Linear Algebra - Homogeneous Cases

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

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

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

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

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

Explanation

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

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

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

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

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

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

Explanation

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

Consider the system of linear equations:

Which of the following gives its complete solution set (in parametric form, if applicable) or its only solution?

Explanation

First, write the matrix of coefficients for the system, seen below:

$\begin{bmatrix} 1&2 &-4 \ 1 & 4 & -2 \ 1 & 0 & -6 \end{matrix}$ $\begin{vmatrix} 0\ 0 \ 0 \end{bmatrix}$

It is necessary to convert this matrix to reduced row-echelon form as follows:

The first step is usually to get a leading "1" in the first row; however, one is already there.

The next two steps, which can be performed simultaneously, is to get a "0" in the first position of the second and third rows by multiplying the first row by the additive inverse of the number already there, and adding it to that row. Therefore, perform the operations

$-R1+ R2 \rightarrow R2$

and

$-R1+ R3 \rightarrow R3$

$\begin{bmatrix} 1&2 &-4 \ -1 +1 &-2+ 4 & -(-4)+ (-2) \ -1+ 1 & -2+ 0 & -(-4) + (-6) \end{matrix}$ $\begin{vmatrix} -0+ 0\ -0+ 0 \ -0 + 0 \end{bmatrix}$

$\begin{bmatrix} 1&2 &-4 \ 0 &2 &2\ 0 & -2 & -2 \end{matrix}$ $\begin{vmatrix} 0\ 0 \ 0 \end{bmatrix}$

The next step is to get a leading "1" in the second rowby multiplying each entry in that row by the multiplicative inverse of the number already there; therefore, perform the operation

$0.5 R2 \rightarrow R2$

$\begin{bmatrix} 1&2 &-4 \0.5\cdot 0 &0.5\cdot 2 &0.5\cdot 2\ 0 & -2 & -2 \end{matrix}$ $\begin{vmatrix} 0\ 0.5\cdot 0 \ 0 \end{bmatrix}$

$\begin{bmatrix} 1&2 &-4 \ 0 &1 &1 \ 0 & -2 & -2 \end{matrix}$ $\begin{vmatrix} 0\ 0 \ 0 \end{bmatrix}$

The next two steps, which can be performed simultaneously, is to get a "0" in the second position of the first and third rows by multiplying the first row by the additive inverse of the number already there, and adding it to that row. Therefore, perform the operations

$-2R2+ R1 \rightarrow R1$

and

$2R2+ R3 \rightarrow R3$

$\begin{bmatrix} -2(0)+1&-2(1)+2 &-2(1)+ (-4 )\ 0 &1 &1 \ 2(0)+0 & 2(1)+ (-2) & 2(1)+ (-2) \end{matrix}$ $\begin{vmatrix}-2(0)+ 0\ 0 \2(0)+ 0 \end{bmatrix}$

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

The matrix is now in reduced row-echelon form - all leading nonzero elements are "1", and go from upper-left to lower-right, and there are no zero rows. This matrix translates as the linear equations

To state the solution set parametrically, rewrite:

Set ; the parametric form of the solution set is

$\begin{align*}&\text{Find nonzero solutions for }\begin{bmatrix}x\y\z\end{bmatrix}\&\text{for the system: }\begin{bmatrix}-5&-1&13\-8&13&3\\end{bmatrix}\begin{bmatrix}x\y\z\end{bmatrix}=\begin{bmatrix}0\0\\end{bmatrix}\end{align*}$

$\begin{align*}&x=\frac{(172z)}{73}\&y=-\frac{(89z)}{73}\end{align*}$

$\begin{align*}&x=-\frac{(516z)}{73}\&y=-\frac{(178z)}{73}\end{align*}$

$\begin{align*}&x=-\frac{(43z)}{73}\&y=-\frac{(445z)}{73}\end{align*}$

$\begin{align*}&x=-\frac{(1204z)}{73}\&y=-\frac{(712z)}{73}\end{align*}$

Explanation

$\begin{align*}&\text{When faced with a system of equations like the}\&\text{one represented by }\&\begin{bmatrix}-5&-1&13\-8&13&3\\end{bmatrix}\begin{bmatrix}x\y\z\end{bmatrix}=\begin{bmatrix}0\0\\end{bmatrix}\&\text{We can manipulate coefficients to eliminate}\&\text{variables in each row. That is to say, to reduce it to echelon form.}\&\text{Reducing our matrix }\begin{bmatrix}-5&-1&13\-8&13&3\\end{bmatrix}\&\text{We find the reduced echelon form: }\begin{bmatrix}1&0&-\frac{172}{73}\0&1&-\frac{89}{73}\\end{bmatrix}\&\text{Since each row is equal to zero, we can therefore relate variables to find}\&\text{nontrivial solutions:}\&x=\frac{(172z)}{73}\&y=\frac{(89z)}{73}\end{align*}$

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

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

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

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

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

Explanation

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

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

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

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

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

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

Explanation

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

$\begin{align*}&\text{Find the nontrivial solutions for the system of equations: }\&\begin{bmatrix}-6&-16&2\-3&1&16\\end{bmatrix}\begin{bmatrix}x\y\z\end{bmatrix}=\begin{bmatrix}0\0\\end{bmatrix}\end{align*}$

$\begin{align*}&x=\frac{(43z)}{9}\&y=\frac{(5z)}{3}\end{align*}$

$\begin{align*}&x=-\frac{(86z)}{9}\&y=5z\end{align*}$

$\begin{align*}&x=-\frac{(215z)}{9}\&y=\frac{(25z)}{3}\end{align*}$

$\begin{align*}&x=-\frac{(344z)}{9}\&y=\frac{(5z)}{24}\end{align*}$

Explanation

$\begin{align*}&\text{When faced with a system of equations like the}\&\text{one represented by }\&\begin{bmatrix}-6&-16&2\-3&1&16\\end{bmatrix}\begin{bmatrix}x\y\z\end{bmatrix}=\begin{bmatrix}0\0\\end{bmatrix}\&\text{We can manipulate coefficients to eliminate}\&\text{variables in each row. That is to say, to reduce it to echelon form.}\&\text{Reducing our matrix }\begin{bmatrix}-6&-16&2\-3&1&16\\end{bmatrix}\&\text{We find the reduced echelon form: }\begin{bmatrix}1&0&-\frac{43}{9}\0&1&\frac{5}{3}\\end{bmatrix}\&\text{Since each row is equal to zero, we can therefore relate variables to find}\&\text{nontrivial solutions:}\&x=\frac{(43z)}{9}\&y=-\frac{(5z)}{3}\end{align*}$

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

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

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

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

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

Explanation

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

$\begin{align*}&\text{Find nontrivial solutions of }\begin{bmatrix}x\y\z\end{bmatrix}\&\text{for the function: }\begin{bmatrix}11&16&-15\12&-13&-19\\end{bmatrix}\begin{bmatrix}x\y\z\end{bmatrix}=\begin{bmatrix}0\0\\end{bmatrix}\end{align*}$

$\begin{align*}&x=\frac{(499z)}{335}\&y=\frac{(29z)}{335}\end{align*}$

$\begin{align*}&x=-\frac{(499z)}{670}\&y=\frac{(29z)}{670}\end{align*}$

$\begin{align*}&x=-\frac{(499z)}{67}\&y=\frac{(29z)}{1340}\end{align*}$

$\begin{align*}&x=-\frac{(499z)}{2345}\&y=\frac{(29z)}{3015}\end{align*}$

Explanation

$\begin{align*}&\text{When faced with a system of equations like the}\&\text{one represented by }\&\begin{bmatrix}11&16&-15\12&-13&-19\\end{bmatrix}\begin{bmatrix}x\y\z\end{bmatrix}=\begin{bmatrix}0\0\\end{bmatrix}\&\text{We can manipulate coefficients to eliminate}\&\text{variables in each row. That is to say, to reduce it to echelon form.}\&\text{Reducing our matrix }\begin{bmatrix}11&16&-15\12&-13&-19\\end{bmatrix}\&\text{We find the reduced echelon form: }\begin{bmatrix}1&0&-\frac{499}{335}\0&1&\frac{29}{335}\\end{bmatrix}\&\text{Since each row is equal to zero, we can therefore relate variables to find}\&\text{nontrivial solutions:}\&x=\frac{(499z)}{335}\&y=-\frac{(29z)}{335}\end{align*}$

Consider the system of linear equations:

Which of the following expresses the solution set in parametric form?

None of the other choices gives the correct response.

Explanation

First, write the matrix of coefficients for the system, seen below:

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

It is necessary to convert this matrix to reduced row-echelon form as follows:

The first step is usually to get a leading "1" in the first row; however, one is already there.

The next step is to get a "0" in the first position of the second row by multiplying the first row by the additive inverse of the number already there, and adding it to that row. Therefore, perform the operation

$-2R1 + R2 \rightarrow R2$ :

$\begin{bmatrix} 1&2 &3 &1 \-2(1)+ 2& -2(2)+5&-2(3)+( -1 )&- 2(1)+ (-1) \end{matrix}$ $\begin{vmatrix} 0\ -2(0)+0 \end{bmatrix}$

$\begin{bmatrix} 1&2 &3 &1 \0& 1&-7&- 3 \end{matrix}$ $\begin{vmatrix} 0\ 0 \end{bmatrix}$

The next step is usually to get a leading "1" in the second row; however, one is already there.

The next step is to get a "0" in the second position of the first row by multiplying the first row by the additive inverse of the number already there, and adding it to that row. Therefore, perform the operation:

$-2R2 + R1 \rightarrow R1$

$\begin{bmatrix}-2(0)+ 1&-2(1)+2 &-2(-7)+ 3 &-2(-3)+1 \0& 1&-7&- 3 \end{matrix}$ $\begin{vmatrix} -2(0)+ 0\ 0 \end{bmatrix}$

$\begin{bmatrix} 1&0&17 &7 \0& 1&-7&- 3 \end{matrix}$ $\begin{vmatrix} 0\ 0 \end{bmatrix}$

The matrix is now in reduced row-echelon form - all leading nonzero elements are "1", and go from upper-left to lower-right, and there are no zero rows. This matrix translates as the linear equations