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 1 + x 2 + 2 x 3 = 6 x 1 + x 3 = − 2 2 x 1 + x 2 + 3 x 3 = 4 \begin{cases}
x_1+x_2+2x_3=6
\\x_1+x_3=-2
\\ 2x_1+x_2+3x_3=4
\end{cases} ⎩ ⎨ ⎧ x 1 + x 2 + 2 x 3 = 6 x 1 + x 3 = − 2 2 x 1 + x 2 + 3 x 3 = 4
Let x 3 = t x_3=t x 3 = t x 1 = − 2 − t x_1=-2-t x 1 = − 2 − t
Then replace x 3 = t , x 1 = − 2 − t x_3=t, x_1=-2-t x 3 = t , x 1 = − 2 − t
{ ( − 2 − t ) + x 2 + 2 t = 6 2 ( − 2 − t ) + x 2 + 3 t = 4 , x 2 = 8 − t \begin{cases}
(-2-t)+x_2+2t=6
\\ 2(-2-t)+x_2+3t=4
\end{cases}, \quad
x_2= 8-t { ( − 2 − t ) + x 2 + 2 t = 6 2 ( − 2 − t ) + x 2 + 3 t = 4 , x 2 = 8 − t
The general solution of the linear system Ax=b is
[ x 1 x 2 x 3 ] = [ − 2 − t 8 − t t ] = [ − 2 8 0 ] + t [ − 1 − 1 0 ] , t ∈ R \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} ⎣ ⎡ x 1 x 2 x 3 ⎦ ⎤ = ⎣ ⎡ − 2 − t 8 − t t ⎦ ⎤ = ⎣ ⎡ − 2 8 0 ⎦ ⎤ + t ⎣ ⎡ − 1 − 1 0 ⎦ ⎤ , t ∈ R
The general solution of the linear system Ax=0 is
[ x 1 x 2 x 3 ] = t [ − 1 − 1 0 ] , t ∈ R \begin{bmatrix}
x_1
\\ x_2
\\ x_3
\end{bmatrix} =
t\begin{bmatrix}-1 \\ -1 \\ 0\end{bmatrix}, \quad t\in \mathbb{R} ⎣ ⎡ x 1 x 2 x 3 ⎦ ⎤ = t ⎣ ⎡ − 1 − 1 0 ⎦ ⎤ , t ∈ R
2. Find a basis for the null space and row space of A.
A = [ 1 − 1 3 5 − 4 − 2 7 − 6 4 ] A =\begin{bmatrix} 1&-1&3
\\ 5&-4&-2
\\ 7&-6&4
\end{bmatrix} A = ⎣ ⎡ 1 5 7 − 1 − 4 − 6 3 − 2 4 ⎦ ⎤
We reduce the matrix A as follows
A = [ 1 − 1 3 5 − 4 − 2 7 − 6 4 ] → R 3 − 7 R 1 R 2 − 5 R 1 → [ 1 − 1 3 0 1 − 17 0 1 − 17 ] → R 1 + R 2 R 3 − R 2 → [ 1 0 − 14 0 1 − 17 0 0 0 ] 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} A = ⎣ ⎡ 1 5 7 − 1 − 4 − 6 3 − 2 4 ⎦ ⎤ → R 3 − 7 R 1 R 2 − 5 R 1 → ⎣ ⎡ 1 0 0 − 1 1 1 3 − 17 − 17 ⎦ ⎤ → R 1 + R 2 R 3 − R 2 → ⎣ ⎡ 1 0 0 0 1 0 − 14 − 17 0 ⎦ ⎤
The last matrix is in reduced row echelon form.
a) basis for null space
We see that the general solution of Ax=0 is
{ x 1 = 14 x 3 x 2 = 17 x 3 \begin{cases}
x_1=14x_3
\\ x_2=17x_3
\end{cases} { x 1 = 14 x 3 x 2 = 17 x 3
where x 3 x_3 x 3
The vector form solution to Ax=0 is
[ x 1 x 2 x 3 ] = x 3 [ 14 17 1 ] , x 3 ∈ R \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} ⎣ ⎡ x 1 x 2 x 3 ⎦ ⎤ = x 3 ⎣ ⎡ 14 17 1 ⎦ ⎤ , x 3 ∈ R
The null space of the matrix A is given by
N ( A ) = { x ∈ R 3 ∣ x = x 3 [ 14 17 1 ] , x 3 ∈ R } = S p a n { [ 14 17 1 ] } 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} N ( A ) = { x ∈ R 3 ∣ x = x 3 ⎣ ⎡ 14 17 1 ⎦ ⎤ , x 3 ∈ R } = Sp an ⎩ ⎨ ⎧ ⎣ ⎡ 14 17 1 ⎦ ⎤ ⎭ ⎬ ⎫
We can see, that vector [ 14 17 1 ] \begin{bmatrix}14 \\ 17 \\ 1\end{bmatrix} ⎣ ⎡ 14 17 1 ⎦ ⎤
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, { [ 1 0 − 14 ] , [ 0 1 − 17 ] } \begin{Bmatrix}
\begin{bmatrix}1 \\ 0 \\ -14\end{bmatrix},\begin{bmatrix}0 \\ 1 \\ -17\end{bmatrix}
\end{Bmatrix} ⎩ ⎨ ⎧ ⎣ ⎡ 1 0 − 14 ⎦ ⎤ , ⎣ ⎡ 0 1 − 17 ⎦ ⎤ ⎭ ⎬ ⎫
3. A matrix in row echelon form is given. By inspection, find bases for the row and column spaces of the matrix A.
A = ( 1 0 2 0 0 1 0 0 0 ) A=\begin{pmatrix} 1&0&2
\\ 0&0&1
\\ 0&0&0
\end{pmatrix} A = ⎝ ⎛ 1 0 0 0 0 0 2 1 0 ⎠ ⎞
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 { [ 1 0 2 ] , [ 0 0 1 ] } \begin{Bmatrix}
\begin{bmatrix}1 \\ 0 \\ 2\end{bmatrix},\begin{bmatrix}0 \\ 0 \\ 1\end{bmatrix}
\end{Bmatrix} ⎩ ⎨ ⎧ ⎣ ⎡ 1 0 2 ⎦ ⎤ , ⎣ ⎡ 0 0 1 ⎦ ⎤ ⎭ ⎬ ⎫
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 { [ 1 0 0 ] , [ 2 1 0 ] } \begin{Bmatrix}
\begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix},\begin{bmatrix}2 \\ 1 \\ 0\end{bmatrix}
\end{Bmatrix} ⎩ ⎨ ⎧ ⎣ ⎡ 1 0 0 ⎦ ⎤ , ⎣ ⎡ 2 1 0 ⎦ ⎤ ⎭ ⎬ ⎫
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