Linear Equations - Linear Algebra

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Question

$A = \begin{bmatrix} e & 0 & 0 & 0 \ 0 & \pi & 0 & 0 \ 0 & 0 & -\pi & 0 \ 0 & 0 & 0 & -e \end{bmatrix}$

True or false: is an example of a matrix in row-echelon form.

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Answer

A matrix is in row-echelon form if and only if it fits three conditions:

Any rows comprising only zeroes are at the bottom.

Any leading nonzero entries are 1's..

Each leading 1 is to the right of the one immediately above.

All four rows have leading nonzero entries, but none of them are 1's. The matrix violates the conditions of a matrix in row-echelon form.

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Question

Find the inverse using row operations

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

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Answer

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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Question

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

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Answer

$\begin{bmatrix} 3 &0 &4 &1 &0 &0 \2 &-2 &5 &0 &1 &0 \ 4& 3& 2& 0& 0& 1\end{bmatrix} \sim$ 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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Question

Change the following matrix into reduced row echelon form.

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

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Answer

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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Question

Change the following matrix into reduced row echelon form.

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

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Answer

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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Question

Is the following matrix in reduced row echelon form? Why or why not.

$\begin{bmatrix} 1& 0& 0& 7\ 0& 1& 0& 5\ 0& 0& 1& 4 \end{bmatrix}$

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Answer

A matrix is in reduced row echelon form if

all nonzero rows are above any all zero rows

the left most nonzero entry in each row (the leading entry) is .

the leading entry is in a column to the right of the leading entry in the column above it

all entries in a column below the leading entry are zero

This matrix meets all four of these criteria.

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Question

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

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Answer

$\begin{align}&\text{Reduced echelon form is a matrix form in which the leading coefficient}\&\text{of each nonzero row is always to the right of the leading coefficient}\&\text{of the row above it, each leading coefficient is the only nonzero}\&\text{value in its column, and each row is scaled down to give the}\&\text{leading coefficients values of one. Beginning with our matrix,}\end{align}$

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

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Question

$\begin{align}&\text{Decide if there is a unique solution for the equation: }\&\begin{bmatrix}-6&-7&-19\-8&19&-14\0&5&2\end{bmatrix}\begin{bmatrix}x\y\z\end{bmatrix}=\begin{bmatrix}-21\-36\-149\end{bmatrix}\end{align}$

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Answer

$\begin{align}&\text{To determine whether or not the system of equations represented}\&\text{by a matrix equation has a unique solution, one should calculate}\&\text{the determinant. If the determinant is a nonzero value, then}\&\text{the system of equations has a unique solution. Otherwise, if}\&\text{the determinant is zero, the system of equation may have no}\&\text{solution, or it may have an infinite number of them. To refresh,}\&\text{to calculate a matrix of the given dimensions we use the formula:}\&det\begin{vmatrix} a&b&c \ d&e&f\g&h&i \end{vmatrix}=a(ei-fh)-b(di-fg)+c(dh-eg)\&det\begin{bmatrix}-6&-7&-19\-8&19&-14\0&5&2\end{bmatrix}=0\&\text{There is not a unique solution.}\end{align}$

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Question

$\begin{align}&\text{Is there a unique solution for the equation }\&\begin{bmatrix}-20&12\7&16\end{bmatrix}\begin{bmatrix}x\y\end{bmatrix}=\begin{bmatrix}-171\149\end{bmatrix}?\end{align}$

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Answer

$\begin{align}&\text{To determine whether or not the system of equations represented}\&\text{by a matrix equation has a unique solution, one should calculate}\&\text{the determinant. If the determinant is a nonzero value, then}\&\text{the system of equations has a unique solution. Otherwise, if}\&\text{the determinant is zero, the system of equation may have no}\&\text{solution, or it may have an infinite number of them. To refresh,}\&\text{to calculate a matrix of the given dimensions we use the formula:}\&det\begin{vmatrix} a&b \ c&d \end{vmatrix}=ad-bc\&det\begin{bmatrix}-20&12\7&16\end{bmatrix}=-404\&\text{There is a unique solution.}\end{align}$

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Question

$\begin{align}&\text{Determine whether or not the equation }\&\begin{bmatrix}6&-8\3&12\end{bmatrix}\begin{bmatrix}x\y\end{bmatrix}=\begin{bmatrix}-188\-101\end{bmatrix}\&\text{has a unique solution.}\end{align}$

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Answer

$\begin{align}&\text{To determine whether or not the system of equations represented}\&\text{by a matrix equation has a unique solution, one should calculate}\&\text{the determinant. If the determinant is a nonzero value, then}\&\text{the system of equations has a unique solution. Otherwise, if}\&\text{the determinant is zero, the system of equation may have no}\&\text{solution, or it may have an infinite number of them. To refresh,}\&\text{to calculate a matrix of the given dimensions we use the formula:}\&det\begin{vmatrix} a&b \ c&d \end{vmatrix}=ad-bc\&det\begin{bmatrix}6&-8\3&12\end{bmatrix}=96\&\text{There is a unique solution.}\end{align}$

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Question

$\begin{align}&\text{Is there a unique solution for the equation }\&\begin{bmatrix}0&-8&2\-5&3&2\20&12&-14\end{bmatrix}\begin{bmatrix}x\y\z\end{bmatrix}=\begin{bmatrix}-163\36\22\end{bmatrix}?\end{align}$

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Answer

$\begin{align}&\text{To determine whether or not the system of equations represented}\&\text{by a matrix equation has a unique solution, one should calculate}\&\text{the determinant. If the determinant is a nonzero value, then}\&\text{the system of equations has a unique solution. Otherwise, if}\&\text{the determinant is zero, the system of equation may have no}\&\text{solution, or it may have an infinite number of them. To refresh,}\&\text{to calculate a matrix of the given dimensions we use the formula:}\&det\begin{vmatrix} a&b&c \ d&e&f\g&h&i \end{vmatrix}=a(ei-fh)-b(di-fg)+c(dh-eg)\&det\begin{bmatrix}0&-8&2\-5&3&2\20&12&-14\end{bmatrix}=0\&\text{There is not a unique solution.}\end{align}$

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Question

$\begin{align}&\text{Is there a unique solution for the equation }\&\begin{bmatrix}13&13&-2\12&12&12\-16&-16&-11\end{bmatrix}\begin{bmatrix}x\y\z\end{bmatrix}=\begin{bmatrix}28\125\107\end{bmatrix}?\end{align}$

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Answer

$\begin{align}&\text{To determine whether or not the system of equations represented}\&\text{by a matrix equation has a unique solution, one should calculate}\&\text{the determinant. If the determinant is a nonzero value, then}\&\text{the system of equations has a unique solution. Otherwise, if}\&\text{the determinant is zero, the system of equation may have no}\&\text{solution, or it may have an infinite number of them. To refresh,}\&\text{to calculate a matrix of the given dimensions we use the formula:}\&det\begin{vmatrix} a&b&c \ d&e&f\g&h&i \end{vmatrix}=a(ei-fh)-b(di-fg)+c(dh-eg)\&det\begin{bmatrix}13&13&-2\12&12&12\-16&-16&-11\end{bmatrix}=0\&\text{There is not a unique solution.}\end{align}$

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Question

$\begin{align}&\text{Decide if there is a unique solution for the equation: }\&\begin{bmatrix}1&11&14\-1&10&-12\11&-5&16\end{bmatrix}\begin{bmatrix}x\y\z\end{bmatrix}=\begin{bmatrix}59\4\-32\end{bmatrix}\end{align}$

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Answer

$\begin{align}&\text{To determine whether or not the system of equations represented}\&\text{by a matrix equation has a unique solution, one should calculate}\&\text{the determinant. If the determinant is a nonzero value, then}\&\text{the system of equations has a unique solution. Otherwise, if}\&\text{the determinant is zero, the system of equation may have no}\&\text{solution, or it may have an infinite number of them. To refresh,}\&\text{to calculate a matrix of the given dimensions we use the formula:}\&det\begin{vmatrix} a&b&c \ d&e&f\g&h&i \end{vmatrix}=a(ei-fh)-b(di-fg)+c(dh-eg)\&det\begin{bmatrix}1&11&14\-1&10&-12\11&-5&16\end{bmatrix}=-2646\&\text{There is a unique solution.}\end{align}$

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Question

$\begin{align}&\text{Determine whether or not the equation }\&\begin{bmatrix}9&-12\-3&4\end{bmatrix}\begin{bmatrix}x\y\end{bmatrix}=\begin{bmatrix}-57\182\end{bmatrix}\&\text{has a unique solution.}\end{align}$

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Answer

$\begin{align}&\text{To determine whether or not the system of equations represented}\&\text{by a matrix equation has a unique solution, one should calculate}\&\text{the determinant. If the determinant is a nonzero value, then}\&\text{the system of equations has a unique solution. Otherwise, if}\&\text{the determinant is zero, the system of equation may have no}\&\text{solution, or it may have an infinite number of them. To refresh,}\&\text{to calculate a matrix of the given dimensions we use the formula:}\&det\begin{vmatrix} a&b \ c&d \end{vmatrix}=ad-bc\&det\begin{bmatrix}9&-12\-3&4\end{bmatrix}=0\&\text{There is not a unique solution.}\end{align}$

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Question

$\begin{align}&\text{Determine whether or not the equation }\&\begin{bmatrix}16&-11\-17&6\end{bmatrix}\begin{bmatrix}x\y\end{bmatrix}=\begin{bmatrix}-8\153\end{bmatrix}\&\text{has a unique solution.}\end{align}$

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Answer

$\begin{align}&\text{To determine whether or not the system of equations represented}\&\text{by a matrix equation has a unique solution, one should calculate}\&\text{the determinant. If the determinant is a nonzero value, then}\&\text{the system of equations has a unique solution. Otherwise, if}\&\text{the determinant is zero, the system of equation may have no}\&\text{solution, or it may have an infinite number of them. To refresh,}\&\text{to calculate a matrix of the given dimensions we use the formula:}\&det\begin{vmatrix} a&b \ c&d \end{vmatrix}=ad-bc\&det\begin{bmatrix}16&-11\-17&6\end{bmatrix}=-91\&\text{There is a unique solution.}\end{align}$

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Question

$\begin{align}&\text{Determine whether or not the equation }\&\begin{bmatrix}2&-5&-12\8&7&1\-10&-2&11\end{bmatrix}\begin{bmatrix}x\y\z\end{bmatrix}=\begin{bmatrix}-71\-38\-50\end{bmatrix}\&\text{has a unique solution.}\end{align}$

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Answer

$\begin{align}&\text{To determine whether or not the system of equations represented}\&\text{by a matrix equation has a unique solution, one should calculate}\&\text{the determinant. If the determinant is a nonzero value, then}\&\text{the system of equations has a unique solution. Otherwise, if}\&\text{the determinant is zero, the system of equation may have no}\&\text{solution, or it may have an infinite number of them. To refresh,}\&\text{to calculate a matrix of the given dimensions we use the formula:}\&det\begin{vmatrix} a&b&c \ d&e&f\g&h&i \end{vmatrix}=a(ei-fh)-b(di-fg)+c(dh-eg)\&det\begin{bmatrix}2&-5&-12\8&7&1\-10&-2&11\end{bmatrix}=0\&\text{There is not a unique solution.}\end{align}$

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Question

$\begin{align}&\text{Decide if there is a unique solution for the equation: }\&\begin{bmatrix}19&-12\9&18\end{bmatrix}\begin{bmatrix}x\y\end{bmatrix}=\begin{bmatrix}-126\-163\end{bmatrix}\end{align}$

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Answer

$\begin{align}&\text{To determine whether or not the system of equations represented}\&\text{by a matrix equation has a unique solution, one should calculate}\&\text{the determinant. If the determinant is a nonzero value, then}\&\text{the system of equations has a unique solution. Otherwise, if}\&\text{the determinant is zero, the system of equation may have no}\&\text{solution, or it may have an infinite number of them. To refresh,}\&\text{to calculate a matrix of the given dimensions we use the formula:}\&det\begin{vmatrix} a&b \ c&d \end{vmatrix}=ad-bc\&det\begin{bmatrix}19&-12\9&18\end{bmatrix}=450\&\text{There is a unique solution.}\end{align}$

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Question

$\begin{align}&\text{Decide if there is a unique solution for the equation: }\&\begin{bmatrix}3&-14&-20\5&0&-3\-6&8&14\end{bmatrix}\begin{bmatrix}x\y\z\end{bmatrix}=\begin{bmatrix}125\155\-70\end{bmatrix}\end{align}$

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Answer

$\begin{align}&\text{To determine whether or not the system of equations represented}\&\text{by a matrix equation has a unique solution, one should calculate}\&\text{the determinant. If the determinant is a nonzero value, then}\&\text{the system of equations has a unique solution. Otherwise, if}\&\text{the determinant is zero, the system of equation may have no}\&\text{solution, or it may have an infinite number of them. To refresh,}\&\text{to calculate a matrix of the given dimensions we use the formula:}\&det\begin{vmatrix} a&b&c \ d&e&f\g&h&i \end{vmatrix}=a(ei-fh)-b(di-fg)+c(dh-eg)\&det\begin{bmatrix}3&-14&-20\5&0&-3\-6&8&14\end{bmatrix}=0\&\text{There is not a unique solution.}\end{align}$

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Question

$\begin{align}&\text{Determine whether or not the equation }\&\begin{bmatrix}-16&-4&-9\-13&-13&3\20&-18&9\end{bmatrix}\begin{bmatrix}x\y\z\end{bmatrix}=\begin{bmatrix}179\-130\-184\end{bmatrix}\&\text{has a unique solution.}\end{align}$

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Answer

$\begin{align}&\text{To determine whether or not the system of equations represented}\&\text{by a matrix equation has a unique solution, one should calculate}\&\text{the determinant. If the determinant is a nonzero value, then}\&\text{the system of equations has a unique solution. Otherwise, if}\&\text{the determinant is zero, the system of equation may have no}\&\text{solution, or it may have an infinite number of them. To refresh,}\&\text{to calculate a matrix of the given dimensions we use the formula:}\&det\begin{vmatrix} a&b&c \ d&e&f\g&h&i \end{vmatrix}=a(ei-fh)-b(di-fg)+c(dh-eg)\&det\begin{bmatrix}-16&-4&-9\-13&-13&3\20&-18&9\end{bmatrix}=-4146\&\text{There is a unique solution.}\end{align}$

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Question

Solve the following system by reducing its corresponding matrix.

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Answer

The corresponding matrix is

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

We add times row one to row two.

$\begin{pmatrix} 1& 2& 0\ 2+(-2)& 4+(-4)& 0+0 \end{pmatrix}=\begin{pmatrix} 1& 2&0 \ 0&0 &0 \end{pmatrix}$

This yields the solution

or

and is a free variable.

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