Determine conditions on the scalars $a, b$ so that the following set $S$ of vectors is linearly dependent.
\begin{align*}
S=\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\},
\end{align*}
where
\[\mathbf{v}_1=\begin{bmatrix}
1 \\
3 \\
1
\end{bmatrix}, \mathbf{v}_2=\begin{bmatrix}
1 \\
a \\
4
\end{bmatrix}, \mathbf{v}_3=\begin{bmatrix}
0 \\
2 \\
b
\end{bmatrix}.\]  

Solution.

Let us consider the linear combination
\[x_1\mathbf{v}_1+x_2\mathbf{v}_2+x_3\mathbf{v}_3=\mathbf{0}.\] We want to find conditions on $a, b$ so that there is $(x_1, x_2, x_3) \neq (0, 0, 0)$ satisfying this linear combination.
Since the linear combination can be written as
\[A\mathbf{x}=\mathbf{0},\] where
\[A=[\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3]=\begin{bmatrix}
1 & 1 & 0 \\
3 &a &2 \\
1 & 4 & b
\end{bmatrix}, \text{ and } \mathbf{x}=\begin{bmatrix}
x_1 \\
x_2 \\
x_3
\end{bmatrix},\] our goal is the same as finding conditions on $a, b$ so that $A$ is a singular matrix.

We apply elementary row operations to the augmented matrix $[A\mid \mathbf{0}]$.
\begin{align*}
&[A\mid \mathbf{0}]= \left[\begin{array}{rrr|r}
1 & 1 & 0 & 0 \\
3 &a & 2 & 0 \\
1 & 4 & b & 0
\end{array} \right]\\[10pt] & \xrightarrow{\substack{R_2-3R_1\\R_3-R_1}}
\left[\begin{array}{rrr|r}
1 & 1 & 0 & 0 \\
0 &a-3 & 2 & 0 \\
0 & 3 & b & 0
\end{array} \right] \xrightarrow{R_2\leftrightarrow R_3}
\left[\begin{array}{rrr|r}
1 & 1 & 0 & 0 \\
0 & 3 & b & 0 \\
0 &a-3 & 2 & 0
\end{array} \right]\\[10pt] & \xrightarrow{\frac{1}{3}R_2}
\left[\begin{array}{rrr|r}
1 & 1 & 0 & 0 \\
0 & 1 & b/3 & 0 \\
0 &a-3 & 2 & 0
\end{array} \right] \xrightarrow{R_3-(a-3)R_2}
\left[\begin{array}{rrr|r}
1 & 1 & 0 & 0 \\
0 & 1 & b/3 & 0 \\
0 & 0 & 2-(a-3)b/3 & 0
\end{array} \right].
\end{align*}
The last matrix is in echelon form.

Let $c=2-(a-3)b/3$ be the number in $(3,3)$-entry.
If $c\neq 0$, then it is easy to see that we can further reduce the matrix to
\[ \left[\begin{array}{rrr|r}
1 & 0 & 0 & 0 \\
0 &1 & 0 & 0 \\
0 & 0 & 1 & 0
\end{array} \right]\] and the only solution is $x_1=x_2=x_3=0$.

On the other hand, if $c=0$, then the last row becomes a zero row and thus we see that $x_3$ is a free variable. Hence the system has a nonzero solution.

Therefore, the vectors $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3$ are linearly dependent if and only if $c=0$.
Hence the condition we are looking for is
\[c=2-(a-3)b/3=0,\] or equivalently,
\[(a-3)b=6.\]

Related Question.

The solution is given in the post ↴
Determine a Condition on $a, b$ so that Vectors are Linearly Dependent

The post Determine Conditions on Scalars so that the Set of Vectors is Linearly Dependent first appeared on Problems in Mathematics.

Solve the following system of linear equations using Gauss-Jordan elimination.
\begin{align*}
6x+8y+6z+3w &=-3 \\
6x-8y+6z-3w &=3\\
8y \,\,\,\,\,\,\,\,\,\,\,- 6w &=6
\end{align*}

We use the following notation.

Elementary row operations.

The three elementary row operations on a matrix are defined as follows.

(1) Interchanging two rows:

$R_i \leftrightarrow R_j$ interchanges rows $i$ and $j$.

(2) Multiplying a row by a non-zero scalar (a number):

$tR_i$ multiplies row $i$ by the non-zero scalar (number) $t$.

(3) Adding a multiple of one row to another row:

$R_j+tR_i$ adds $t$ times row $i$ to row $j$.

Solution.

The augmented matrix of the system is
\[A=\left[ \begin{array}{rrrr|r}
6 & 8 & 6 & 3 & -3 \\
6 & -8 & 6 & -3 & 3\\
0 & 8 & 0 & -6 & 6
\end{array} \right].\] We apply elementary row operations as follows to reduce the system to a matrix in reduced row echelon form.

\[A \xrightarrow{R_2 – R_1}
\left[\begin{array}{rrrr|r}
6 & 8 & 6 & 3 & -3 \\
0 & -16 & 0 & -6 & 6\\
0 & 8 & 0 & -6 & 6
\end{array}\right] \xrightarrow[R_2+2R_3]{R_1-R_3}
\left[\begin{array}{rrrr|r}
6 & 0 & 6 & 9 & -9 \\
0 & 0 & 0 & -18 & 18\\
0 & 8 & 0 & -6 & 6
\end{array}\right] \] \[\xrightarrow[\frac{1}{2}R_3]{\frac{1}{3}R_1, \frac{-1}{18}R_2}
\left[\begin{array}{rrrr|r}
2 & 0 & 2 & 3 & -3 \\
0 & 0 & 0 & 1 & -1\\
0 & 4 & 0 & -3 & 3
\end{array}\right] \xrightarrow[R_3+3R_2]{R_1-3R_2}
\left[\begin{array}{rrrr|r}
2 & 0 & 2 & 0 & 0 \\
0 & 0 & 0 & 1 & -1\\
0 & 4 & 0 & 0 & 0
\end{array}\right] \]

\[
\xrightarrow[\frac{1}{4}R_3]{\frac{1}{2}R_1}
\left[\begin{array}{rrrr|r}
1 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 1 & -1\\
0 & 1 & 0 & 0 & 0\end{array}\right] \xrightarrow{R_2 \leftrightarrow R_3}
\left[\begin{array}{rrrr|r}
1 & 0 & 1 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & -1
\end{array}\right] \] The last matrix is in reduced row echelon form.
The corresponding system of linear equations is
\begin{align*}
x+z &=0\\
y&=0 \\
w&=-1
\end{align*}

Let $z=t$ be a free variable. Then the solution is $(x,y,z,w)=(-t,0,t,-1)$ for any number $t$.

Similar Problem.

Another similar problem is Solving a system of linear equations using Gaussian elimination.
The instruction of the problem says to use Gaussian elimination, but try to solve it using Gauss-Jordan elimination as well.

The post Solve a System of Linear Equations by Gauss-Jordan Elimination first appeared on Problems in Mathematics.

Solve the following system of linear equations using Gaussian elimination.
\begin{align*}
x+2y+3z &=4 \\
5x+6y+7z &=8\\
9x+10y+11z &=12
\end{align*}

Elementary row operations

The three elementary row operations on a matrix are defined as follows.

Solution.

First, the augmented matrix of the system is
\[A=\left[ \begin{array}{rrr|r}
1 & 2 & 3 & 4 \\
5 & 6 & 7 & 8 \\
9 & 10 & 11 & 12
\end{array} \right].\] We apply elementary row operations as follows to reduce the system to row echelon form.

\[A \xrightarrow{R_3 -9R_1}
\left[\begin{array}{rrr|r}
1 & 2 & 3 & 4 \\
5 & 6 & 7 & 8 \\
0 & -8 & -16 & -24
\end{array}\right] \xrightarrow{-\frac{1}{8}R_3}
\left[\begin{array}{rrr|r}
1 & 2 & 3 & 4 \\
5 & 6 & 7 & 8 \\
0 & 1 & 2 & 3
\end{array}\right] \] \[\xrightarrow{R_2-5R_1}
\left[\begin{array}{rrr|r}
1 & 2 & 3 & 4 \\
0 & -4 & -8 & -12 \\
0 & 1 & 2 & 3
\end{array}\right] \xrightarrow{-\frac{1}{4} R_2}
\left[\begin{array}{rrr|r}
1 & 2 & 3 & 4 \\
0 & 1 & 2 & 3 \\
0 & 1 & 2 & 3
\end{array}\right] \] \[\xrightarrow{R_3-R_2}
\left[\begin{array}{rrr|r}
1 & 2 & 3 & 4 \\
0 & 1 & 2 & 3 \\
0 & 0 & 0 & 0
\end{array}\right] \] The last matrix is in row echelon form.

The corresponding system of linear equations of it is
\begin{align*}
x+2y+3z &=4\\
y+2z&=3 \\
0z&=0
\end{align*}
The last equation $0z=0$ means that $z$ can be any number.
(More systematically, the variables corresponding to leading $1$’s in the echelon form matrix are dependent variables, and the rests are independent (free) variables.)

So let us say that $t$ is a value for $z$, namely $z=t$.
Then from the second equation, we have $y=-2t+3$.
From the first equation, we have
\[x=-2y-3z+4=-2(-2t+3)-3t+4=t-2.\] Thus the solution set is
\[(x,y,z)=(t-2, -2t+3, t)\] for any $t$.

Comment.

You may want to check whether the answer is correct by substituting this solution to the original equations.

Also, if you further reduce the matrix into reduced row echelon form, the last system becomes simpler (and simplest in a sense).
This procedure, to reduce a matrix until reduced row echelon form, is called the Gauss-Jordan elimination.

Related Question.

For a similar problem, you may want to check out Solve a system of linear equations by Gauss-Jordan elimination.

The post Solving a System of Linear Equations Using Gaussian Elimination first appeared on Problems in Mathematics.