Question

use the Gaussian Elimination method to find the value of a so that the system of equations.
x+(a+4)y+(4a+2) z=0
2x+3ay+(3a+4) z=0
x+2(a+1)y+(3a+4) z=0
has (i) a unique solution ;
(ii) infinitely many solutions.
Further, find the solution set in each case.

Answer 1

Answer on Question #68345 – Math – Linear Algebra

Question

use the Gaussian Elimination method to find the value of $a$ so that the system of equations

$$\left\{ \begin{array}{c} x + (a + 4) y + (4 a + 2) z = 0 \\ 2 x + 3 a y + (3 a + 4) z = 0 \\ x + 2 (a + 1) y + (3 a + 4) z = 0 \end{array} \right.$$

$x + (a + 4)y + (4a + 2)z = 0$

$2x + 3ay + (3a + 4)z = 0$

$x + 2(a + 1)y + (3a + 4)z = 0$

has (i) a unique solution ;

(ii) infinitely many solutions.

Further, find the solution set in each case.

Solution

$$\left\{ \begin{array}{c} x + (a + 4) y + (4 a + 2) z = 0 \quad (R1) \\ 2 x + 3 a y + (3 a + 4) z = 0 \quad (R2) \\ x + 2 (a + 1) y + (3 a + 4) z = 0 \quad (R3) \end{array} \right.$$ \left\{ \begin{array}{c} x + (a + 4) y + (4 a + 2) z = 0 \qquad ((R1)' \leftarrow R1) \\ (a - 8) y - 5 a z = 0 \qquad ((R2)' \leftarrow R2 - 2R1) \\ (a - 2) y + (- a + 2) z = 0 \qquad ((R3)' \leftarrow R3 - R1) \end{array} \right. \\ \end{array} \right. $$\left\{ \begin{array}{c} x + (a + 4) y + (4 a + 2) z = 0 \qquad ((R1)'' \leftarrow (R1)') \\ (a - 8) y - 5 a z = 0 \qquad ((R2)'' \leftarrow (R2)') \\ \frac {4 a ^ {2} - 16}{a - 8} z = 0 \qquad ((R3)'' \leftarrow (R3)' - (R2)' \cdot \frac {a - 2}{a - 8}) \end{array} \right.$$

The matrix $A$ for this linear system will be as follows:

$$A = \left( \begin{array}{ccc} 1 & a + 4 & 4 a + 2 \\ 0 & a - 8 & -5a \\ 0 & 0 & \frac{4a^2 - 16}{a - 8} \end{array} \right)$$

The determinant of the matrix $A$ is given by

$$\det (A) = \left| \begin{array}{ccc} 1 & a + 4 & 4 a + 2 \\ 0 & a - 8 & -5a \\ 0 & 0 & \frac{4a^2 - 16}{a - 8} \end{array} \right| = 1 \cdot (a - 8) \cdot \frac{4a^2 - 16}{a - 8} = 4a^2 - 16 = 4(a - 2)(a + 2).$$

Next,

$$\left\{ \begin{array}{c} x + \left(4 a + 2 + \frac{5a(a + 4)}{a - 8}\right) z = 0 \quad ((R1)''' \leftarrow (R1)'' - \frac{a + 4}{a - 8} \cdot (R2)'') \\ y - \frac{5a}{a - 8} z = 0 \quad ((R2)''' \leftarrow (R2)'' \cdot \frac{1}{a - 8}) \\ \frac{4a^2 - 16}{a - 8} z = 0 \quad ((R3)''' \leftarrow (R3)'' \end{array} \right.$$

The criterion says that the system has the unique trivial solution only if $\det(A)$ is not equal zero and infinitely many solutions otherwise.

So:

I. Unique solution $x = y = z = 0$ if $a \neq 2$ and $a \neq -2$;

II. Infinitely many solutions $z = C$, where $C$ is an arbitrary real constant,

$$y = \frac{5az}{a - 8},$$ $$x = -\left(4a + 2 + \frac{5a(a + 4)}{a - 8}\right)z = -\frac{9a^2 - 10a - 16}{a - 8}z \text{ if } a = 2 \text{ or } a = -2.$$

If $a = 2$, then $z = C$, where $C$ is arbitrary real constant, $y = -\frac{5z}{3} = -\frac{5C}{3}$, $x = 0$.

If $a = -2$, then $z = D$, where $D$ is arbitrary real constant, $y = -\frac{10z}{-10} = z = D$,

$$x = \frac{-40z}{-10} = 4z = 4D.$$

Answer provided by https://www.AssignmentExpert.com