Consider the determinants `Delta=|{:(2,-1,3),(1,1,1),(1,-1,1):}|,Delta'=|{:(2,0,-2),(-2,-1,1),(-4,1,3):}|`
STATEMENT-1 `Delta'=Delta^2`.
STATEMENT-2 : The determinant formed by the cofactors of the elements of a determinant of order 3 is equal in value to the square of the original determinant.

Answer

To solve the problem, we need to evaluate the determinants given and verify the statements regarding them.

### Step 1: Calculate the determinant \( \Delta \)

Given:
\[ \Delta = \begin{vmatrix} 2 & -1 & 3 \\ 1 & 1 & 1 \\ 1 & -1 & 1 \end{vmatrix} \]

Using the formula for the determinant of a 3x3 matrix:
\[ \Delta = a(ei - fh) - b(di - fg) + c(dh - eg) \]
where the matrix is:
\[ \begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix} \]

Substituting the values:
\[ \Delta = 2 \begin{vmatrix} 1 & 1 \\ -1 & 1 \end{vmatrix} - (-1) \begin{vmatrix} 1 & 1 \\ 1 & 1 \end{vmatrix} + 3 \begin{vmatrix} 1 & 1 \\ 1 & -1 \end{vmatrix} \]

Calculating the 2x2 determinants:
1. \( \begin{vmatrix} 1 & 1 \\ -1 & 1 \end{vmatrix} = (1)(1) - (1)(-1) = 1 + 1 = 2 \)
2. \( \begin{vmatrix} 1 & 1 \\ 1 & 1 \end{vmatrix} = (1)(1) - (1)(1) = 0 \)
3. \( \begin{vmatrix} 1 & 1 \\ 1 & -1 \end{vmatrix} = (1)(-1) - (1)(1) = -1 - 1 = -2 \)

Now substituting back:
\[ \Delta = 2(2) - (-1)(0) + 3(-2) = 4 + 0 - 6 = -2 \]

### Step 2: Calculate \( \Delta^2 \)

Now we find \( \Delta^2 \):
\[ \Delta^2 = (-2)^2 = 4 \]

### Step 3: Calculate the determinant \( \Delta' \)

Given:
\[ \Delta' = \begin{vmatrix} 2 & 0 & -2 \\ -2 & -1 & 1 \\ -4 & 1 & 3 \end{vmatrix} \]

Using the same determinant formula:
\[ \Delta' = 2 \begin{vmatrix} -1 & 1 \\ 1 & 3 \end{vmatrix} - 0 + (-2) \begin{vmatrix} -2 & -1 \\ -4 & 1 \end{vmatrix} \]

Calculating the 2x2 determinants:
1. \( \begin{vmatrix} -1 & 1 \\ 1 & 3 \end{vmatrix} = (-1)(3) - (1)(1) = -3 - 1 = -4 \)
2. \( \begin{vmatrix} -2 & -1 \\ -4 & 1 \end{vmatrix} = (-2)(1) - (-1)(-4) = -2 - 4 = -6 \)

Now substituting back:
\[ \Delta' = 2(-4) + 0 + (-2)(-6) = -8 + 0 + 12 = 4 \]

### Step 4: Verify the statements

1. **Statement 1**: \( \Delta' = \Delta^2 \)

We found \( \Delta' = 4 \) and \( \Delta^2 = 4 \). Thus, Statement 1 is **true**.

2. **Statement 2**: The determinant formed by the cofactors of the elements of a determinant of order 3 is equal in value to the square of the original determinant.

This is a known property of determinants: the determinant of the cofactor matrix (adjugate) of a matrix \( A \) is equal to \( \det(A)^{n-1} \) for an \( n \times n \) matrix. For \( n = 3 \), this means \( \det(\text{adj}(A)) = \det(A)^2 \). Thus, Statement 2 is also **true**.

### Conclusion

Both statements are true.