The number of solutions of equations `x + y - z=0, 3x - y - z= 0,x - 3y + z = 0` is

Options

A. 0
B. 1
C. 2
D. infinite

Answer

To determine the number of solutions for the system of equations:

1. \( x + y - z = 0 \)
2. \( 3x - y - z = 0 \)
3. \( x - 3y + z = 0 \)

we will use the concept of determinants.

### Step 1: Write the system in matrix form
We can express the system of equations in the form \( A\mathbf{x} = \mathbf{0} \), where \( A \) is the coefficient matrix and \( \mathbf{x} \) is the vector of variables.

The coefficient matrix \( A \) is:
\[ A = \begin{bmatrix} 1 & 1 & -1 \\ 3 & -1 & -1 \\ 1 & -3 & 1 \end{bmatrix} \]

### Step 2: Calculate the determinant of \( A \)
To find the number of solutions, we first need to calculate the determinant \( D \) of the matrix \( A \).

\[ D = \begin{vmatrix} 1 & 1 & -1 \\ 3 & -1 & -1 \\ 1 & -3 & 1 \end{vmatrix} \]

Using the determinant formula for a 3x3 matrix:
\[ D = a(ei - fh) - b(di - fg) + c(dh - eg) \]
where \( a, b, c \) are the elements of the first row, and \( d, e, f, g, h, i \) are the elements of the remaining rows.

Calculating the determinant:
\[ D = 1 \begin{vmatrix} -1 & -1 \\ -3 & 1 \end{vmatrix} - 1 \begin{vmatrix} 3 & -1 \\ 1 & 1 \end{vmatrix} - 1 \begin{vmatrix} 3 & -1 \\ 1 & -3 \end{vmatrix} \]

Calculating each of these 2x2 determinants:
1. \( \begin{vmatrix} -1 & -1 \\ -3 & 1 \end{vmatrix} = (-1)(1) - (-1)(-3) = -1 - 3 = -4 \)

2. \( \begin{vmatrix} 3 & -1 \\ 1 & 1 \end{vmatrix} = (3)(1) - (-1)(1) = 3 + 1 = 4 \)

3. \( \begin{vmatrix} 3 & -1 \\ 1 & -3 \end{vmatrix} = (3)(-3) - (-1)(1) = -9 + 1 = -8 \)

Putting it all together:
\[ D = 1(-4) - 1(4) - 1(-8) = -4 - 4 + 8 = 0 \]

### Step 3: Check the conditions for solutions
Since \( D = 0 \), we need to check the determinants formed by replacing the columns with the constant terms (which are all zero in this case).

Let’s denote:
- \( D_1 \) as the determinant formed by replacing the first column with the constants,
- \( D_2 \) as the determinant formed by replacing the second column with the constants,
- \( D_3 \) as the determinant formed by replacing the third column with the constants.

Since all the constant terms are zero, we will find that:
\[ D_1 = D_2 = D_3 = 0 \]

### Conclusion
Since \( D = 0 \) and \( D_1 = D_2 = D_3 = 0 \), by the criteria of Cramer’s rule, the system has infinitely many solutions.

### Final Answer
The number of solutions of the given equations is **infinite**.

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