Day 2:

Row reduction and elementary matrices

Bases

Definition 4. Let \(V\) be a subspace. A basis for \(V\) is a collection of vectors \(\{v_{1},v_{2},\ldots,v_{r}\}\subset V\) with the following properties:

  1. If \(v\in V\), then there are scalars \(a_{1},\ldots,a_{r}\) such that \[a_{1}v_{1} + a_{2}v_{2} + \cdots + a_{r}v_{r} = v\]
  2. If \[b_{1}v_{1}+b_{2}v_{2}+\cdots+b_{r}v_{r} = 0\] for some scalars \(b_{1},\ldots,b_{r}\), then \(b_{1}=b_{2}=\cdots=b_{r}=0\).

If a sequence of vectors satisfy Property 1, then we say that \(\{v_{1},v_{2},\ldots,v_{r}\}\) spans \(V\).

If a sequence of vectors satisfy Property 2, then we say that \(\{v_{1},v_{2},\ldots,v_{r}\}\) is linearly independent.

Linear Dependence

Definition 5. Let \(v_{1},v_{2},\ldots,v_{r}\) be vectors in \(\R^M\). If there are scalars \(a_{1},a_{2},\ldots,a_{r}\) at least one of which is not zero such that \[a_{1}v_{1}+a_{2}v_{2}+\cdots+a_{r}v_{r} = 0,\] then we say that the vectors \(v_{1},v_{2},\ldots,v_{r}\) are (linearly) dependent.

Examples. 1) \(\left[\begin{matrix}1\\ 0\end{matrix}\right]\) and \(\left[\begin{matrix}2\\ 0\end{matrix}\right]\) are dependent since \(2\left[\begin{matrix}1\\ 0\end{matrix}\right]+(-1)\left[\begin{matrix}2\\ 0\end{matrix}\right]=0\)

2) \(\left[\begin{matrix}1\\ 0\\ -1\end{matrix}\right],\left[\begin{matrix}1\\ 1\\ 2\end{matrix}\right],\left[\begin{matrix}2.5\\ 1\\ 0.5\end{matrix}\right]\) are dependent since \[3\!\left[\begin{matrix}1\\ 0\\ -1\end{matrix}\right]\!\!+2\!\left[\begin{matrix}1\\ 1\\ 2\end{matrix}\right]\!-2\!\left[\begin{matrix}2.5\\ 1\\ 0.5\end{matrix}\right]\! =\! \left[\begin{matrix} 0\\ 0\\ 0\end{matrix}\right] \]

3) The zero vector is dependent. It is the only vector that is dependent.

Finding Dependence

Given vectors \(v_{1},v_{2},\ldots,v_{r}\), how can we figure out whether they are dependent?

  • Find scalars \(a_{1},a_{2},\ldots,a_{r}\) so that \(a_{1}v_{1} +a_{2}v_{2}+\cdots + a_{r}v_{r} = 0\)
  • Set \[A = \left[\begin{matrix} v_{1} & v_{2} & \cdots & v_{r}\end{matrix}\right]\] Find \(b\in\R^{r}\) so that \(Ab=0\).
  • With \(A\) as above, solve \(Ax=0\).
  • Find the reduced row echelon form of \(A\)...

Equivalent problems:

Row reduction

Two matrices \(A\) and \(B\) are row equivalent if \(B\) can be obtained from \(A\) by some sequence of row operations. If \(A\) and \(B\) are row equivalent, then we write \(A\sim B\).

 

The row operations are the following:

  1. Swap two rows
  2. Replace row \(R_{1}\) with \(\beta R_{1}\) for some nonzero scalar \(\beta\).
  3. Given two rows \(R_{1}\) and \(R_{2}\) we can replace \(R_{1}\) with \(R_{1}+\beta R_{2}\) for some scalar \(\beta\).

Row reduction

Two matrices \(A\) and \(B\) are row equivalent if \(B\) can be obtained from \(A\) by some sequence of row operations. If \(A\) and \(B\) are row equivalent, then we write \(A\sim B\).

 

The row operations are the following:

  1. Swap two rows
  2. Replace row \(R_{1}\) with \(\beta R_{1}\) for some nonzero scalar \(\beta\).
  3. Given two rows \(R_{1}\) and \(R_{2}\) we can replace \(R_{1}\) with \(R_{1}+\beta R_{2}\) for some scalar \(\beta\).

\[\begin{bmatrix} a & b & c\\ d & e & f\end{bmatrix} \sim \begin{bmatrix} d & e & f\\ a & b & c\end{bmatrix}\]
 

Row reduction

Two matrices \(A\) and \(B\) are row equivalent if \(B\) can be obtained from \(A\) by some sequence of row operations. If \(A\) and \(B\) are row equivalent, then we write \(A\sim B\).

 

The row operations are the following:

  1. Swap two rows
  2. Replace row \(R_{1}\) with \(\beta R_{1}\) for some nonzero scalar \(\beta\).
  3. Given two rows \(R_{1}\) and \(R_{2}\) we can replace \(R_{1}\) with \(R_{1}+\beta R_{2}\) for some scalar \(\beta\).

\[\begin{bmatrix} a & b & c\\ d & e & f\end{bmatrix} \sim \begin{bmatrix} \beta a & \beta b & \beta c\\ d & e & f\end{bmatrix}\]

Row reduction

Two matrices \(A\) and \(B\) are row equivalent if \(B\) can be obtained from \(A\) by some sequence of row operations. If \(A\) and \(B\) are row equivalent, then we write \(A\sim B\).

 

The row operations are the following:

  1. Swap two rows
  2. Replace row \(R_{1}\) with \(\beta R_{1}\) for some nonzero scalar \(\beta\).
  3. Given two rows \(R_{1}\) and \(R_{2}\) we can replace \(R_{1}\) with \(R_{1}+\beta R_{2}\) for some scalar \(\beta\).

\[\begin{bmatrix} a & b & c\\ d & e & f\end{bmatrix} \sim \begin{bmatrix} a+\beta d & b+\beta e & c+\beta f\\ d & e & f\end{bmatrix}\]

Example. Consider the matrix

\[\begin{bmatrix}  0 & 2 & 4\\ 1 & 0 & 1\\ 0 & 0 & 1\end{bmatrix}\]

Multiply row \(2\) by \(\frac{1}{2}\):

\[\begin{bmatrix} 1 & 0 & 1\\ 0 & 2 & 4\\ 0 & 0 & 1\end{bmatrix}\sim\begin{bmatrix} 1 & 0 & 1\\ 0 & 1 & 2\\ 0 & 0 & 1\end{bmatrix}\]

Replace (Row 1) by (Row 1)-(Row 3):

\[\begin{bmatrix} 1 & 0 & 1\\ 0 & 1 & 2\\ 0 & 0 & 1\end{bmatrix}\sim \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 2\\ 0 & 0 & 1\end{bmatrix}\]

Swap (Row 1) and (Row 2):

\[\begin{bmatrix}  0 & 2 & 4\\ 1 & 0 & 1\\ 0 & 0 & 1\end{bmatrix}\sim \begin{bmatrix} 1 & 0 & 1\\ 0 & 2 & 4\\ 0 & 0 & 1\end{bmatrix}\]

Row reduction

\[A=\left[\begin{matrix} a_{11} & a_{12} & \cdots & a_{1n}\\ a_{21} & a_{22} & & a_{2n}\\ \vdots & & \ddots & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn}\end{matrix}\right]\quad B=\left[\begin{matrix} b_{11} & b_{12} & \cdots & b_{1k}\\ b_{21} & b_{22} & & b_{2k}\\ \vdots & & \ddots & \vdots \\ b_{n1} & b_{n2} & \cdots & b_{nk}\end{matrix}\right]\]

\[ A=\left[\begin{matrix} a_{11} & a_{12} & \cdots & a_{1n}\\ a_{21} & a_{22} & & a_{2n}\\ \vdots & & \ddots & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn}\end{matrix}\right]\quad B=\left[\begin{matrix} \left[\begin{matrix}b_{11}\\ b_{21}\\ \vdots\\ b_{n1}\end{matrix}\right] & \left[\begin{matrix}b_{12}\\ b_{22}\\ \vdots\\ b_{n2}\end{matrix}\right] & \cdots & \left[\begin{matrix}b_{1k}\\ b_{2k}\\ \vdots\\ b_{nk}\end{matrix}\right] \end{matrix}\right] \]

\[ A=\left[\begin{matrix} a_{11} & a_{12} & \cdots & a_{1n}\\ a_{21} & a_{22} & & a_{2n}\\ \vdots & & \ddots & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn}\end{matrix}\right]\quad B=\left[\begin{matrix} b_{1} & b_{2} & \cdots & b_{k}\end{matrix}\right] \]

 

where  \(b_{1}=\left[\begin{matrix}b_{11}\\ b_{21}\\ \vdots\\ b_{n1}\end{matrix}\right],\ b_{2}=\left[\begin{matrix}b_{12}\\ b_{22}\\ \vdots\\ b_{n2}\end{matrix}\right],\ldots, b_{k}=\left[\begin{matrix}b_{1k}\\ b_{2k}\\ \vdots\\ b_{nk}\end{matrix}\right]\)

\[AB=\left[\begin{matrix} Ab_{1} & Ab_{2} & \cdots & Ab_{k}\end{matrix}\right]\]

Matrix multiplication (Columnwise)

Then we define the product as follows:

\[A=\left[\begin{matrix} a_{11} & a_{12} & \cdots & a_{1n}\\ a_{21} & a_{22} & & a_{2n}\\ \vdots & & \ddots & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn}\end{matrix}\right]\quad B=\left[\begin{matrix} b_{11} & b_{12} & \cdots & b_{1k}\\ b_{21} & b_{22} & & b_{2k}\\ \vdots & & \ddots & \vdots \\ b_{n1} & b_{n2} & \cdots & b_{nk}\end{matrix}\right]\]

\(A=\left[\begin{matrix} a_{11} & a_{12} & \cdots & a_{1n}\\ a_{21} & a_{22} & & a_{2n}\\ \vdots & & \ddots & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn}\end{matrix}\right]\quad B=\left[\begin{array}{ccc} R_{1}\\ R_{2}\\ \vdots \\ R_{m}\end{array}\right]\)

 

where \(R_{j} = \left[\begin{matrix} b_{j1} & b_{j2} & \cdots & b_{jk}\end{matrix}\right]\) for each \(j=1,2,\ldots,n\)

\(A=\left[\begin{matrix} a_{11} & a_{12} & \cdots & a_{1n}\\ a_{21} & a_{22} & & a_{2n}\\ \vdots & & \ddots & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn}\end{matrix}\right]\quad B=\left[\begin{array}{ccc} [\begin{matrix} b_{11} & b_{12} & \cdots & b_{1k}\end{matrix}]\\ [\begin{matrix} b_{21} & b_{22} & \cdots & b_{2k}\end{matrix}]\\ \vdots \\ [\begin{matrix} b_{n1} & b_{n2} & \cdots & b_{nk}\end{matrix}]\end{array}\right]\)

\[AB=\left[\begin{matrix} a_{11} & a_{12} & \cdots & a_{1n}\\ a_{21} & a_{22} & & a_{2n}\\ \vdots & & \ddots & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn}\end{matrix}\right]\left[\begin{matrix}R_{1}\\ R_{2}\\ \vdots \\ R_{n}\end{matrix}\right] = \left[\begin{matrix} a_{11}R_{1}+ a_{12}R_{2} + \cdots + a_{1n}R_{n}\\ a_{21}R_{1} + a_{22}R_{2} + \cdots + a_{2n}R_{n}\\ \vdots \\ a_{m1}R_{1} + a_{m2}R_{2} \cdots + a_{mn}R_{n}\end{matrix}\right]\]

Each row of \(AB\) is a linear combination of rows of \(B\).

Matrix multiplication (Rowwise)

Matrix multiplication (Rowwise)

Example. Consider the product

\[\begin{bmatrix}  1 & -2 & 3\\ 1 & 0 & 0\\ 0 & 0 & 4\end{bmatrix}\begin{bmatrix}  0 & 2 & 4\\ 1 & 0 & 1\\ 0 & 0 & 1\end{bmatrix}\]

Row 1 in the product is the linear combination:

\[1\cdot \begin{bmatrix} 0 & 2 & 4\end{bmatrix} + (-2)\cdot \begin{bmatrix} 1 & 0 & 1\end{bmatrix} + 3\cdot \begin{bmatrix} 0 & 0 & 1\end{bmatrix} = \begin{bmatrix} -2 & 2 & 5\end{bmatrix}\]

Row 2 in the product is the linear combination:

\[1\cdot \begin{bmatrix} 0 & 2 & 4\end{bmatrix} + 0\cdot \begin{bmatrix} 1 & 0 & 1\end{bmatrix} + 0\cdot \begin{bmatrix} 0 & 0 & 1\end{bmatrix} = \begin{bmatrix} 0 & 2 & 4\end{bmatrix}\]

Row 3 in the product is the linear combination:

\[0\cdot \begin{bmatrix} 0 & 2 & 4\end{bmatrix} + 0\cdot \begin{bmatrix} 1 & 0 & 1\end{bmatrix} + 4\cdot \begin{bmatrix} 0 & 0 & 1\end{bmatrix} = \begin{bmatrix} 0 & 0 & 4\end{bmatrix}\]

Hence:

\[\begin{bmatrix}  1 & -2 & 3\\ 1 & 0 & 0\\ 0 & 0 & 4\end{bmatrix}\begin{bmatrix}  0 & 2 & 4\\ 1 & 0 & 1\\ 0 & 0 & 1\end{bmatrix} = \begin{bmatrix} -2 & 2 & 5\\ 0 & 2 & 4\\ 0 & 0 & 4\end{bmatrix}\]

Elementary Matrices and Row Operations

Two matrices \(A\) and \(B\) are row equivalent if \(B\) can be obtained from \(A\) by some sequence of row operations. The row operations are the following:

  1. Swap two rows
  2. Replace row \(R_{1}\) with \(aR_{1}\) for some nonzero scalar \(a\).
  3. Given two rows \(R_{1}\) and \(R_{2}\) we can replace \(R_{1}\) with \(R_{1}+aR_{2}\) for some scalar \(a\).

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

Example: 

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

Swap rows 2 and 3

\(\sim\)

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

\(=\)

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

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

Elementary Matrices and Row Operations

Two matrices \(A\) and \(B\) are row equivalent if \(B\) can be obtained from \(A\) by some sequence of row operations. The row operations are the following:

  1. Swap two rows
  2. Replace row \(R_{1}\) with \(aR_{1}\) for some nonzero scalar \(a\).
  3. Given two rows \(R_{1}\) and \(R_{2}\) we can replace \(R_{1}\) with \(R_{1}+aR_{2}\) for some scalar \(a\).

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

Example: 

\(\begin{bmatrix} 2 & 2 & 1 & 0\\ 0 & -6 & 3 & 9\\ 1 & 0 & 1 & 2\end{bmatrix}\)

Replace row 2 with 3 times row 2

\(\sim\)

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

\(=\)

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

\(\begin{bmatrix} 2 & 2 & 1 & 0\\ 0 & -6 & 3 & 9\\ 1 & 0 & 1 & 2\end{bmatrix}\)

Elementary Matrices and Row Operations

Two matrices \(A\) and \(B\) are row equivalent if \(B\) can be obtained from \(A\) by some sequence of row operations. The row operations are the following:

  1. Swap two rows
  2. Replace row \(R_{1}\) with \(aR_{1}\) for some nonzero scalar \(a\).
  3. Given two rows \(R_{1}\) and \(R_{2}\) we can replace \(R_{1}\) with \(R_{1}+aR_{2}\) for some scalar \(a\).

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

Example: 

\(\begin{bmatrix} 0 & 2 & -1 & -4\\ 0 & -6 & 3 & 9\\ 1 & 0 & 1 & 2\end{bmatrix}\)

Replace row 1 with row 1 plus (-2) times row 3

\(\sim\)

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

\(=\)

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

\(\begin{bmatrix} 0 & 2 & -1 & -4\\ 0 & -6 & 3 & 9\\ 1 & 0 & 1 & 2\end{bmatrix}\)

Example. Consider the matrix

\[\begin{bmatrix}  0 & 2 & 4\\ 1 & 0 & 1\\ 0 & 0 & 1\end{bmatrix}\]

Multiply row \(2\) by \(\frac{1}{2}\):

\[\begin{bmatrix} 1 & 0 & 0\\ 0 & \frac{1}{2} & 0\\ 0 & 0 & 1\end{bmatrix}\begin{bmatrix} 1 & 0 & 1\\ 0 & 2 & 4\\ 0 & 0 & 1\end{bmatrix} = \begin{bmatrix} 1 & 0 & 1\\ 0 & 1 & 2\\ 0 & 0 & 1\end{bmatrix}\]

Replace (Row 1) by (Row 1)-(Row 3):

\[\begin{bmatrix} 1 & 0 & -1\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}\begin{bmatrix} 1 & 0 & 1\\ 0 & 1 & 2\\ 0 & 0 & 1\end{bmatrix} = \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 2\\ 0 & 0 & 1\end{bmatrix}\]

Swap (Row 1) and (Row 2):

\[\begin{bmatrix} 0 & 1 & 0\\ 1 & 0 & 0\\ 0 & 0 & 1\end{bmatrix}\begin{bmatrix}  0 & 2 & 4\\ 1 & 0 & 1\\0 & 0 & 1\end{bmatrix} = \begin{bmatrix} 1 & 0 & 1\\ 0 & 2 & 4\\ 0 & 0 & 1\end{bmatrix}\]

Elementary Matrices and Row Operations

Swap (Row 1) and (Row 3),

Multiply row \(2\) by \(\frac{1}{2}\),

Replace (Row 1) by (Row 1)-(Row 3):

\[\underbrace{\begin{bmatrix} 1 & 0 & -1\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}\begin{bmatrix} 1 & 0 & 0\\ 0 & \frac{1}{2} & 0\\ 0 & 0 & 1\end{bmatrix}\begin{bmatrix} 0 & 1 & 0\\ 1 & 0 & 0\\ 0 & 0 & 1\end{bmatrix}}\begin{bmatrix}  0 & 2 & 4\\ 1 & 0 & 1\\0 & 0 & 1\end{bmatrix} = \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 2\\ 0 & 0 & 1\end{bmatrix}\]

Elementary Matrices and Row Operations

Example. Consider the matrix

\[\begin{bmatrix}  0 & 2 & 4\\ 1 & 0 & 1\\ 0 & 0 & 1\end{bmatrix}\]

\[\begin{bmatrix} 0 & 1 & -1\\ \frac{1}{2} & 0 & 0\\ 0 & 0 & 1\end{bmatrix}\begin{bmatrix}  0 & 2 & 4\\ 1 & 0 & 1\\0 & 0 & 1\end{bmatrix} = \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 2\\ 0 & 0 & 1\end{bmatrix}\]

Elementary Matrices and Row Operations

Note that every row operation can be undone by another row operation:

Row operation

  1. Swap two rows
  2. Replace \(R_{1}\) with \(aR_{1}\), \((a\neq 0)\)
  3.  Replace \(R_{1}\) with \(R_{1}+aR_{2}\) 

Inverse row operation

  1. Swap two rows
  2. Replace \(R_{1}\) with \(\frac{1}{a}R_{1}\)
  3. Replace \(R_{1}\) with \(R_{1}-aR_{2}\)

Elementary Matrices and Row Operations

Swap (Row 1) and (Row 2),

Multiply (Row \(2\)) by \(\frac{1}{2}\),

Replace (Row 1) by (Row 1)-(Row 3):

\[\begin{bmatrix} 0 & 1 & -1\\ \frac{1}{2} & 0 & 0\\ 0 & 0 & 1\end{bmatrix}\begin{bmatrix}  0 & 2 & 4\\ 1 & 0 & 1\\0 & 0 & 1\end{bmatrix} = \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 2\\ 0 & 0 & 1\end{bmatrix}\]

\[\begin{bmatrix} 1 & 0 & -1\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}\begin{bmatrix} 1 & 0 & 0\\ 0 & \frac{1}{2} & 0\\ 0 & 0 & 1\end{bmatrix}\begin{bmatrix} 0 & 1 & 0\\ 1 & 0 & 0\\ 0 & 0 & 1\end{bmatrix}\begin{bmatrix}  0 & 2 & 4\\ 1 & 0 & 1\\0 & 0 & 1\end{bmatrix} = \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 2\\ 0 & 0 & 1\end{bmatrix}\]

Replace (Row 1) by (Row 1)+(Row 3),

Multiply (Row \(2\)) by \(2\),

Swap (Row 1) and (Row 2):

\[\begin{bmatrix} 0 & 1 & 0\\ 1 & 0 & 0\\ 0 & 0 & 1\end{bmatrix}\begin{bmatrix} 1 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 1\end{bmatrix}\begin{bmatrix} 1 & 0 & 1\\ 0 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 2\\ 0 & 0 & 1\end{bmatrix} = \begin{bmatrix}  0 & 2 & 4\\ 1 & 0 & 1\\0 & 0 & 1\end{bmatrix}\]

\[\begin{bmatrix} 0 & 2 & 0\\ 1 & 0 & 1\\ 0 & 0 & 1\end{bmatrix}\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 2\\ 0 & 0 & 1\end{bmatrix} = \begin{bmatrix}  0 & 2 & 4\\ 1 & 0 & 1\\0 & 0 & 1\end{bmatrix}\]

Elementary Matrices and Row Operations

To summarize: 

 

If \(A\sim B\) then there is a matrix \(R\), which is a product of elementary matrices, such that \(B=RA\).

 

Moreover, there is a matrix \(S\), which is also a product of elementary matrices such that \(A=SB\).

 

 

Next time we will prove:

 

Theorem. If \(A\) and \(B\) are \(M\times N\) matrices such that \(A\sim B\), then

\[\{x\in\mathbb{R}^{N} : Ax=0\} = \{x\in\mathbb{R}^{N} : Bx=0\}.\]

RREF

A matrix is in reduced row echelon form (RREF) if it satisfies all of the following:

  1. All nonzero rows are above any rows of all zeros
  2. The leading entry in each nonzero row is strictly to the right of the leading entry of the row above it.
  3. The leading entry of each nonzero row is \(1\).
  4. Each column containing a leading \(1\) (also known as a pivot) has all zeros in the rest of the column.

a) \(\begin{bmatrix} 1 & 0 & 2\\ 0 & 1 & 0\end{bmatrix}\)

g) \(\begin{bmatrix} 0 & 1 & 2 & 0\\ 0 & 0 & 0 & 1\end{bmatrix}\)

b) \(\begin{bmatrix} 0 & 1 & 1\\ 1 & 0 & 0\end{bmatrix}\)

f) \(\begin{bmatrix} 1 & 2 & 0\\ 0 & 0 & 2\end{bmatrix}\)

c) \(\begin{bmatrix} 1 & 0 & 1\\ 0 & 0 & 0\\ 0 & 1 & 2 \end{bmatrix}\)

d) \(\begin{bmatrix} 1 & 0 & 1\\ 0 & 1 & 2\\ 0 & 0 & 1\\ 0 & 0 & 0\end{bmatrix}\)

Which of the following are in RREF?

X

X

X

X

End Day 2