# Consider the following homogeneous system of linear equations: x - y + (\lambda - 1)z = 0 }] y +...

## Question:

Consider the following homogeneous system of linear equations:

{eq}x - y + (\lambda - 1)z = 0 {/eq}

y + 2z = 0

{eq}(1 - \lambda)x + 4y - 2z = 0 {/eq}

(i) Write this system in the form of a matrix equation.

(ii) For which values of {eq}\lambda {/eq} does this system have exactly one solution? What is this solution?

(iii) For which values of {eq}\lambda {/eq} does this system have infinitely many solutions? Find one such non-trivial solution.

## Solution of a Homogeneous System of Equations

Given a homogeneous system of three linear equations in three variables, we find restrictions on the coefficients so that the system has a unique solution and then conditions on the coefficients for which the system has an infinite number of solutions. Then we find one such solution for the latter case. The concepts used include evaluating the determinant of a 3 x 3 matrix and using substitution to solve the system.

## Answer and Explanation: 1

(i) The following matrix equation is used to represent the system of three linear equations in x, y and z:

{eq}\left( \begin{array}{ccc} 1 & -1 & \lambda-1 \\ 0 & 1 & 2 \\ 1-\lambda & 4 & -2 \end{array} \right) \left( \begin{array}{c} x \\ y \\ z \end{array} \right) = \left( \begin{array}{c} 0 \\ 0 \\ 0 \end{array} \right) {/eq}

(ii) For the above system to have exactly one solution which is the zero solution or {eq}(x,y,z)=(0,0,0)
{/eq} the determinant of the coefficient matrix must be non-zero so that its inverse exists. To that end we calculate the determinant of the coefficient matrix using the first column with minors and co-factors for expansion we get the determinant **(det)** equal to

{eq}1 \; det \; \left( \begin{array}{cc} 1 & 2 \\ 4 & -2 \end{array} \right) + (1-\lambda) \; det \; \left( \begin{array}{cc} -1 & \lambda-1 \\ 1 & 2 \end{array} \right) {/eq}

Using that {eq}det \; \left( \begin{array}{cc} a & b \\ c & d \end{array} \right) = a \cdot d - b \cdot c {/eq} we get the above expression equal to

{eq}1(-2-8)+(1- \lambda) \left[ (-1)(2)-1(\lambda-1) \right] = -10 + \lambda^2-1 = \lambda^2-11. {/eq}

Hence for the determinant to be non-zero, {eq}\lambda \ne \pm \sqrt {11}
{/eq} in which case the system will have one unique solution **(x, y, z) = (0, 0, 0).**

(iii) On the flip side from part (ii) above, the lambda values of plus or minus square root of 11 will give us infinitely many solutions. To find one such solution we let {eq}\lambda = \sqrt {11} {/eq} in our system of equations to get the following system:

{eq}\begin{array}{l} x - y + (\sqrt {11} - 1) z = 0 \qquad (1) \\ 0x + y + 2z = 0 \qquad (2) \\ (1-\sqrt {11})x +4y - 2z = 0 \qquad (3) \end{array} {/eq}

From (2) we get {eq}y=-2z. {/eq} Setting z = 1 gives us y = -2(1) = -2. Substitute in (3) above to get

{eq}\displaystyle (1-\sqrt {11})x +4(-2) - 2(1) = 0 \implies x = \frac {10}{1-\sqrt {11}}=-4.32. {/eq}

**One such solution is (x, y, z) = (-4.32, -2, 1).**

#### Learn more about this topic:

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Chapter 1 / Lesson 8Learn the three ways to solve two-variable equations. Study how to solve it algebraically with substitution or elimination and graphing on a coordinate plane.