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Linear Algebra : Homogeneous Cases

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Example Questions

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:

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

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

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

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

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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Example Question #2 : Homogeneous Cases

$\begin{align*}&\text{Find nonzero solutions for }\begin{bmatrix}x\\y\\z\end{bmatrix}\\&\text{for the system: }\begin{bmatrix}4&7&3\\17&6&1\\\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0\\0\\\end{bmatrix}\end{align*}$

Possible Answers:

$\begin{align*}&x=-\frac{(33z)}{95}\\&y=\frac{(94z)}{95}\end{align*}$

$\begin{align*}&x=\frac{(11z)}{95}\\&y=\frac{(47z)}{95}\end{align*}$

$\begin{align*}&x=-\frac{(77z)}{95}\\&y=\frac{(47z)}{665}\end{align*}$

$\begin{align*}&x=-\frac{(11z)}{570}\\&y=\frac{(47z)}{570}\end{align*}$

Correct answer:

$\begin{align*}&x=\frac{(11z)}{95}\\&y=\frac{(47z)}{95}\end{align*}$

Explanation:

$\begin{align*}&\text{When faced with a system of equations like the}\\&\text{one represented by }\\&\begin{bmatrix}4&7&3\\17&6&1\\\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0\\0\\\end{bmatrix}\\&\text{We can manipulate coefficients to eliminate}\\&\text{variables in each row. That is to say, to reduce it to echelon form.}\\&\text{Reducing our matrix }\begin{bmatrix}4&7&3\\17&6&1\\\end{bmatrix}\\&\text{We find the reduced echelon form: }\begin{bmatrix}1&0&-\frac{11}{95}\\0&1&\frac{47}{95}\\\end{bmatrix}\\&\text{Since each row is equal to zero, we can therefore relate variables to find}\\&\text{nontrivial solutions:}\\&x=\frac{(11z)}{95}\\&y=-\frac{(47z)}{95}\end{align*}$

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Example Question #3 : Homogeneous Cases

$\begin{align*}&\text{Find nontrivial solutions of }\begin{bmatrix}x\\y\\z\end{bmatrix}\\&\text{for the function: }\begin{bmatrix}-14&0&3\\-13&-7&9\\\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0\\0\\\end{bmatrix}\end{align*}$

$\begin{align*}&x=\frac{(3z)}{14}\\&y=-\frac{(87z)}{98}\end{align*}$

Possible Answers:

$\begin{align*}&x=-\frac{(3z)}{7}\\&y=-\frac{(261z)}{98}\end{align*}$

$\begin{align*}&x=-\frac{(12z)}{7}\\&y=-\frac{(348z)}{49}\end{align*}$

$\begin{align*}&x=-\frac{(3z)}{56}\\&y=-\frac{(29z)}{196}\end{align*}$

Correct answer:

$\begin{align*}&x=-\frac{(3z)}{7}\\&y=-\frac{(261z)}{98}\end{align*}$

Explanation:

$\begin{align*}&\text{When faced with a system of equations like the}\\&\text{one represented by }\\&\begin{bmatrix}-14&0&3\\-13&-7&9\\\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0\\0\\\end{bmatrix}\\&\text{We can manipulate coefficients to eliminate}\\&\text{variables in each row. That is to say, to reduce it to echelon form.}\\&\text{Reducing our matrix }\begin{bmatrix}-14&0&3\\-13&-7&9\\\end{bmatrix}\\&\text{We find the reduced echelon form: }\begin{bmatrix}1&0&-\frac{3}{14}\\0&1&-\frac{87}{98}\\\end{bmatrix}\\&\text{Since each row is equal to zero, we can therefore relate variables to find}\\&\text{nontrivial solutions:}\\&x=\frac{(3z)}{14}\\&y=\frac{(87z)}{98}\end{align*}$

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Example Question #2 : Homogeneous Cases

$\begin{align*}&\text{Find the nontrivial solutions for the system of equations: }\\&\begin{bmatrix}-2&13&12\\-9&-18&-18\\\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0\\0\\\end{bmatrix}\end{align*}$

Possible Answers:

$\begin{align*}&x=\frac{(6z)}{17}\\&y=\frac{(8z)}{17}\end{align*}$

$\begin{align*}&x=-\frac{(2z)}{17}\\&y=\frac{(16z)}{17}\end{align*}$

$\begin{align*}&x=\frac{(2z)}{119}\\&y=\frac{(112z)}{17}\end{align*}$

$\begin{align*}&x=\frac{(10z)}{17}\\&y=\frac{(8z)}{51}\end{align*}$

Correct answer:

$\begin{align*}&x=-\frac{(2z)}{17}\\&y=\frac{(16z)}{17}\end{align*}$

Explanation:

$\begin{align*}&\text{When faced with a system of equations like the}\\&\text{one represented by }\\&\begin{bmatrix}-2&13&12\\-9&-18&-18\\\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0\\0\\\end{bmatrix}\\&\text{We can manipulate coefficients to eliminate}\\&\text{variables in each row. That is to say, to reduce it to echelon form.}\\&\text{Reducing our matrix }\begin{bmatrix}-2&13&12\\-9&-18&-18\\\end{bmatrix}\\&\text{We find the reduced echelon form: }\begin{bmatrix}1&0&\frac{2}{17}\\0&1&\frac{16}{17}\\\end{bmatrix}\\&\text{Since each row is equal to zero, we can therefore relate variables to find}\\&\text{nontrivial solutions:}\\&x=-\frac{(2z)}{17}\\&y=-\frac{(16z)}{17}\end{align*}$

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Example Question #4 : Homogeneous Cases

$\begin{align*}&\text{Find nonzero solutions for }\begin{bmatrix}x\\y\\z\end{bmatrix}\\&\text{for the system: }\begin{bmatrix}-7&-2&-11\\-19&6&-14\\\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0\\0\\\end{bmatrix}\end{align*}$

Possible Answers:

$\begin{align*}&x=\frac{(47z)}{280}\\&y=\frac{(111z)}{560}\end{align*}$

$\begin{align*}&x=\frac{(47z)}{80}\\&y=\frac{(333z)}{80}\end{align*}$

$\begin{align*}&x=\frac{(47z)}{200}\\&y=\frac{(111z)}{16}\end{align*}$

$\begin{align*}&x=-\frac{(47z)}{40}\\&y=\frac{(111z)}{80}\end{align*}$

Correct answer:

$\begin{align*}&x=-\frac{(47z)}{40}\\&y=\frac{(111z)}{80}\end{align*}$

Explanation:

$\begin{align*}&\text{When faced with a system of equations like the}\\&\text{one represented by }\\&\begin{bmatrix}-7&-2&-11\\-19&6&-14\\\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0\\0\\\end{bmatrix}\\&\text{We can manipulate coefficients to eliminate}\\&\text{variables in each row. That is to say, to reduce it to echelon form.}\\&\text{Reducing our matrix }\begin{bmatrix}-7&-2&-11\\-19&6&-14\\\end{bmatrix}\\&\text{We find the reduced echelon form: }\begin{bmatrix}1&0&\frac{47}{40}\\0&1&\frac{111}{80}\\\end{bmatrix}\\&\text{Since each row is equal to zero, we can therefore relate variables to find}\\&\text{nontrivial solutions:}\\&x=-\frac{(47z)}{40}\\&y=-\frac{(111z)}{80}\end{align*}$

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Example Question #81 : Linear Equations

$\begin{align*}&\text{Find the nontrivial solutions for the system of equations: }\\&\begin{bmatrix}-1&16&-18\\0&5&10\\\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0\\0\\\end{bmatrix}\end{align*}$

Possible Answers:

$\begin{align*}&x=-50z\\&y=2z\end{align*}$

$\begin{align*}&x=25z\\&y=6z\end{align*}$

$\begin{align*}&x=200z\\&y=\frac{(2z)}{5}\end{align*}$

$\begin{align*}&x=350z\\&y=\frac{(2z)}{7}\end{align*}$

Correct answer:

$\begin{align*}&x=-50z\\&y=2z\end{align*}$

Explanation:

$\begin{align*}&\text{When faced with a system of equations like the}\\&\text{one represented by }\\&\begin{bmatrix}-1&16&-18\\0&5&10\\\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0\\0\\\end{bmatrix}\\&\text{We can manipulate coefficients to eliminate}\\&\text{variables in each row. That is to say, to reduce it to echelon form.}\\&\text{Reducing our matrix }\begin{bmatrix}-1&16&-18\\0&5&10\\\end{bmatrix}\\&\text{We find the reduced echelon form: }\begin{bmatrix}1&0&50\\0&1&2\\\end{bmatrix}\\&\text{Since each row is equal to zero, we can therefore relate variables to find}\\&\text{nontrivial solutions:}\\&x=-50z\\&y=-2z\end{align*}$

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Example Question #82 : Linear Equations

$\begin{align*}&\text{Find the nontrivial solutions for the system of equations: }\\&\begin{bmatrix}8&-13&12\\1&2&-12\\\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0\\0\\\end{bmatrix}\end{align*}$

Possible Answers:

$\begin{align*}&x=-\frac{(396z)}{29}\\&y=-\frac{(324z)}{29}\end{align*}$

$\begin{align*}&x=-\frac{(44z)}{87}\\&y=-\frac{(864z)}{29}\end{align*}$

$\begin{align*}&x=-\frac{(132z)}{145}\\&y=-\frac{(27z)}{29}\end{align*}$

$\begin{align*}&x=\frac{(132z)}{29}\\&y=-\frac{(108z)}{29}\end{align*}$

Correct answer:

$\begin{align*}&x=\frac{(132z)}{29}\\&y=-\frac{(108z)}{29}\end{align*}$

Explanation:

$\begin{align*}&\text{When faced with a system of equations like the}\\&\text{one represented by }\\&\begin{bmatrix}8&-13&12\\1&2&-12\\\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0\\0\\\end{bmatrix}\\&\text{We can manipulate coefficients to eliminate}\\&\text{variables in each row. That is to say, to reduce it to echelon form.}\\&\text{Reducing our matrix }\begin{bmatrix}8&-13&12\\1&2&-12\\\end{bmatrix}\\&\text{We find the reduced echelon form: }\begin{bmatrix}1&0&-\frac{132}{29}\\0&1&-\frac{108}{29}\\\end{bmatrix}\\&\text{Since each row is equal to zero, we can therefore relate variables to find}\\&\text{nontrivial solutions:}\\&x=\frac{(132z)}{29}\\&y=\frac{(108z)}{29}\end{align*}$

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Example Question #7 : Homogeneous Cases

$\begin{align*}&\text{Find nonzero solutions for }\begin{bmatrix}x\\y\\z\end{bmatrix}\\&\text{for the system: }\begin{bmatrix}-19&-13&18\\18&-2&13\\\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0\\0\\\end{bmatrix}\end{align*}$

Possible Answers:

$\begin{align*}&x=\frac{(665z)}{272}\\&y=-\frac{(571z)}{1632}\end{align*}$

$\begin{align*}&x=-\frac{(133z)}{272}\\&y=-\frac{(571z)}{272}\end{align*}$

$\begin{align*}&x=\frac{(133z)}{34}\\&y=-\frac{(571z)}{2176}\end{align*}$

$\begin{align*}&x=\frac{(133z)}{136}\\&y=-\frac{(571z)}{544}\end{align*}$

Correct answer:

$\begin{align*}&x=-\frac{(133z)}{272}\\&y=-\frac{(571z)}{272}\end{align*}$

Explanation:

$\begin{align*}&\text{When faced with a system of equations like the}\\&\text{one represented by }\\&\begin{bmatrix}-19&-13&18\\18&-2&13\\\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0\\0\\\end{bmatrix}\\&\text{We can manipulate coefficients to eliminate}\\&\text{variables in each row. That is to say, to reduce it to echelon form.}\\&\text{Reducing our matrix }\begin{bmatrix}-19&-13&18\\18&-2&13\\\end{bmatrix}\\&\text{We find the reduced echelon form: }\begin{bmatrix}1&0&\frac{133}{272}\\0&1&-\frac{571}{272}\\\end{bmatrix}\\&\text{Since each row is equal to zero, we can therefore relate variables to find}\\&\text{nontrivial solutions:}\\&x=-\frac{(133z)}{272}\\&y=\frac{(571z)}{272}\end{align*}$

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Example Question #1 : Homogeneous Cases

$\begin{align*}&\text{Find the nontrivial solutions for the system of equations: }\\&\begin{bmatrix}8&15&-1\\-2&13&13\\\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0\\0\\\end{bmatrix}\end{align*}$

Possible Answers:

$\begin{align*}&x=-\frac{(624z)}{67}\\&y=\frac{(204z)}{67}\end{align*}$

$\begin{align*}&x=-\frac{(208z)}{67}\\&y=\frac{(51z)}{134}\end{align*}$

$\begin{align*}&x=-\frac{(936z)}{67}\\&y=\frac{(17z)}{201}\end{align*}$

$\begin{align*}&x=\frac{(104z)}{67}\\&y=\frac{(51z)}{67}\end{align*}$

Correct answer:

$\begin{align*}&x=\frac{(104z)}{67}\\&y=\frac{(51z)}{67}\end{align*}$

Explanation:

$\begin{align*}&\text{When faced with a system of equations like the}\\&\text{one represented by }\\&\begin{bmatrix}8&15&-1\\-2&13&13\\\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0\\0\\\end{bmatrix}\\&\text{We can manipulate coefficients to eliminate}\\&\text{variables in each row. That is to say, to reduce it to echelon form.}\\&\text{Reducing our matrix }\begin{bmatrix}8&15&-1\\-2&13&13\\\end{bmatrix}\\&\text{We find the reduced echelon form: }\begin{bmatrix}1&0&-\frac{104}{67}\\0&1&\frac{51}{67}\\\end{bmatrix}\\&\text{Since each row is equal to zero, we can therefore relate variables to find}\\&\text{nontrivial solutions:}\\&x=\frac{(104z)}{67}\\&y=-\frac{(51z)}{67}\end{align*}$

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Linear Algebra : Homogeneous Cases

Study concepts, example questions & explanations for Linear Algebra

All Linear Algebra Resources

Example Questions

Example Question #1 : Homogeneous Cases

Example Question #1 : Homogeneous Cases

Example Question #2 : Homogeneous Cases

Example Question #3 : Homogeneous Cases

Example Question #2 : Homogeneous Cases

Example Question #4 : Homogeneous Cases

Example Question #81 : Linear Equations

Example Question #82 : Linear Equations

Example Question #7 : Homogeneous Cases

Example Question #1 : Homogeneous Cases

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