Question

The values of \(\lambda\) such that \((x,y,z) \ne (0,0,0)\) and \[ (\hat{i} + \hat{j} + 3\hat{k})x + (3\hat{i} - 3\hat{j} + \hat{k})y + (-4\hat{i} + 5\hat{j})z = \lambda(x\hat{i} + y\hat{j} + z\hat{k}) \]

• A. \(0,1\)
• B. \(-1,1\)
• C. \(-1,0\)
• D. \(-2,0\)

Hint:

For vector equations of form \(A\vec{X}=\lambda \vec{X}\), directly form matrix and solve determinant.

Answer 1

The Correct Option is C

To determine the values of \(\lambda\) such that \(x, y, z\) are not all zero, we start by analyzing the given vector equation:

\((\hat{i} + \hat{j} + 3\hat{k})x + (3\hat{i} - 3\hat{j} + \hat{k})y + (-4\hat{i} + 5\hat{j})z = \lambda(x\hat{i} + y\hat{j} + z\hat{k})\)

We can express the above equation as:

\(x\hat{i} + y\hat{j} + 3x\hat{k} + 3y\hat{i} - 3y\hat{j} + y\hat{k} - 4z\hat{i} + 5z\hat{j} = \lambda x\hat{i} + \lambda y\hat{j} + \lambda z\hat{k}\)

By equating the coefficients of \(\hat{i}, \hat{j}, \hat{k}\) from both sides, we obtain:

1. \(x + 3y - 4z = \lambda x\)
2. \(x - 3y + 5z = \lambda y\)
3. \(3x + y = \lambda z\)

We aim to find non-trivial solutions by forming a system of equations expressed in matrix form:

\(1 - \lambda\)	3	-4	0
1	\(-3 - \lambda\)	5	0
3	1	\(-\lambda\)	0

We need the determinant of this matrix to be zero for non-trivial solutions.

Calculate the determinant:

1. The determinant function for 3x3 matrices is: \(a(ei − fh) − b(di − fg) + c(dh − eg)\).
2. Apply this with our matrix terms:
   1. \((1 - \lambda)(- \lambda(-\lambda -3) - (5)) - 3(1(-\lambda) - 3(5)) - 4(1(-\lambda -3) - (-3)(3))\)

Simplifying each term gives:

1. \((1 - \lambda)(\lambda^2 + 3\lambda - 5)\)
2. \(- 3(-3\lambda + 15)\)
3. \(- 4(-\lambda -3 + 9)\)

Further simplification:

The determinant simplifies to \((\lambda + 1)(\lambda - 1)\).

Setting this to zero gives the solutions \(\lambda = 0\) and \(\lambda = -1\).

Thus the values of \(\lambda\) for which there are non-trivial solutions are \(-1, 0\).