Understanding the Problem: Determinant and AP

We are asked to find the value of a given 3x3 determinant. The elements of the first two rows are simple linear expressions involving \(x\), specifically consecutive integers starting from \(x+1\). The elements of the third row involve \(x\) and terms \(a, b, c\), where it is stated that \(a, b, c\) are in Arithmetic Progression (AP).

Key Property of Arithmetic Progression (AP)

If three numbers \(a, b, c\) are in Arithmetic Progression (AP), it means that the difference between consecutive terms is constant. This property can be expressed as:

\[ b - a = c - b \]

Rearranging this equation gives us a crucial relationship between \(a, b,\) and \(c\):

\[ 2b = a + c \]

or equivalently,

\[ a + c - 2b = 0 \]

This relationship will be very useful in simplifying the determinant.

Evaluating the Determinant

The given determinant is:

\[ \Delta = \left|\begin{array}{lll} x+1 & x+2 & x+3 \\ x+2 & x+3 & x+4 \\ x+a & x+b & x+c \end{array}\right| \]

To simplify the determinant, we can use row and column operations. These operations do not change the value of the determinant.

Step 1: Apply Row Operations

Let's perform the following row operations:

• \(R_2 \leftarrow R_2 - R_1\) (Subtract the first row from the second row)
• \(R_3 \leftarrow R_3 - R_1\) (Subtract the first row from the third row)

Applying \(R_2 \leftarrow R_2 - R_1\):

• \((x+2) - (x+1) = 1\)
• \((x+3) - (x+2) = 1\)
• \((x+4) - (x+3) = 1\)

The new second row becomes \([1 \ 1 \ 1]\).

Applying \(R_3 \leftarrow R_3 - R_1\):

• \((x+a) - (x+1) = x+a-x-1 = a-1\)
• \((x+b) - (x+2) = x+b-x-2 = b-2\)
• \((x+c) - (x+3) = x+c-x-3 = c-3\)

The new third row becomes \([a-1 \ b-2 \ c-3]\).

The determinant now becomes:

\[ \Delta = \left|\begin{array}{lll} x+1 & x+2 & x+3 \\ 1 & 1 & 1 \\ a-1 & b-2 & c-3 \end{array}\right| \]

Step 2: Apply Column Operations

Now, let's apply column operations on the modified determinant to introduce zeros, which simplifies expansion. Let's perform:

• \(C_2 \leftarrow C_2 - C_1\) (Subtract the first column from the second column)
• \(C_3 \leftarrow C_3 - C_1\) (Subtract the first column from the third column)

Applying \(C_2 \leftarrow C_2 - C_1\):

• \((x+2) - (x+1) = 1\)
• \(1 - 1 = 0\)
• \((b-2) - (a-1) = b-2-a+1 = b-a-1\)

The new second column is \([1 \ 0 \ b-a-1]^T\).

Applying \(C_3 \leftarrow C_3 - C_1\):

• \((x+3) - (x+1) = 2\)
• \(1 - 1 = 0\)
• \((c-3) - (a-1) = c-3-a+1 = c-a-2\)

The new third column is \([2 \ 0 \ c-a-2]^T\).

The determinant now becomes:

\[ \Delta = \left|\begin{array}{lll} x+1 & 1 & 2 \\ 1 & 0 & 0 \\ a-1 & b-a-1 & c-a-2 \end{array}\right| \]

Step 3: Expand the Determinant

The determinant can be easily expanded along the second row \((R_2)\) because it contains two zeros. The formula for expanding along the second row is:

\[ \Delta = -1 \cdot C_{21} + 0 \cdot C_{22} + 0 \cdot C_{23} \]

where \(C_{ij}\) is the cofactor of the element in the \(i\)-th row and \(j\)-th column. We only need to calculate the cofactor \(C_{21}\) (the cofactor of the element '1' in the second row, first column).

The minor \(M_{21}\) is the determinant of the 2x2 matrix obtained by removing the 2nd row and 1st column:

\[ M_{21} = \left|\begin{array}{ll} 1 & 2 \\ b-a-1 & c-a-2 \end{array}\right| \]

The cofactor \(C_{21}\) is given by \((-1)^{2+1} M_{21} = -M_{21}\).

Calculate the 2x2 determinant \(M_{21}\):

\[ M_{21} = (1)(c-a-2) - (2)(b-a-1) \]

\[ M_{21} = c-a-2 - (2b-2a-2) \]

\[ M_{21} = c-a-2 - 2b+2a+2 \]

Combine like terms:

\[ M_{21} = (a+c) - 2b + (-2+2) \]

\[ M_{21} = a+c - 2b \]

Now substitute \(M_{21}\) back into the determinant expansion:

\[ \Delta = -1 \cdot C_{21} = -1 \cdot (-M_{21}) = M_{21} \]

\[ \Delta = a+c - 2b \]

Step 4: Use the AP Property

We know that \(a, b, c\) are in AP, which means \(a + c - 2b = 0\).

Substitute this into the expression for \(\Delta\):

\[ \Delta = (a+c) - 2b = 0 \]

Thus, the value of the determinant is 0.

Conclusion

By applying row and column operations and utilizing the property that \(a, b, c\) are in AP (\(a+c=2b\)), we find that the value of the determinant is 0.

Revision Table: Determinants and AP

Additional Information: More on Determinant Properties

Here are some other useful properties of determinants:

• If two rows or two columns of a matrix are identical, the determinant is 0.
• If two rows or two columns are proportional, the determinant is 0.
• If a row or column consists entirely of zeros, the determinant is 0.
• Swapping two rows or two columns multiplies the determinant by -1.
• Multiplying a single row or column by a scalar \(k\) multiplies the determinant by \(k\).
• The determinant of a matrix is equal to the determinant of its transpose (\(\det(A) = \det(A^T)\)).
• The determinant of a product of matrices is the product of their determinants (\(\det(AB) = \det(A)\det(B)\)).

In our problem, the row/column operations we used helped reveal a hidden structure related to the AP property, ultimately leading to a simplified determinant that evaluates to zero. Specifically, the operations helped create a dependency between the third row (or column after transformations) and the AP condition, causing a linear combination of terms to become zero.