Question

$\begin{vmatrix} x+1 & 3 & 5\\ 2 & x+2 & 5\\ 2 & 3 & x+4 \end{vmatrix} = 0$, then x will be equal to:

• A. -1,-9
• B. 1,-9
• C. -1,9
• D. 1,9

Answer 1

Answer: (B)

1,-9

Answer 2

To solve the given determinant equation, we will use properties of determinants to simplify it before expansion.

The given equation is: $\begin{vmatrix} x+1 & 3 & 5\\ 2 & x+2 & 5\\ 2 & 3 & x+4 \end{vmatrix} = 0$

Let $D$ denote the determinant.

Step 1: Apply column operation $C_1 \to C_1 + C_2 + C_3$

This operation adds the elements of the second and third columns to the first column.

$D = \begin{vmatrix} (x+1)+3+5 & 3 & 5\\ 2+(x+2)+5 & x+2 & 5\\ 2+3+(x+4) & 3 & x+4 \end{vmatrix}$

$D = \begin{vmatrix} x+9 & 3 & 5\\ x+9 & x+2 & 5\\ x+9 & 3 & x+4 \end{vmatrix}$

Step 2: Take $(x+9)$ common from the first column ($C_1$)

$D = (x+9) \begin{vmatrix} 1 & 3 & 5\\ 1 & x+2 & 5\\ 1 & 3 & x+4 \end{vmatrix}$

Step 3: Apply row operations $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_1$

These operations will create zeros in the first column, making the determinant easier to evaluate.

$D = (x+9) \begin{vmatrix} 1 & 3 & 5\\ 1-1 & (x+2)-3 & 5-5\\ 1-1 & 3-3 & (x+4)-5 \end{vmatrix}$

$D = (x+9) \begin{vmatrix} 1 & 3 & 5\\ 0 & x-1 & 0\\ 0 & 0 & x-1 \end{vmatrix}$

Step 4: Evaluate the determinant

The resulting matrix is an upper triangular matrix. The determinant of a triangular matrix is the product of its diagonal elements.

$D = (x+9) \cdot [1 \cdot (x-1) \cdot (x-1)]$ $D = (x+9)(x-1)^2$

Step 5: Set the determinant to zero and solve for $x$

Given $D=0$: $(x+9)(x-1)^2 = 0$

This equation holds true if either of the factors is zero:

1. $x+9 = 0 \implies x = -9$
2. $(x-1)^2 = 0 \implies x-1 = 0 \implies x = 1$

Thus, the values of $x$ for which the determinant is zero are $1$ and $-9$.