Question

If a, b, c, are in A.P, then the determinant $\begin{vmatrix} x+2 & x+3 & x+2a \\ x+3 & x+4 & x+2b \\ x+4 & x+5 & x+2c \end{vmatrix}$ is

• A. 0
• B. 1
• C. x
• D. 2x

Answer 1

Answer: (A)

0

Answer 2

To evaluate the given determinant: $\Delta = \begin{vmatrix} x+2 & x+3 & x+2a \\ x+3 & x+4 & x+2b \\ x+4 & x+5 & x+2c \end{vmatrix}$

We are given that a, b, c are in A.P. (Arithmetic Progression). This means that the common difference is constant, so $b-a = c-b$, which implies $2b = a+c$.

Let's apply row operations to simplify the determinant.

Step 1: Apply Row Operations $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_1$. This operation will simplify the first two columns, which are in A.P. $\Delta = \begin{vmatrix} x+2 & x+3 & x+2a \\ (x+3)-(x+2) & (x+4)-(x+3) & (x+2b)-(x+2a) \\ (x+4)-(x+2) & (x+5)-(x+3) & (x+2c)-(x+2a) \end{vmatrix}$ Simplifying the elements: $\Delta = \begin{vmatrix} x+2 & x+3 & x+2a \\ 1 & 1 & 2b-2a \\ 2 & 2 & 2c-2a \end{vmatrix}$

Step 2: Apply Row Operation $R_3 \to R_3 - 2R_2$. This operation will simplify the third row further. $\Delta = \begin{vmatrix} x+2 & x+3 & x+2a \\ 1 & 1 & 2b-2a \\ 2-2(1) & 2-2(1) & (2c-2a)-2(2b-2a) \end{vmatrix}$ Simplifying the elements: $\Delta = \begin{vmatrix} x+2 & x+3 & x+2a \\ 1 & 1 & 2b-2a \\ 0 & 0 & 2c-2a-4b+4a \end{vmatrix}$ $\Delta = \begin{vmatrix} x+2 & x+3 & x+2a \\ 1 & 1 & 2b-2a \\ 0 & 0 & 2a+2c-4b \end{vmatrix}$

Step 3: Use the A.P. condition ($2b = a+c$). The element in the third row, third column is $2a+2c-4b$. Substitute $a+c = 2b$ into this expression: $2a+2c-4b = 2(a+c) - 4b = 2(2b) - 4b = 4b - 4b = 0$.

So, the determinant becomes: $\Delta = \begin{vmatrix} x+2 & x+3 & x+2a \\ 1 & 1 & 2b-2a \\ 0 & 0 & 0 \end{vmatrix}$

Step 4: Evaluate the determinant. A fundamental property of determinants states that if any row (or column) of a matrix consists entirely of zeros, then the value of its determinant is zero. In this case, the third row of the determinant consists entirely of zeros. Therefore, $\Delta = 0$.