Value of determinant \[ \begin{vmatrix} a - b & b - c & c - a \\ b - c & c - a & a - b \\ c - a & a - b & b - c \end{vmatrix} \] is:
0
We are asked to find the value of the determinant:
\( \Delta = \begin{vmatrix} a - b & b - c & c - a \\ b - c & c - a & a - b \\ c - a & a - b & b - c \end{vmatrix} \)
To evaluate this determinant, we can use properties of determinants. Let's consider applying column operations. Adding the second and third columns to the first column (operation \( C_1 \to C_1 + C_2 + C_3 \)) often simplifies determinants with this kind of cyclic pattern.
Let's perform the operation \( C_1 \to C_1 + C_2 + C_3 \). The new elements in the first column will be:
After applying the column operation, the determinant becomes:
\( \Delta = \begin{vmatrix} 0 & b - c & c - a \\ 0 & c - a & a - b \\ 0 & a - b & b - c \end{vmatrix} \)
A fundamental property of determinants states that if any column (or row) of a determinant consists entirely of zeros, then the value of the determinant is zero.
In this modified determinant, the first column consists entirely of zeros. Therefore, the value of the determinant is 0.
\( \Delta = 0 \)
This method using column operations is much simpler than expanding the determinant directly.
| Property | Description |
|---|---|
| Row/Column Swap | Swapping two rows or two columns changes the sign of the determinant. |
| Scalar Multiplication | Multiplying a row or column by a scalar \( k \) multiplies the determinant by \( k \). |
| Row/Column Addition | Adding a multiple of one row (or column) to another row (or column) does not change the value of the determinant. |
| Zero Row/Column | If a determinant has a row or column of all zeros, its value is 0. |
| Identical Rows/Columns | If a determinant has two identical rows or columns, its value is 0. |
The given matrix has a specific structure where the elements in each row and column follow a cyclic pattern: \( (x, y, z) \), \( (y, z, x) \), \( (z, x, y) \), where \( x=a-b, y=b-c, z=c-a \). Notice that the sum of these elements is \( x+y+z = (a-b) + (b-c) + (c-a) = 0 \).
Matrices with this type of structure are related to circulant matrices. For the specific pattern where the sum of elements in the base sequence \( (x, y, z) \) is zero, the determinant will indeed be zero. This is because, as we showed, the sum of elements in each row (or column) is zero. Applying the operation \( R_1 \to R_1 + R_2 + R_3 \) (or \( C_1 \to C_1 + C_2 + C_3 \) as done in the solution) would result in a row (or column) of zeros, leading to a determinant of zero.
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(A) \(\begin{bmatrix} 1 & -1 \\ -1 & 5 \end{bmatrix}\)
(B) \(\begin{bmatrix} 13 & -1 \\ -1 & 15 \end{bmatrix}\)
(C) \(\begin{bmatrix} 16 & -1 \\ -11 & 15 \end{bmatrix}\)
(D) \(\begin{bmatrix} 6 & -12 \\ 11 & 15 \end{bmatrix}\)
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