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

(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).