Row Operations and Equivalent Systems

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Linear Algebra › Row Operations and Equivalent Systems

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

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

CORRECT

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

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

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

Explanation

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

$\begin{align*}&\text{Which of the following matrices is not equivalent to}\&\begin{bmatrix}4&14&-11&18\-8&-15&7&12\\end{bmatrix}?\end{align*}$

$\begin{bmatrix}7&17&-8&21\-5&-12&10&15\\end{bmatrix}$

CORRECT

$\begin{bmatrix}2&7&-\frac{11}{2}&9\-4&-\frac{15}{2}&\frac{7}{2}&6\\end{bmatrix}$

$\begin{bmatrix}2&7&-\frac{11}{2}&9\48&90&-42&-72\\end{bmatrix}$

$\begin{bmatrix}12&29&-18&6\-8&-15&7&12\\end{bmatrix}$

Explanation

$\begin{align*}&\text{Two matrices are equivalent if one can be turned into another}\&\text{via row operations: scaling a row by multiplication of some}\&\text{constant, addition/subtraction of one row with another, or a}\&\text{combination of these processes..}\&\text{The matrix: }\begin{bmatrix}2&7&-\frac{11}{2}&9\-4&-\frac{15}{2}&\frac{7}{2}&6\\end{bmatrix}\text{ is our entire initial matrix scaled by }\frac{1}{2}.\&\text{The matrix }\begin{bmatrix}2&7&-\frac{11}{2}&9\48&90&-42&-72\\end{bmatrix}\text{ can be formed by scaling the first and second rows of}\&\text{the orignal by }\frac{1}{2}\text{ and }-6.\&\text{Finally, subtracting one row from another creates }\begin{bmatrix}12&29&-18&6\-8&-15&7&12\\end{bmatrix}.\&\text{No row operations, however, will create }\begin{bmatrix}7&17&-8&21\-5&-12&10&15\\end{bmatrix}.\end{align*}$

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

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

CORRECT

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

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

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

Explanation

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

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

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

CORRECT

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

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

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

Explanation

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

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

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

CORRECT

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

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

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

Explanation

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

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

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

CORRECT

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

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

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

Explanation

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

Which matrix is equivalent to

$\begin{pmatrix} 3&6 \ 5&7 \end{pmatrix}$

$\begin{pmatrix} 3&6 \ 8&13 \end{pmatrix}$

CORRECT

$\begin{pmatrix} 3&6 \ 8&1 \end{pmatrix}$

$\begin{pmatrix} 5&13 \ 4&2 \end{pmatrix}$

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

Explanation

By adding row one to row two we get

$\begin{pmatrix} 3&6 \ 5+3&7+6 \end{pmatrix}$

$=\begin{pmatrix} 3&6 \ 8&13 \end{pmatrix}$

Which matrix is equivalent to

$\begin{pmatrix} 1& 2&5 &-3 \ 2&4 &6 &8 \end{pmatrix}$

$\begin{pmatrix} 1& 2&5 &-3 \ 1&2 &1 &11 \end{pmatrix}$

CORRECT

$\begin{pmatrix} 1& 2&5 &-3 \ 3&6 &11 &11 \end{pmatrix}$

$\begin{pmatrix} 1& 2&5 &3 \ 2&4 &6 &8 \end{pmatrix}$

$\begin{pmatrix} 3& -2&11 &5 \ 2&4 &6 &8 \end{pmatrix}$

Explanation

By adding times row one to row two we get

$\begin{pmatrix} 1& 2&5 &-3 \ 2+(-1)&4+(-2) &6+(-5) &8+3 \end{pmatrix}$

$=\begin{pmatrix} 1& 2&5 &-3 \ 1&2 &1 &11 \end{pmatrix}$

Which matrix is equivalent to

$\begin{pmatrix} 3&6 \ 9&4 \end{pmatrix}$

$\begin{pmatrix} 12&10 \ 9&4 \end{pmatrix}$

CORRECT

$\begin{pmatrix} 3&9 \ 6&4 \end{pmatrix}$

$\begin{pmatrix} 4&9 \ 6&3 \end{pmatrix}$

$\begin{pmatrix} 4&6 \ 9&3 \end{pmatrix}$

Explanation

We add row two to row one to get

$\begin{pmatrix} 3+9&6+4 \ 9&4 \end{pmatrix}=\begin{pmatrix} 12&10 \ 9&4 \end{pmatrix}$

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

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

CORRECT

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

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

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

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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