The number of solutions of the following equations `x_(2)-x_(3)=1,-x_(1)+2x_(3)=-2, x_(1)-2x_(2)=3` is

Options

A. zero
B. one
C. two
D. infinite

Answer

To find the number of solutions for the given system of equations, we will first rewrite the equations in a standard form and then represent them in matrix form. The equations are:

1. \( x_2 - x_3 = 1 \)
2. \( -x_1 + 2x_3 = -2 \)
3. \( x_1 - 2x_2 = 3 \)

### Step 1: Rewrite the equations in standard form

We can rewrite the equations as:

1. \( 0x_1 + 1x_2 - 1x_3 = 1 \)
2. \( -1x_1 + 0x_2 + 2x_3 = -2 \)
3. \( 1x_1 - 2x_2 + 0x_3 = 3 \)

### Step 2: Form the augmented matrix

The augmented matrix for the system of equations can be written as:

\[ \begin{bmatrix} 0 & 1 & -1 & | & 1 \\ -1 & 0 & 2 & | & -2 \\ 1 & -2 & 0 & | & 3 \end{bmatrix} \]

### Step 3: Calculate the determinant of the coefficient matrix

The coefficient matrix is:

\[ A = \begin{bmatrix} 0 & 1 & -1 \\ -1 & 0 & 2 \\ 1 & -2 & 0 \end{bmatrix} \]

To find the determinant of \( A \), we can use the formula for the determinant of a 3x3 matrix:

\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]

Where the matrix is represented as:

\[ \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix} \]

Substituting the values from our matrix:

\[ \text{det}(A) = 0(0 \cdot 0 - 2 \cdot -2) - 1(-1 \cdot 0 - 2 \cdot 1) + (-1)(-1 \cdot -2 - 0 \cdot 1) \]

Calculating each term:

1. The first term is \( 0 \).
2. The second term is \( -1(0 + 2) = -2 \).
3. The third term is \( -1(2 - 0) = -2 \).

Thus, we have:

\[ \text{det}(A) = 0 - 2 - 2 = -4 \]

### Step 4: Determine the number of solutions

Since the determinant of the coefficient matrix \( A \) is not equal to zero (\( \text{det}(A) \neq 0 \)), the system of equations has a unique solution.

### Conclusion

The number of solutions of the given system of equations is **1 (unique solution)**.

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