Question

If $\begin{vmatrix} x + 2 & 8 & 9 \\ 4 & x + 6 & 9 \\ 4 & 8 & x + 7 \end{vmatrix} = (x - 2)^{2}(ax + b),$ then the values of $a$ and $b$ respectively are

• A. 2 and 19
• B. 1 and 21
• C. 1 and 19
• D. 1 and -19
• E. -1 and 19

Hint:

Math Tip: When diagonal elements contain the variable '$x$' and off-diagonal elements are constants, always check if summing the rows ($C_1 \to C_1+C_2+C_3$) or columns ($R_1 \to R_1+R_2+R_3$) yields a uniform linear factor. This immediately simplifies the determinant!

Answer 1

The Correct Option is C

Concept:
Determinants - Row/Column Operations and Factorization.
Step 1: Observe row/column sums to find a common factor.
Notice that the sum of the elements in each row is the same:

• Row 1 sum: $(x+2) + 8 + 9 = x + 19$
• Row 2 sum: $4 + (x+6) + 9 = x + 19$
• Row 3 sum: $4 + 8 + (x+7) = x + 19$

Apply the column operation: $C_1 \rightarrow C_1 + C_2 + C_3$. $$ \Delta = \begin{vmatrix} x+19 & 8 & 9 \\ x+19 & x+6 & 9 \\ x+19 & 8 & x+7 \end{vmatrix} $$
Step 2: Factor out the common term from $C_1$.
Take $(x+19)$ common from the first column: $$ \Delta = (x+19) \begin{vmatrix} 1 & 8 & 9 \\ 1 & x+6 & 9 \\ 1 & 8 & x+7 \end{vmatrix} $$
Step 3: Create zeros in $C_1$ to simplify expansion.
Apply row operations: $R_2 \rightarrow R_2 - R_1$ and $R_3 \rightarrow R_3 - R_1$. $$ \Delta = (x+19) \begin{vmatrix} 1 & 8 & 9 \\ 0 & (x+6)-8 & 9-9 \\ 0 & 8-8 & (x+7)-9 \end{vmatrix} $$ $$ \Delta = (x+19) \begin{vmatrix} 1 & 8 & 9 \\ 0 & x-2 & 0 \\ 0 & 0 & x-2 \end{vmatrix} $$
Step 4: Expand the determinant.
Expand along $C_1$: $$ \Delta = (x+19) \cdot 1 \cdot \left[ (x-2)(x-2) - (0)(0) \right] $$ $$ \Delta = (x+19)(x-2)^2 $$
Step 5: Compare with the given expression.
We are given that $\Delta = (x-2)^2(ax+b)$.
By equating our result to the given expression: $$ (x-2)^2(x+19) = (x-2)^2(ax+b) $$ Comparing the terms inside the linear factor, we get: $$ ax + b = 1x + 19 $$ Therefore, $a = 1$ and $b = 19$.