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 \(\displaystyle 1\).
• 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 (\(\displaystyle aR_n+bR_m\) where \(\displaystyle a\) and \(\displaystyle b\) are constants).

\(\displaystyle \begin{bmatrix} 2& -4& 7\\ 3& -2& 9 \end{bmatrix}\)

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

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

We will make the first number in the \(\displaystyle 2nd\) row \(\displaystyle 0\) using the row operation \(\displaystyle R_2=r_2-3r_1\)

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

We now make the leading entry of the \(\displaystyle 2nd\) row \(\displaystyle 1\) using the row operation \(\displaystyle R_2=r_2/4\)

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

The matrix is now in echelon form.

Next we change the \(\displaystyle 2nd\) term in the \(\displaystyle 1st\) row to \(\displaystyle 0\) using the row operation \(\displaystyle R_1=r_1+2r_2\)

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

\(\displaystyle \begin{bmatrix} 1& 0& 11/4\\ 0& 1& -3/8\end{bmatrix}\)

The matrix is now in reduced row echelon form.

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 \(\displaystyle 1\).
• 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 (\(\displaystyle aR_n+bR_m\) where \(\displaystyle a\) and \(\displaystyle b\) are constants).

The leading entry in the \(\displaystyle 1st\) row is \(\displaystyle 1\), so no operation is needed

\(\displaystyle \begin{bmatrix} 1& -4& 3\\ 5& -4& 9 \end{bmatrix}\)

We will make the first number in the \(\displaystyle 2nd\) row \(\displaystyle 0\) using the row operation \(\displaystyle R_2=r_2-5r_1\)

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

We now make the leading entry of the \(\displaystyle 2nd\) row \(\displaystyle 1\) using the row operation \(\displaystyle R_2=r_2/16\)

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

The matrix is now in echelon form.

Next we change the \(\displaystyle 2nd\) term in the \(\displaystyle 1st\) row to \(\displaystyle 0\) using the row operation \(\displaystyle R_1=r_1+4r_2\)

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

\(\displaystyle \begin{bmatrix} 1&0& 3/2\\ 0& 1& -3/8 \end{bmatrix}\)

The matrix is now in reduced row echelon form.

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 \(\displaystyle 1\).
• 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 (\(\displaystyle aR_n+bR_m\) where \(\displaystyle a\) and \(\displaystyle b\) are constants).

The leading entry in the \(\displaystyle 1st\) row is \(\displaystyle 1\), so no operation is needed

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

We will make the first number in the \(\displaystyle 2nd\) row \(\displaystyle 0\) using the row operation \(\displaystyle R_2=r_2-5r_1\)

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

We now make the leading entry of the \(\displaystyle 2nd\) row \(\displaystyle 1\) using the row operation \(\displaystyle R_2=-r_2/14\)

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

Next we will make the first number in the \(\displaystyle 3rd\) row \(\displaystyle 0\) using the row operation \(\displaystyle R_3=r_3-7r_1\)

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

Next we make the \(\displaystyle 2nd\) number in the \(\displaystyle 3rd\) row \(\displaystyle 0\) using the row operation \(\displaystyle R_3=r_3+5r_2\)

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

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

We now make the leading entry of the \(\displaystyle 3rd\) row \(\displaystyle 1\) using the row operation \(\displaystyle R_3=-14r_3/109\)

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

The matrix is now in echelon form.

Next we change the \(\displaystyle 3rd\) term in the \(\displaystyle 2nd\) row to \(\displaystyle 0\) using the row operation \(\displaystyle R_2=r_2-(9/14)r_3\)

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

Next we change the \(\displaystyle 2nd\) term in the \(\displaystyle 1st\) row to \(\displaystyle 0\) using the row operation \(\displaystyle R_1=r_1-2r_2\)

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

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

Last we change the \(\displaystyle 3rd\) term in the \(\displaystyle 1st\) row to \(\displaystyle 0\) using the row operation \(\displaystyle R_1=r_1-3r_3\)

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

The matrix is now in reduced row echelon form.

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 \(\displaystyle 1\).
• 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 (\(\displaystyle aR_n+bR_m\) where \(\displaystyle a\) and \(\displaystyle b\) are constants).

The leading entry in the \(\displaystyle 1st\) row is \(\displaystyle 1\), so no operation is needed

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

We will make the first number in the \(\displaystyle 2nd\) row \(\displaystyle 0\) using the row operation \(\displaystyle R_2=r_2-5r_1\)

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

We now make the leading entry of the \(\displaystyle 2nd\) row \(\displaystyle 1\) using the row operation \(\displaystyle R_2=-r_2/14\)

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

Next we will make the first number in the \(\displaystyle 3rd\) row \(\displaystyle 0\) using the row operation \(\displaystyle R_3=r_3-7r_1\)

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

Next we make the \(\displaystyle 2nd\) number in the \(\displaystyle 3rd\) row \(\displaystyle 0\) using the row operation \(\displaystyle R_3=r_3+5r_2\)

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

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

We now make the leading entry of the \(\displaystyle 3rd\) row \(\displaystyle 1\) using the row operation \(\displaystyle R_3=-14r_3/109\)

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

The matrix is now in echelon form.

Next we change the \(\displaystyle 3rd\) term in the \(\displaystyle 2nd\) row to \(\displaystyle 0\) using the row operation \(\displaystyle R_2=r_2-(9/14)r_3\)

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

Next we change the \(\displaystyle 2nd\) term in the \(\displaystyle 1st\) row to \(\displaystyle 0\) using the row operation \(\displaystyle R_1=r_1-2r_2\)

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

\(\displaystyle \begin{bmatrix} 1& 0& 3\\ 0& 1& 0\\ 0& 0& 1 \end{bmatrix}\)

Last we change the \(\displaystyle 3rd\) term in the \(\displaystyle 1st\) row to \(\displaystyle 0\) using the row operation \(\displaystyle R_1=r_1-3r_3\)

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

The matrix is now in reduced row echelon form.

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

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

Taking the product \(\displaystyle AB\) has the same effect on \(\displaystyle B\):

\(\displaystyle B = \begin{bmatrix} 1 & 1 & -1 & -1 \\ 2 & 1 & 3 & -3 \\ 1 & -1 & -2 & 2 \\ 1 & 2 & 1 & 2 \end{bmatrix}\),

so

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

\(\displaystyle = \begin{bmatrix} 1+ \log_{2}3 & 1 + \log_{2}9 & -1 + \log_{2}3& -1 + \log_{2}9 \\ 2 & 1 & 3 & -3 \\ 1 & -1 & -2 & 2 \\ 1 & 2 & 1 & 2 \end{bmatrix}\)

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

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. \(\displaystyle A\) 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.

A matrix in row echelon form must fit three conditions:

1) Any rows comprising only zeroes must be concentrated at the bottom. \(\displaystyle A\) 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. \(\displaystyle A\) fits this condition also.

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

\(\displaystyle A\) is in row-echelon form.

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

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

Taking the product \(\displaystyle AB\) has the same effect on \(\displaystyle B\):

\(\displaystyle B = \begin{bmatrix} 1 & 2 & -2 & -1 \\ 2 & 1 & 3 & -3 \\ 1 & -1 & -2 & 2 \\ 1 & 2 & 1 & 2 \end{bmatrix}\)

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

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

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

the correct choice.

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, \(\displaystyle A\) violates this condition, and it is not in row-echelon form.

An LU-decomposition of a matrix \(\displaystyle A\) expresses the matrix as a product of a Lower-triangular matrix \(\displaystyle L\) and an Upper-triangular matrix \(\displaystyle U\). 

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, \(\displaystyle L\) and \(\displaystyle U\) are lower- and upper-triangular, respectively:

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

\(\displaystyle L\) is lower-triangular, as the elements above its main diagonal are all zero.

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

\(\displaystyle U\) is upper-triangular, as the elements below its main diagonal are all zero.

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

\(\displaystyle L U = \begin{bmatrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ - 1 & -2 & 1 \end{bmatrix}\begin{bmatrix} 2 & 3 & 1 \\ 0 & -2 & 1 \\ 0 & 0 & -1 \end{bmatrix}\) 

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

\(\displaystyle = \begin{bmatrix} 2 & 3 & 1 \\ 6 & 7 & 4 \\ -2 & 1 & -4 \end{bmatrix}\)

\(\displaystyle \ne A\)

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