Answer to Question #203667 in Linear Algebra for Rama

Question

Using the the concept of rank of the matrix find whether the following system of equations

are consistent or inconsistent,

i.

4y + z = 0

12x - 5y - 3z = 34

-6x + 4z = 8

ii.

5x - 3y + z = 7

2x + 3y - z = 0

8x + 9y - 3z = 2:

iii.

-8x + 2z = 1

6y + 4z = 3

12x + 2y = 2

Answer 1

a. $Ax=b$ is inconsistent (i.e., no solution exists) if and only if $\text{rank}[A]<\text{rank}[A|b].$

b. $Ax=b$ has an unique solution if and only if $\text{rank}[A]=\text{rank}[A|b]=n.$

c. $Ax=b$ has infinitely many solutions if and only if $\text{rank}[A]=\text{rank}[A|b]<n.$

i.

$$A=\begin{bmatrix} 0 & 4 & 1 \\ 12 & -5 & -3 \\ -6 & 0 & 4 \end{bmatrix}$$

Swap rows 1and 2

$$\begin{bmatrix} 12 & -5 & -3 \\ 0 & 4 & 1 \\ -6 & 0 & 4 \end{bmatrix}$$

$R_3=R_3+(1/2)R_1$

$$\begin{bmatrix} 12 & -5 & -3 \\ 0 & 4 & 1 \\ 0 & -5/2 & 5/2 \end{bmatrix}$$

$R_3=R_3+(5/8)R_2$

The rank of a matrix is the number of nonzero rows in the reduced matrix, so the rank is 3.

Since $\text{rank}[A]=\text{rank}[A|b]=3=n,$ then $Ax=b$ has an unique solution.

ii.

$$A=\begin{bmatrix} 5 & -3 & 1 \\ 2 & 3 & -1 \\ 8 & 9 & -3 \end{bmatrix}$$

$R_1=R_1/5$

$$\begin{bmatrix} 1 & -3/5 & 1/5 \\ 2 & 3 & -1 \\ 8 & 9 & -3 \end{bmatrix}$$

$R_2=R_2-2R_1$

$$\begin{bmatrix} 1 & -3/5 & 1/5 \\ 0 & 21/5 & -7/5 \\ 8 & 9 & -3 \end{bmatrix}$$

$R_3=R_3-8R_1$

$$\begin{bmatrix} 1 & -3/5 & 1/5 \\ 0 & 21/5 & -7/5 \\ 0 & 69/5 & -23/5 \end{bmatrix}$$

$R_2=(5/21)R_2$

$$\begin{bmatrix} 1 & -3/5 & 1/5 \\ 0 & 1 & -1/3 \\ 0 & 69/5 & -23/5 \end{bmatrix}$$

$R_1=R_1+(3/5)R_2$

$$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & -1/3 \\ 0 & 69/5 & -23/5 \end{bmatrix}$$

$R_3=R_3-(69/5)R_2$

$$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & -1/3 \\ 0 & 0 & 0 \end{bmatrix}$$

The rank of a matrix is the number of nonzero rows in the reduced matrix, so the rank is 2.

$$\begin{bmatrix} 5 & -3 & 1 & & 7 \\ 2 & 3 & -1 & & 0 \\ 8 & 9 & -3 & & 2 \end{bmatrix}$$

$R_1=R_1/5$

$$\begin{bmatrix} 1 & -3/5 & 1/5 & & 7/5 \\ 2 & 3 & -1 & & 0 \\ 8 & 9 & -3 & & 2 \end{bmatrix}$$

$R_2=R_2-2R_1$

$$\begin{bmatrix} 1 & -3/5 & 1/5 & & 7/5 \\ 0 & 21/5 & -7/5 & & -14/5 \\ 8 & 9 & -3 & & 2 \end{bmatrix}$$

$R_3=R_3-8R_1$

$$\begin{bmatrix} 1 & -3/5 & 1/5 & & 7/5 \\ 0 & 21/5 & -7/5 & & -14/5 \\ 0 & 69/5 & -23/5 & & -46/5 \end{bmatrix}$$

$R_2=(5/21)R_2$

$$\begin{bmatrix} 1 & -3/5 & 1/5 & & 7/5 \\ 0 & 1 & -1/3 & & -2/3 \\ 0 & 69/5 & -23/5 & & -46/5 \end{bmatrix}$$

$R_1=R_1+(3/5)R_2$

$$\begin{bmatrix} 1 & 0 & 0 & & 1 \\ 0 & 1 & -1/3 & & -2/3 \\ 0 & 69/5 & -23/5 & & -46/5 \end{bmatrix}$$

$R_3=R_3-(69/5)R_2$

$$\begin{bmatrix} 1 & 0 & 0 & & 1 \\ 0 & 1 & -1/3 & & -2/3 \\ 0 & 0 & 0 & & 0 \end{bmatrix}$$

The rank of a matrix is the number of nonzero rows in the reduced matrix, so the rank is 2.

Since $\text{rank}[A]=\text{rank}[A|b]=2<3=n,$ then $Ax=b$ has infinitely many solutions.

iii.

$$A=\begin{bmatrix} -8 & 0 & 2 \\ 0 & 6 & 4 \\ 12 & 2 & 0 \end{bmatrix}$$

$R_1=-R_1/8$

$$\begin{bmatrix} 1 & 0 & -1/4 \\ 0 & 6 & 4 \\ 12 & 2 & 0 \end{bmatrix}$$

$R_3=R_3-12R_1$

$$\begin{bmatrix} 1 & 0 & -1/4 \\ 0 & 6 & 4 \\ 0 & 2 & 3 \end{bmatrix}$$

$R_2=R_2/6$

$$\begin{bmatrix} 1 & 0 & -1/4 \\ 0 & 1 & 2/3 \\ 0 & 2 & 3 \end{bmatrix}$$

$R_3=R_3-2R_2$

$$\begin{bmatrix} 1 & 0 & -1/4 \\ 0 & 1 & 2/3 \\ 0 & 0 & 5/3 \end{bmatrix}$$

$R_3=(3/5)R_3$

$$\begin{bmatrix} 1 & 0 & -1/4 \\ 0 & 1 & 2/3 \\ 0 & 0 & 1 \end{bmatrix}$$

$R_1=R_1+(1/4)R_3$

$$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 2/3 \\ 0 & 0 & 1 \end{bmatrix}$$

$R_2=R_2-(2/3)R_3$

$$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$

The rank of a matrix is the number of nonzero rows in the reduced matrix, so the rank is 3.

Since $\text{rank}[A]=\text{rank}[A|b]=3=n,$ then $Ax=b$ has an unique solution.