Question

The value of the determinant $\begin{vmatrix}1 & 1 & 1 \\ 1+x & 1 & 1 \\ 1 & 1+y & 1\end{vmatrix}$.

• A. 0 $x^2+y^2$
• B. 0 0
• C. 0 $xy$
• D. 0 $1+x+y$

Answer 1

Answer: (C)

$xy$

Answer 2

To find the value of the determinant $\begin{vmatrix}1 & 1 & 1 \\ 1+x & 1 & 1 \\ 1 & 1+y & 1\end{vmatrix}$, we can use elementary row operations to simplify it.

Let the given determinant be $D$. $D = \begin{vmatrix}1 & 1 & 1 \\ 1+x & 1 & 1 \\ 1 & 1+y & 1\end{vmatrix}$

Apply the row operations:

1. $R_2 \rightarrow R_2 - R_1$ (Subtract the first row from the second row)
2. $R_3 \rightarrow R_3 - R_1$ (Subtract the first row from the third row)

After applying $R_2 \rightarrow R_2 - R_1$: The new second row elements will be: $(1+x) - 1 = x$ $1 - 1 = 0$ $1 - 1 = 0$ So, the second row becomes $(x, 0, 0)$.

After applying $R_3 \rightarrow R_3 - R_1$: The new third row elements will be: $1 - 1 = 0$ $(1+y) - 1 = y$ $1 - 1 = 0$ So, the third row becomes $(0, y, 0)$.

The determinant now simplifies to: $D = \begin{vmatrix}1 & 1 & 1 \\ x & 0 & 0 \\ 0 & y & 0\end{vmatrix}$

Now, expand the determinant along the second row ($R_2$) because it contains two zeros, which simplifies the calculation. The expansion formula for a $3 \times 3$ determinant along $R_2$ is: $D = a_{21}C_{21} + a_{22}C_{22} + a_{23}C_{23}$ where $C_{ij}$ is the cofactor of the element $a_{ij}$, given by $C_{ij} = (-1)^{i+j}M_{ij}$, and $M_{ij}$ is the minor.

For our determinant: $a_{21} = x$, $a_{22} = 0$, $a_{23} = 0$.

$D = x \cdot (-1)^{2+1} \begin{vmatrix}1 & 1 \\ y & 0\end{vmatrix} + 0 \cdot (-1)^{2+2} \begin{vmatrix}1 & 1 \\ 0 & 0\end{vmatrix} + 0 \cdot (-1)^{2+3} \begin{vmatrix}1 & 1 \\ 0 & y\end{vmatrix}$

$D = x \cdot (-1) (1 \cdot 0 - 1 \cdot y) + 0 + 0$ $D = -x (0 - y)$ $D = -x (-y)$ $D = xy$

The value of the determinant is $xy$.