14\\\end{bmatrix}$

$\begin{bmatrix}22&-10&-21&22\\-5&14&19&-15\\\end{bmatrix}$

Correct answer:

$\begin{bmatrix}18&5&-1&8\\-4&15&20&-14\\\end{bmatrix}$

Explanation:

$\begin{align*}&\text{Two matrices are equivalent if one can be turned into another}\\&\text{via row operations: scaling a row by multiplication of some}\\&\text{constant, addition/subtraction of one row with another, or a}\\&\text{combination of these processes.}\\&\text{The matrix: }\begin{bmatrix}-68&-16&8&-28\\20&-56&-76&60\\\end{bmatrix}\text{ is our entire initial matrix scaled by }-4.\\&\text{The matrix }\begin{bmatrix}34&8&-4&14\\-30&84&114&-90\\\end{bmatrix}\text{ can be formed by scaling the first and second rows of}\\&\text{the orignal by }2\text{ and }6.\\&\text{Finally, subtracting one row from another creates }\begin{bmatrix}22&-10&-21&22\\-5&14&19&-15\\\end{bmatrix}.\\&\text{No row operations, however, will create }\begin{bmatrix}18&5&-1&8\\-4&15&20&-14\\\end{bmatrix}.\end{align*}$

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Example Question #11 : Row Operations And Equivalent Systems

$\begin{align*}&\text{Which of the following matrices is not equivalent to}\\&\begin{bmatrix}2&10&20\\6&6&-11\\\end{bmatrix}?\end{align*}$

Possible Answers:

$\begin{bmatrix}8&40&80\\24&24&-44\\\end{bmatrix}$

$\begin{bmatrix}-\frac{2}{3}&-\frac{10}{3}&-\frac{20}{3}\\\frac{3}{2}&\frac{3}{2}&-\frac{11}{4}\\\end{bmatrix}$

$\begin{bmatrix}-4&4&31\\6&6&-11\\\end{bmatrix}$

$\begin{bmatrix}1&9&19\\5&5&-12\\\end{bmatrix}$

Correct answer:

$\begin{bmatrix}1&9&19\\5&5&-12\\\end{bmatrix}$

Explanation:

$\begin{align*}&\text{Two matrices are equivalent if one can be turned into another}\\&\text{via row operations: scaling a row by multiplication of some}\\&\text{constant, addition/subtraction of one row with another, or a}\\&\text{combination of these processes.}\\&\text{The matrix: }\begin{bmatrix}8&40&80\\24&24&-44\\\end{bmatrix}\text{ is our entire initial matrix scaled by }4.\\&\text{The matrix }\begin{bmatrix}-\frac{2}{3}&-\frac{10}{3}&-\frac{20}{3}\\\frac{3}{2}&\frac{3}{2}&-\frac{11}{4}\\\end{bmatrix}\text{ can be formed by scaling the first and second rows of}\\&\text{the orignal by }-\frac{1}{3}\text{ and }\frac{1}{4}.\\&\text{Finally, subtracting one row from another creates }\begin{bmatrix}-4&4&31\\6&6&-11\\\end{bmatrix}.\\&\text{No row operations, however, will create }\begin{bmatrix}1&9&19\\5&5&-12\\\end{bmatrix}.\end{align*}$

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Example Question #13 : Row Operations And Equivalent Systems

$\begin{align*}&\text{Among the choices given, find the matrix that is not }A=\begin{bmatrix}-13&-15&16&-14\\8&16&-13&3\\\end{bmatrix}\\&\text{in an equivalent configuration.}\end{align*}$

Possible Answers:

$\begin{bmatrix}-\frac{13}{3}&-5&\frac{16}{3}&-\frac{14}{3}\\-32&-64&52&-12\\\end{bmatrix}$

$\begin{bmatrix}-21&-31&29&-17\\8&16&-13&3\\\end{bmatrix}$

$\begin{bmatrix}52&60&-64&56\\-32&-64&52&-12\\\end{bmatrix}$

$\begin{bmatrix}-16&-18&13&-17\\5&13&-16&0\\\end{bmatrix}$

Correct answer:

$\begin{bmatrix}-16&-18&13&-17\\5&13&-16&0\\\end{bmatrix}$

Explanation:

$\begin{align*}&\text{Two matrices are equivalent if one can be turned into another}\\&\text{via row operations: scaling a row by multiplication of some}\\&\text{constant, addition/subtraction of one row with another, or a}\\&\text{combination of these processes.}\\&\text{The matrix: }\begin{bmatrix}52&60&-64&56\\-32&-64&52&-12\\\end{bmatrix}\text{ is our entire initial matrix scaled by }-4.\\&\text{The matrix }\begin{bmatrix}-\frac{13}{3}&-5&\frac{16}{3}&-\frac{14}{3}\\-32&-64&52&-12\\\end{bmatrix}\text{ can be formed by scaling the first and second rows of}\\&\text{the orignal by }\frac{1}{3}\text{ and }-4.\\&\text{Finally, subtracting one row from another creates }\begin{bmatrix}-21&-31&29&-17\\8&16&-13&3\\\end{bmatrix}.\\&\text{No row operations, however, will create }\begin{bmatrix}-16&-18&13&-17\\5&13&-16&0\\\end{bmatrix}.\end{align*}$

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Example Question #71 : Linear Equations

$\begin{align*}&\text{Which of the following matrices is not equivalent to}\\&\begin{bmatrix}-3&-7&-15\\15&-15&18\\\end{bmatrix}?\end{align*}$

Possible Answers:

$\begin{bmatrix}-3&-7&-15\\18&-8&33\\\end{bmatrix}$

$\begin{bmatrix}-2&-6&-14\\16&-14&19\\\end{bmatrix}$

$\begin{bmatrix}9&21&45\\-45&45&-54\\\end{bmatrix}$

$\begin{bmatrix}-1&-\frac{7}{3}&-5\\-60&60&-72\\\end{bmatrix}$

Correct answer:

$\begin{bmatrix}-2&-6&-14\\16&-14&19\\\end{bmatrix}$

Explanation:

$\begin{align*}&\text{Two matrices are equivalent if one can be turned into another}\\&\text{via row operations: scaling a row by multiplication of some}\\&\text{constant, addition/subtraction of one row with another, or a}\\&\text{combination of these processes.}\\&\text{The matrix: }\begin{bmatrix}9&21&45\\-45&45&-54\\\end{bmatrix}\text{ is our entire initial matrix scaled by }-3.\\&\text{The matrix }\begin{bmatrix}-1&-\frac{7}{3}&-5\\-60&60&-72\\\end{bmatrix}\text{ can be formed by scaling the first and second rows of}\\&\text{the orignal by }\frac{1}{3}\text{ and }-4.\\&\text{Finally, subtracting one row from another creates }\begin{bmatrix}-3&-7&-15\\18&-8&33\\\end{bmatrix}.\\&\text{No row operations, however, will create }\begin{bmatrix}-2&-6&-14\\16&-14&19\\\end{bmatrix}.\end{align*}$

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Example Question #1 : Homogeneous Cases

Solve the system by reducing its corresponding matrix.

x_1+3x_2=0

4x_1+12x_2=0

Possible Answers:

x_1=-3x_2 and x_2 is a free variable.

x_1=4x_2 and x_2 is a free variable.

x_1=-x_2 and x_2 is a free variable.

Correct answer:

x_1=-3x_2 and x_2 is a free variable.

Explanation:

The corresponding matrix is

$\begin{pmatrix} 1& 3& 0\\ 4& 12& 0 \end{pmatrix}$

We add times row one to row two.

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

This yields the solution

x_1+3x_2=0

x_1=-3x_2 and x_2 is a free variable.

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Example Question #1 : Homogeneous Cases

$\begin{align*}&\text{Find the nontrivial solutions for the system of equations: }\\&\begin{bmatrix}-10&2&-11\\-20&11&-20\\\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0\\0\\\end{bmatrix}\end{align*}$

Possible Answers:

$\begin{align*}&x=-\frac{(729z)}{70}\\&y=\frac{(18z)}{7}\end{align*}$

$\begin{align*}&x=-\frac{(81z)}{70}\\&y=\frac{(2z)}{7}\end{align*}$

$\begin{align*}&x=-\frac{(243z)}{35}\\&y=-\frac{(8z)}{7}\end{align*}$

$\begin{align*}&x=-\frac{(81z)}{35}\\&y=\frac{(6z)}{7}\end{align*}$

Correct answer:

$\begin{align*}&x=-\frac{(81z)}{70}\\&y=\frac{(2z)}{7}\end{align*}$

Explanation:

$\begin{align*}&\text{When faced with a system of equations like the}\\&\text{one represented by }\\&\begin{bmatrix}-10&2&-11\\-20&11&-20\\\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0\\0\\\end{bmatrix}\\&\text{We can manipulate coefficients to eliminate}\\&\text{variables in each row. That is to say, to reduce it to echelon form.}\\&\text{Reducing our matrix }\begin{bmatrix}-10&2&-11\\-20&11&-20\\\end{bmatrix}\\&\text{We find the reduced echelon form: }\begin{bmatrix}1&0&\frac{81}{70}\\0&1&\frac{2}{7}\\\end{bmatrix}\\&\text{Since each row is equal to zero, we can therefore relate variables to find}\\&\text{nontrivial solutions:}\\&x=-\frac{(81z)}{70}\\&y=-\frac{(2z)}{7}\end{align*}$

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Linear Algebra : Linear Algebra

Study concepts, example questions & explanations for Linear Algebra

All Linear Algebra Resources

Example Questions

Example Question #65 : Linear Equations

Example Question #71 : Linear Equations

Example Question #3 : Row Operations And Equivalent Systems

Example Question #71 : Linear Equations

Example Question #632 : Linear Algebra

Example Question #11 : Row Operations And Equivalent Systems

Example Question #13 : Row Operations And Equivalent Systems

Example Question #71 : Linear Equations

Example Question #1 : Homogeneous Cases

Example Question #1 : Homogeneous Cases

All Linear Algebra Resources

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