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Linear Algebra : Linear Equations

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Example Question #41 : Reduced Row Echelon Form And Row Operations

Write the following matrix in reduced row echelon form.

$\begin{bmatrix} 2& -4& 7\\ 3& -2& 9 \end{bmatrix}$

Possible Answers:

$\begin{bmatrix} 1& -2& 7/4\\ 0& 1& -3/8\end{bmatrix}$

$\begin{bmatrix} 1& 0& 11/41\\ 0& 1& -13/8\end{bmatrix}$

$\begin{bmatrix} 1& 0& 11/4\\ 0& 1& -3/8\end{bmatrix}$

$\begin{bmatrix} 1& 0& 7/2\\ 0& 1& -5/8\end{bmatrix}$

Correct answer:

$\begin{bmatrix} 1& 0& 11/4\\ 0& 1& -3/8\end{bmatrix}$

Explanation:

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

We will use row operations to transform this matrix. There are three kinds of row operations that can be performed. Rows can be switched, a row can be multiplied by a constant and rows can be linearly combined ( aR_n+bR_m where and are constants).

$\begin{bmatrix} 2& -4& 7\\ 3& -2& 9 \end{bmatrix}$

We make the leading entry in the 1st row , using the row operation R_1=r_1/2

$\begin{bmatrix} 2/2& -4/2& 7/2\\ 3& -2& 9 \end{bmatrix}$ $\Rightarrow$ $\begin{bmatrix} 1& -2& 7/2\\ 3& -2& 9 \end{bmatrix}$

We will make the first number in the 2nd row using the row operation R_2=r_2-3r_1

$\begin{bmatrix} 1& -2& 7/2\\ 3-3(1)& -2-3(-2)& 9-3(7/2) \end{bmatrix}$ $\Rightarrow$ $\begin{bmatrix} 1& -2& 7/2\\ 0& 4& -3/2\end{bmatrix}$

We now make the leading entry of the 2nd row using the row operation R_2=r_2/4

$\begin{bmatrix} 1& -2& 7/2\\ 0& 4/4& (-3/2)/4\end{bmatrix}$ $\Rightarrow$ $\begin{bmatrix} 1& -2& 7/2\\ 0& 1& -3/8\end{bmatrix}$

The matrix is now in echelon form.

Next we change the 2nd term in the 1st row to using the row operation R_1=r_1+2r_2

$\begin{bmatrix} 1+2(0)& -2+2(1)& 7/2+2(-3/8)\\ 0& 1& -3/8\end{bmatrix}$ $\Rightarrow$

$\begin{bmatrix} 1& 0& 11/4\\ 0& 1& -3/8\end{bmatrix}$

The matrix is now in reduced row echelon form.

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Example Question #42 : Linear Equations

Write the following matrix in reduced row echelon form.

$\begin{bmatrix} 1& -4& 3\\ 5& -4& 9 \end{bmatrix}$

Possible Answers:

$\begin{bmatrix} 1&0& 1\\ 0& 1& 1 \end{bmatrix}$

$\begin{bmatrix} 1& -4& 3\\ 0& 1& -3/8 \end{bmatrix}$

$\begin{bmatrix} 1&0& 2\\ 0& 1& -8 \end{bmatrix}$

$\begin{bmatrix} 1&0& 3/2\\ 0& 1& -3/8 \end{bmatrix}$

Correct answer:

$\begin{bmatrix} 1&0& 3/2\\ 0& 1& -3/8 \end{bmatrix}$

Explanation:

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

The leading entry in the 1st row is , so no operation is needed

$\begin{bmatrix} 1& -4& 3\\ 5& -4& 9 \end{bmatrix}$

We will make the first number in the 2nd row using the row operation R_2=r_2-5r_1

$\begin{bmatrix} 1& -4& 3\\ 5-5(1)& -4-5(-4)& 9-5(3) \end{bmatrix}$ $\Rightarrow$ $\begin{bmatrix} 1& -4& 3\\ 0& 16& -6 \end{bmatrix}$

We now make the leading entry of the 2nd row using the row operation R_2=r_2/16

$\begin{bmatrix} 1& -4& 3\\ 0/16& 16/16& -6/16 \end{bmatrix}$ $\Rightarrow$ $\begin{bmatrix} 1& -4& 3\\ 0& 1& -3/8 \end{bmatrix}$

The matrix is now in echelon form.

Next we change the 2nd term in the 1st row to using the row operation R_1=r_1+4r_2

$\begin{bmatrix} 1+4(0)& -4+4(1)& 3+4(-3/8)\\ 0& 1& -3/8 \end{bmatrix}$ $\Rightarrow$

$\begin{bmatrix} 1&0& 3/2\\ 0& 1& -3/8 \end{bmatrix}$

The matrix is now in reduced row echelon form.

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Example Question #43 : Linear Equations

Write the following matrix in reduced row echelon form.

$\begin{bmatrix} 1& 2& 3&1\\ 5& -4& 6&1\\ 7& 9& 10&-10 \end{bmatrix}$

Possible Answers:

$\begin{bmatrix} 1& 2&1& -3\\ 0& 1& 0& 1\\ 0& 0& 1& 8\end{bmatrix}$

$\begin{bmatrix} 1& 2& 3&1\\ 0& 1& 9/14&4/14\\ 0& 0& 1&2 \end{bmatrix}$

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

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

Correct answer:

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

Explanation:

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

The leading entry in the 1st row is , so no operation is needed

$\begin{bmatrix} 1& 2& 3&1\\ 5& -4& 6&1\\ 7& 9& 10&-10 \end{bmatrix}$

We will make the first number in the 2nd row using the row operation R_2=r_2-5r_1

$\begin{bmatrix} 1& 2& 3&1\\ 5-5(1)& -4-5(2)& 6-5(3) &1-5(1)\\ 7& 9& 10 &-10 \end{bmatrix}$ $\Rightarrow$ $\begin{bmatrix} 1& 2& 3&1\\ 0& -14& -9&-4\\ 7& 9& 10 &-10\end{bmatrix}$

We now make the leading entry of the 2nd row using the row operation R_2=-r_2/14

$\begin{bmatrix} 1& 2& 3&1\\ 0& 14/14& 9/14&4/14\\ 7& 9& 10&-10 \end{bmatrix}$ $\Rightarrow$ $\begin{bmatrix} 1& 2& 3&1\\ 0& 1& 9/14& 4/14\\ 7& 9& 10&-10 \end{bmatrix}$

Next we will make the first number in the 3rd row using the row operation R_3=r_3-7r_1

$\begin{bmatrix} 1& 2& 3&1\\ 0& 1& 9/14&4/14\\ 7-7(1)& 9-7(2)& 10-7(3)&-10-7(1) \end{bmatrix}$ $\Rightarrow$ $\begin{bmatrix} 1& 2& 3&1\\ 0& 1& 9/14&4/14\\ 0& -5& -11&-17 \end{bmatrix}$

Next we make the 2nd number in the 3rd row using the row operation R_3=r_3+5r_2

$\begin{bmatrix} 1& 2& 3&1\\ 0& 1& 9/14&4/14\\ 0& -5+5*(1)& -11+5*(9/14) &-17+5*(4/14)\end{bmatrix}$ $\Rightarrow$

$\begin{bmatrix} 1& 2& 3&1\\ 0& 1& 9/14&4/14\\ 0& 0& -109/14&-218/14 \end{bmatrix}$

We now make the leading entry of the 3rd row using the row operation R_3=-14r_3/109

$\begin{bmatrix} 1& 2& 3&1\\ 0& 1& 9/14&4/14\\ 0& 0& -109/14*(-14/109) &-218/14*(-14/109)\end{bmatrix}$ $\Rightarrow$ $\begin{bmatrix} 1& 2& 3&1\\ 0& 1& 9/14&4/14\\ 0& 0& 1&2 \end{bmatrix}$

The matrix is now in echelon form.

Next we change the 3rd term in the 2nd row to using the row operation R_2=r_2-(9/14)r_3

$\begin{bmatrix} 1& 2& 3&1\\ 0-(9/14)*0& 1-(9/14)*0& 9/14-(9/14)*1&4/14-(9/14)*2\\ 0& 0& 1&2 \end{bmatrix}$ $\Rightarrow$ $\begin{bmatrix} 1& 2& 3& 1\\ 0& 1& 0& -1\\ 0& 0& 1& 2 \end{bmatrix}$

Next we change the 2nd term in the 1st row to using the row operation R_1=r_1-2r_2

$\begin{bmatrix} 1-(2)0& 2-(2)1& 3-(2)0& 1-(2)(-1)\\ 0& 1& 0& -1\\ 0& 0& 1 & 2\end{bmatrix}$ $\Rightarrow$

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

Last we change the 3rd term in the 1st row to using the row operation R_1=r_1-3r_3

$\begin{bmatrix} 1-(3)0& 0-(3)0& 3-(3)1& 3-(3)2\\ 0& 1& 0& -1\\ 0& 0& 1& 2 \end{bmatrix}$ $\Rightarrow$ $\begin{bmatrix} 1& 0& 0& -3\\ 0& 1& 0& -1\\ 0& 0& 1& 2\end{bmatrix}$

The matrix is now in reduced row echelon form.

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Example Question #41 : Linear Equations

Write the following matrix in reduced row echelon form.

$\begin{bmatrix} 1& 2& 3\\ 5& -4& 6\\ 7& 9& 10 \end{bmatrix}$

Possible Answers:

$\begin{bmatrix} 1& 0& 3\\ 0& 1& 0\\ 0& 0& 1 \end{bmatrix}$

$\begin{bmatrix} 1& 2& 3\\ 0& 1& 9/14\\ 0& 0& 1 \end{bmatrix}$

$\begin{bmatrix} 1& 0& 0\\ 0& 1& 0\\ 0& 0& 1 \end{bmatrix}$

$\begin{bmatrix} 1& 0& 0\\ 0& 2& 1\\ 0& 0& 1 \end{bmatrix}$

Correct answer:

$\begin{bmatrix} 1& 0& 0\\ 0& 1& 0\\ 0& 0& 1 \end{bmatrix}$

Explanation:

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

The leading entry in the 1st row is , so no operation is needed

$\begin{bmatrix} 1& 2& 3\\ 5& -4& 6\\ 7& 9& 10 \end{bmatrix}$

We will make the first number in the 2nd row using the row operation R_2=r_2-5r_1

$\begin{bmatrix} 1& 2& 3\\ 5-5(1)& -4-5(2)& 6-5(3)\\ 7& 9& 10 \end{bmatrix}$ $\Rightarrow$ $\begin{bmatrix} 1& 2& 3\\ 0& -14& -9\\ 7& 9& 10 \end{bmatrix}$

We now make the leading entry of the 2nd row using the row operation R_2=-r_2/14

$\begin{bmatrix} 1& 2& 3\\ 0& 14/14& 9/14\\ 7& 9& 10 \end{bmatrix}$ $\Rightarrow$ $\begin{bmatrix} 1& 2& 3\\ 0& 1& 9/14\\ 7& 9& 10 \end{bmatrix}$

Next we will make the first number in the 3rd row using the row operation R_3=r_3-7r_1

$\begin{bmatrix} 1& 2& 3\\ 0& 1& 9/14\\ 7-7(1)& 9-7(2)& 10-7(3) \end{bmatrix}$ $\Rightarrow$ $\begin{bmatrix} 1& 2& 3\\ 0& 1& 9/14\\ 0& -5& -11 \end{bmatrix}$

Next we make the 2nd number in the 3rd row using the row operation R_3=r_3+5r_2

$\begin{bmatrix} 1& 2& 3\\ 0& 1& 9/14\\ 0& -5+5*(1)& -11+5*(9/14) \end{bmatrix}$ $\Rightarrow$

$\begin{bmatrix} 1& 2& 3\\ 0& 1& 9/14\\ 0& 0& -109/14 \end{bmatrix}$

We now make the leading entry of the 3rd row using the row operation R_3=-14r_3/109

$\begin{bmatrix} 1& 2& 3\\ 0& 1& 9/14\\ 0& 0& -109/14*(-14/109) \end{bmatrix}$ $\Rightarrow$ $\begin{bmatrix} 1& 2& 3\\ 0& 1& 9/14\\ 0& 0& 1 \end{bmatrix}$

The matrix is now in echelon form.

Next we change the 3rd term in the 2nd row to using the row operation R_2=r_2-(9/14)r_3

$\begin{bmatrix} 1& 2& 3\\ 0-(9/14)*0& 1-(9/14)*0& 9/14-(9/14)*1\\ 0& 0& 1 \end{bmatrix}$ $\Rightarrow$ $\begin{bmatrix} 1& 2& 3\\ 0& 1& 0\\ 0& 0& 1 \end{bmatrix}$

Next we change the 2nd term in the 1st row to using the row operation R_1=r_1-2r_2

$\begin{bmatrix} 1-(2)0& 2-(2)1& 3-(2)0\\ 0& 1& 0\\ 0& 0& 1 \end{bmatrix}$ $\Rightarrow$

$\begin{bmatrix} 1& 0& 3\\ 0& 1& 0\\ 0& 0& 1 \end{bmatrix}$

Last we change the 3rd term in the 1st row to using the row operation R_1=r_1-3r_3

$\begin{bmatrix} 1-(3)0& 0-(3)0& 3-(3)1\\ 0& 1& 0\\ 0& 0& 1 \end{bmatrix}$ $\Rightarrow$ $\begin{bmatrix} 1& 0& 0\\ 0& 1& 0\\ 0& 0& 1 \end{bmatrix}$

The matrix is now in reduced row echelon form.

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Example Question #45 : Linear Equations

$A= \begin{bmatrix} 1 & 0 & 0 & \log_{2} 3 \\ 0 & 1 & 0 & 0 \\ 0 & 0 &1 & 0 \\ 0 & 0 & 0 &1 \end{bmatrix}$

$B = \begin{bmatrix} 1 & 1 & -1 & -1 \\ 2 & 1 & 3 & -3 \\ 1 & -1 & -2 & 2 \\ 1 & 2 & 1 & 2 \end{bmatrix}$

Evaluate .

Possible Answers:

$\begin{bmatrix} \log_{2} \frac{2}{3} & 1 & -1 & -1 \\ \log_{2} \frac{4}{27} & 1 & 3 & -3 \\ \log_{2} 18 & -1 & -2 & 2 \\ \log_{2} 18 & 2 & 1 & 2 \end{bmatrix}$

$\begin{bmatrix} 1 & 1 & -1 & -1 \\ 2 & 1 & 3 & -3 \\ 1 & -1 & -2 & 2 \\ \log_{2} 6 & \log_{2}12 & \log_{2}\frac{2}{3}& \log_{2}\frac{4}{3} \end{bmatrix}$

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

$\begin{bmatrix} 1 & 1 & -1 & \log_{2} \frac{3}{2} \\ 2 & 1 & 3 & \log_{2} \frac{9}{8} \\ 1 & -1 & -2 & \log_{2} 12 \\ 1 & 2 & 1 & \log_{2}1 2 \end{bmatrix}$

$\begin{bmatrix} \log_{2}6 & \log_{2}18& \log_{2} \frac{3}{2}& \log_{2} \frac{9}{2} \\ 2 & 1 & 3 & -3 \\ 1 & -1 & -2 & 2 \\ 1 & 2 & 1 & 2 \end{bmatrix}$

Correct answer:

$\begin{bmatrix} \log_{2}6 & \log_{2}18& \log_{2} \frac{3}{2}& \log_{2} \frac{9}{2} \\ 2 & 1 & 3 & -3 \\ 1 & -1 & -2 & 2 \\ 1 & 2 & 1 & 2 \end{bmatrix}$

Explanation:

is an elementary matrix, in that it can be formed from the (four-by-four) identity matrix by a single row operation. Since differs from in that the entry $\log_{2} 3$ is in Row 1, Column 4, the row operation is

$(\log_{2}3) R4 + R1 \rightarrow R1$

Taking the product has the same effect on :

$B = \begin{bmatrix} 1 & 1 & -1 & -1 \\ 2 & 1 & 3 & -3 \\ 1 & -1 & -2 & 2 \\ 1 & 2 & 1 & 2 \end{bmatrix}$ ,

$AB = \begin{bmatrix} 1+ \log_{2}3 & 1 +2 \log_{2}3 & -1 + \log_{2}3& -1 + 2 \log_{2}3 \\ 2 & 1 & 3 & -3 \\ 1 & -1 & -2 & 2 \\ 1 & 2 & 1 & 2 \end{bmatrix}$

$= \begin{bmatrix} 1+ \log_{2}3 & 1 + \log_{2}9 & -1 + \log_{2}3& -1 + \log_{2}9 \\ 2 & 1 & 3 & -3 \\ 1 & -1 & -2 & 2 \\ 1 & 2 & 1 & 2 \end{bmatrix}$

$= \begin{bmatrix} \log_{2}6 & \log_{2}18& \log_{2} \frac{3}{2}& \log_{2} \frac{9}{2} \\ 2 & 1 & 3 & -3 \\ 1 & -1 & -2 & 2 \\ 1 & 2 & 1 & 2 \end{bmatrix}$

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Example Question #41 : Reduced Row Echelon Form And Row Operations

$A = \begin{bmatrix} 1 & 2 & 3 & 4 \\ 0 & 1 & 2 & 3 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 2 \\ 0 & 0 & 0 & 1 \end{bmatrix}$

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

Possible Answers:

False

True

Correct answer:

False

Explanation:

A matrix in row echelon form must fit three conditions:

1) Any rows comprising only zeroes must be concentrated at the bottom.

2) The leading nonzero entry in each other row must be a 1.

3) Each leading 1 must be to the right of the leading 1 in the row above it. fits this condition also.

Examine the second and third rows. The leading 1 in the third row is immediately below the leading 1 in the second, violating the third condition. The matrix is not in row echelon form.

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Example Question #42 : Reduced Row Echelon Form And Row Operations

$A= \begin{bmatrix} 1 & e & e^{2} & e^{3} \\ 0 & 1 & 0 & \frac{1}{e} \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{bmatrix}$

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

Possible Answers:

False

True

Correct answer:

True

Explanation:

A matrix in row echelon form must fit three conditions:

1) Any rows comprising only zeroes must be concentrated at the bottom. has one zero row, which is the bottom row, so it fits this condition.

2) The leading nonzero entry in each other row must be a 1. fits this condition also.

3) Each leading 1 must be to the right of the leading 1 in the row above it. fits this condition also.

is in row-echelon form.

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Example Question #48 : Linear Equations

$A= \begin{bmatrix} \log_{3}4 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 &1 & 0 \\ 0 & 0 & 0 &1 \end{bmatrix}$

$B = \begin{bmatrix} 1 & 2 & -2 & -1 \\ 2 & 1 & 3 & -3 \\ 1 & -1 & -2 & 2 \\ 1 & 2 & 1 & 2 \end{bmatrix}$

Evaluate .

Possible Answers:

$\begin{bmatrix} \log_{3}4 & 2 & -2 & -1 \\ 2 & \log_{3}4 & 3 & -3 \\ 1 & -1 & \log_{3} \frac{1}{16} & 2 \\ 1 & 2 & 1 & \log_{3}16 \end{bmatrix}$

$\begin{bmatrix} \log_{3}4 & 2 & -2 & -1 \\ 2 & 1 & 3 & -3 \\ 1 & -1 & -2 & 2 \\ 1 & 2 & 1 & 2 \end{bmatrix}$

$\begin{bmatrix} \log_{3}4 & \log_{3}16 & \log_{3}\frac{1}{16}& \log_{3} \frac{1}{4} \\ \log_{3}16 & 1 & 3 & -3 \\ \log_{3}4 & -1 & -2 & 2 \\ \log_{3}4 & 2 & 1 & 2 \end{bmatrix}$

$\begin{bmatrix} \log_{3}4 & \log_{3}16 & \log_{3}\frac{1}{16}& \log_{3} \frac{1}{4} \\ 2 & 1 & 3 & -3 \\ 1 & -1 & -2 & 2 \\ 1 & 2 & 1 & 2 \end{bmatrix}$

$\begin{bmatrix} \log_{3}4 & 2 & -2 & -1 \\ \log_{3} 16 & 1 & 3 & -3 \\ \log_{3} 4 & -1 & -2 & 2 \\ \log_{3}4 & 2 & 1 & 2 \end{bmatrix}$

Correct answer:

$\begin{bmatrix} \log_{3}4 & \log_{3}16 & \log_{3}\frac{1}{16}& \log_{3} \frac{1}{4} \\ 2 & 1 & 3 & -3 \\ 1 & -1 & -2 & 2 \\ 1 & 2 & 1 & 2 \end{bmatrix}$

Explanation:

is an elementary matrix, in that it can be formed from the (four-by-four) identity matrix by a single row operation. Since differs from in that the entry $\log_{3}4$ is in Row 1 Column 1 rather than 1, the row operation is

$( \log_{3}4)R1 \rightarrow R1$ .

Taking the product has the same effect on :

$B = \begin{bmatrix} 1 & 2 & -2 & -1 \\ 2 & 1 & 3 & -3 \\ 1 & -1 & -2 & 2 \\ 1 & 2 & 1 & 2 \end{bmatrix}$

$AB = \begin{bmatrix} 1 ( \log_{3}4) & 2( \log_{3}4) & -2( \log_{3}4) & -1( \log_{3}4) \\ 2 & 1 & 3 & -3 \\ 1 & -1 & -2 & 2 \\ 1 & 2 & 1 & 2 \end{bmatrix}$

$= \begin{bmatrix} \log_{3}4 & \log_{3} (4^{2} ) & \log_{3}(4^{-2}) & \log_{3}(4^{-1 }) \\ 2 & 1 & 3 & -3 \\ 1 & -1 & -2 & 2 \\ 1 & 2 & 1 & 2 \end{bmatrix}$

$= \begin{bmatrix} \log_{3}4 & \log_{3}16 & \log_{3}\frac{1}{16}& \log_{3} \frac{1}{4} \\ 2 & 1 & 3 & -3 \\ 1 & -1 & -2 & 2 \\ 1 & 2 & 1 & 2 \end{bmatrix}$ ,

the correct choice.

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Example Question #601 : Linear Algebra

$A = \begin{bmatrix} 2 & 3 & 6 \\ 0 & 2 & 1 \\ 0 & 0 & 2 \end{bmatrix}$

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

Possible Answers:

False

True

Correct answer:

False

Explanation:

One of the conditions a matrix must meet in order to be a matrix in row-echelon form is that each row with nonzero entries must have 1 as its first such entry. Since in all three rows, the first nonzero entry is 2, violates this condition, and it is not in row-echelon form.

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Example Question #41 : Linear Equations

$L = \begin{bmatrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ - 1 & -2 & 1 \end{bmatrix}$ ; $U = \begin{bmatrix} 2 & 3 & 1 \\ 0 & -2 & 1 \\ 0 & 0 & -1 \end{bmatrix}$ ; $A = \begin{bmatrix} 2 & 6 & -2 \\ 3 & 7 & 1 \\ 1 & 4 & -4 \end{bmatrix}$

True or false: gives a valid LU-factorization of .

Possible Answers:

False, because is not the correct type of matrix.

True

False, because $LU \ne A$ .

Correct answer:

False, because $LU \ne A$ .

Explanation:

An LU-decomposition of a matrix expresses the matrix as a product of a Lower-triangular matrix and an Upper-triangular matrix .

A matrix is lower-triangular if all entries above the main diagonal are equal to 0; an upper-triangular matrix is analogously defined. As seen below, and are lower- and upper-triangular, respectively:

$L = \begin{bmatrix} 1 & {\color{Red}\textbf{ 0}} &{\color{Red}\textbf{ 0}} \\ 3 & 1 & {\color{Red}\textbf{ 0}} \\ - 1 & -2 & 1 \end{bmatrix}$

is lower-triangular, as the elements above its main diagonal are all zero.

$U = \begin{bmatrix} 2 & 3 & 1 \\ {\color{Red}\textbf{ 0}} & -2 & 1 \\ {\color{Red}\textbf{ 0}} & {\color{Red}\textbf{ 0}} & -1 \end{bmatrix}$

is upper-triangular, as the elements below its main diagonal are all zero.

It remains to be shown that LU = A . Multiply each row of the former by each column of the latter by adding the products of corresponding entries:

$L U = \begin{bmatrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ - 1 & -2 & 1 \end{bmatrix}\begin{bmatrix} 2 & 3 & 1 \\ 0 & -2 & 1 \\ 0 & 0 & -1 \end{bmatrix}$

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

$= \begin{bmatrix} 2 & 3 & 1 \\ 6 & 7 & 4 \\ -2 & 1 & -4 \end{bmatrix}$

$\ne A$

Therefore, does not give a valid LU-factorization of .; the reason is that $LU \ne A$ .

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Linear Algebra : Linear Equations

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Example Question #41 : Reduced Row Echelon Form And Row Operations

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Example Question #41 : Reduced Row Echelon Form And Row Operations

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