The equation `[1 x y][(1,3,1),(0,2,-1),(0,0,1)] [(1),(x),(y)]=[0]` has
i) y=0 (p) rational roots
ii) y=-1 (q) irrational roots
(r) integral roots

Options

A. `{:("(i)",(ii)),((p),"(r)"):}`
B. `{:("(i)",(ii)),((q),"(p)"):}`
C. `{:("(i)",(ii)),((p),"(q)"):}`
D. `{:("(i)",(ii)),((r),"(p)"):}`

Explanation

To solve the given matrix equation

\[ [1 \quad x \quad y] \begin{pmatrix} 1 & 3 & 1 \\ 0 & 2 & -1 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 1 \\ x \\ y \end{pmatrix} = [0] \]

we will analyze the cases for \( y = 0 \) and \( y = -1 \).

### Step 1: Substitute \( y = 0 \)

Substituting \( y = 0 \) into the equation, we have:

\[ [1 \quad x \quad 0] \begin{pmatrix} 1 & 3 & 1 \\ 0 & 2 & -1 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 1 \\ x \\ 0 \end{pmatrix} = [0] \]

### Step 2: Multiply the matrices

Now we will multiply the matrices:

1. First, calculate the product of the second matrix and the column matrix:

\[ \begin{pmatrix} 1 & 3 & 1 \\ 0 & 2 & -1 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 1 \\ x \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \cdot 1 + 3 \cdot x + 1 \cdot 0 \\ 0 \cdot 1 + 2 \cdot x - 1 \cdot 0 \\ 0 \cdot 1 + 0 \cdot x + 1 \cdot 0 \end{pmatrix} = \begin{pmatrix} 1 + 3x \\ 2x \\ 0 \end{pmatrix} \]

2. Next, multiply the row matrix with the result:

\[ [1 \quad x \quad 0] \begin{pmatrix} 1 + 3x \\ 2x \\ 0 \end{pmatrix} = 1(1 + 3x) + x(2x) + 0(0) = 1 + 3x + 2x^2 \]

### Step 3: Set the equation to zero

Now we set the equation to zero:

\[ 2x^2 + 3x + 1 = 0 \]

### Step 4: Solve the quadratic equation

Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

- Here, \( a = 2 \), \( b = 3 \), \( c = 1 \).

Calculating the discriminant:

\[ b^2 - 4ac = 3^2 - 4 \cdot 2 \cdot 1 = 9 - 8 = 1 \]

Now substituting into the quadratic formula:

\[ x = \frac{-3 \pm \sqrt{1}}{2 \cdot 2} = \frac{-3 \pm 1}{4} \]

Calculating the roots:

1. \( x = \frac{-3 + 1}{4} = \frac{-2}{4} = -\frac{1}{2} \)
2. \( x = \frac{-3 - 1}{4} = \frac{-4}{4} = -1 \)

### Conclusion for \( y = 0 \)

Both roots \( -\frac{1}{2} \) and \( -1 \) are rational numbers. Therefore, for \( y = 0 \), we have rational roots.

### Step 5: Substitute \( y = -1 \)

Now, we substitute \( y = -1 \):

\[ [1 \quad x \quad -1] \begin{pmatrix} 1 & 3 & 1 \\ 0 & 2 & -1 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 1 \\ x \\ -1 \end{pmatrix} = [0] \]

### Step 6: Multiply the matrices again

1. First, calculate the product of the second matrix and the column matrix:

\[ \begin{pmatrix} 1 & 3 & 1 \\ 0 & 2 & -1 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 1 \\ x \\ -1 \end{pmatrix} = \begin{pmatrix} 1 + 3x - 1 \\ 2x + 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 3x \\ 2x + 1 \\ 0 \end{pmatrix} \]

2. Next, multiply the row matrix with the result:

\[ [1 \quad x \quad -1] \begin{pmatrix} 3x \\ 2x + 1 \\ 0 \end{pmatrix} = 1(3x) + x(2x + 1) - 1(0) = 3x + 2x^2 + x = 2x^2 + 4x \]

### Step 7: Set the equation to zero

Setting the equation to zero:

\[ 2x^2 + 4x = 0 \]

### Step 8: Factor the equation

Factoring out \( 2x \):

\[ 2x(x + 2) = 0 \]

### Step 9: Solve for \( x \)

The solutions are:

1. \( 2x = 0 \) which gives \( x = 0 \)
2. \( x + 2 = 0 \) which gives \( x = -2 \)

Both roots \( 0 \) and \( -2 \) are integral.

### Conclusion for \( y = -1 \)

For \( y = -1 \), we have integral roots.

### Final Answer

- For \( y = 0 \): Rational roots (p)
- For \( y = -1 \): Integral roots (r)