Question

1. Find the vector form of the general solution of the given linear system Ax = b ; then use that result to find the vector form of the general solution of Ax = 0.

x subscript 1 + x subscript 2 + 2x subscript 3 = 6
x subscript 1 + x subscript 3 = -2
2x subscript 1 + x subscript 2 + 3x subscript = 4

2. Find a basis for the null space and row space of A.
1 -1 3
A = 5 -4 -2
7 -6 4

3. A matrix in row echelon form is given. By inspection, find bases for the row and column spaces of the matrix A.

1 0 2
A = 0 0 1
0 0 0

Answer 1

1. Find the vector form of the general solution of the given linear system Ax = b ; then use that result to find the vector form of the general solution of Ax = 0.

$\begin{cases} x_1+x_2+2x_3=6 \\x_1+x_3=-2 \\ 2x_1+x_2+3x_3=4 \end{cases}$

Let $x_3=t$ , replace this in the second equation: $x_1=-2-t$

Then replace $x_3=t, x_1=-2-t$ in the first and third equations:

$\begin{cases} (-2-t)+x_2+2t=6 \\ 2(-2-t)+x_2+3t=4 \end{cases}, \quad x_2= 8-t$

The general solution of the linear system Ax=b is

$\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} -2-t \\ 8-t \\t \end{bmatrix} = \begin{bmatrix} -2 \\ 8 \\0 \end{bmatrix}+ t\begin{bmatrix}-1 \\ -1 \\ 0\end{bmatrix}, \quad t\in \mathbb{R}$

The general solution of the linear system Ax=0 is

$\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = t\begin{bmatrix}-1 \\ -1 \\ 0\end{bmatrix}, \quad t\in \mathbb{R}$

2. Find a basis for the null space and row space of A.

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

We reduce the matrix A as follows

$A =\begin{bmatrix} 1&-1&3 \\ 5&-4&-2 \\ 7&-6&4 \end{bmatrix} \to ^{R_2-5R_1}_{R_3-7R_1}\to \begin{bmatrix} 1&-1&3 \\ 0&1&-17 \\ 0&1&-17 \end{bmatrix} \to^{R_3-R_2}_{R_1+R_2} \to \begin{bmatrix} 1&0&-14 \\ 0&1&-17 \\ 0&0&0 \end{bmatrix}$

The last matrix is in reduced row echelon form.

a) basis for null space

We see that the general solution of Ax=0 is

$\begin{cases} x_1=14x_3 \\ x_2=17x_3 \end{cases}$

where $x_3$ is free variable.

The vector form solution to Ax=0 is

$\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = x_3\begin{bmatrix}14 \\ 17 \\ 1\end{bmatrix}, \quad x_3\in \mathbb{R}$

The null space of the matrix A is given by

$N(A)=\{ x\in \mathbb{R}^3 | x= x_3\begin{bmatrix}14 \\ 17 \\ 1\end{bmatrix}, \quad x_3\in \mathbb{R}\} = Span \begin{Bmatrix} {\begin{bmatrix}14 \\ 17 \\ 1\end{bmatrix} } \end{Bmatrix}$

We can see, that vector $\begin{bmatrix}14 \\ 17 \\ 1\end{bmatrix}$ is a basis for null space.

b) basis for row space

The nonzero rows of matrix (in reduced row echelon form) is a basis for the row space of A.

Thus, $\begin{Bmatrix} \begin{bmatrix}1 \\ 0 \\ -14\end{bmatrix},\begin{bmatrix}0 \\ 1 \\ -17\end{bmatrix} \end{Bmatrix}$ is a basis for the row space of A.

3. A matrix in row echelon form is given. By inspection, find bases for the row and column spaces of the matrix A.

$A=\begin{pmatrix} 1&0&2 \\ 0&0&1 \\ 0&0&0 \end{pmatrix}$

a) bases for the row space

The nonzero rows of a matrix in row echelon form are linearly independent.

Therefore, the row space has basis $\begin{Bmatrix} \begin{bmatrix}1 \\ 0 \\ 2\end{bmatrix},\begin{bmatrix}0 \\ 0 \\ 1\end{bmatrix} \end{Bmatrix}$

b) bases for the column space

The nonzero columns of a matrix in row echelon form are linearly independent.

Therefore, the column space has basis $\begin{Bmatrix} \begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix},\begin{bmatrix}2 \\ 1 \\ 0\end{bmatrix} \end{Bmatrix}$