Question

The value of the determinant \( \left| \begin{matrix} x+a & y & z \\ x & y+b & z \\ x & y & z+c \end{matrix} \right| = abc \), where \( a, b, c \neq 0 \). The value of the expression \( \frac{x}{a} + \frac{y}{b} + \frac{z}{c} \) is:

• A. \( 0 \)
• B. \( 1 \)
• C. \( -1 \)
• D. \( 2 \)

Hint:

When a question involves algebraic fractions inside determinants, factoring out the denominators from the rows or columns is almost always the fastest path to a solution.

Answer 1

The Correct Option is A

Concept:
Determinants containing linear variable distributions can be simplified using row or column transformations to factor out common expressions.

Step 1: Factor out constants \(a\), \(b\), and \(c\) from columns 1, 2, and 3 respectively.

Taking out factors from the determinant:

\[ D = abc \begin{vmatrix} \frac{x}{a}+1 & \frac{y}{b} & \frac{z}{c} \\ \frac{x}{a} & \frac{y}{b}+1 & \frac{z}{c} \\ \frac{x}{a} & \frac{y}{b} & \frac{z}{c}+1 \end{vmatrix} \]

Step 2: Apply the row operation \(R_1 \rightarrow R_1 + R_2 + R_3\).

After adding the rows:

\[ D = abc \begin{vmatrix} 1+\frac{x}{a}+\frac{y}{b}+\frac{z}{c} & 1+\frac{x}{a}+\frac{y}{b}+\frac{z}{c} & 1+\frac{x}{a}+\frac{y}{b}+\frac{z}{c} \\ \frac{x}{a} & \frac{y}{b}+1 & \frac{z}{c} \\ \frac{x}{a} & \frac{y}{b} & \frac{z}{c}+1 \end{vmatrix} \]

Factor out the common expression from the first row:

\[ D = abc \left(1+\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\right) \begin{vmatrix} 1 & 1 & 1 \\ \frac{x}{a} & \frac{y}{b}+1 & \frac{z}{c} \\ \frac{x}{a} & \frac{y}{b} & \frac{z}{c}+1 \end{vmatrix} \]

Now apply the column operations:

\[ C_2 \rightarrow C_2 - C_1, \qquad C_3 \rightarrow C_3 - C_1 \]

The determinant reduces to:

\[ \begin{vmatrix} 1 & 0 & 0 \\ \frac{x}{a} & 1 & 0 \\ \frac{x}{a} & 0 & 1 \end{vmatrix} = 1 \]

Therefore,

\[ D = abc \left(1+\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\right) \]

Step 3: Use the given condition \(D = abc\).

\[ abc\left(1+\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\right)=abc \]

Since \(a,b,c \neq 0\), divide both sides by \(abc\):

\[ 1+\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1 \] \[ \therefore \quad \frac{x}{a}+\frac{y}{b}+\frac{z}{c}=0 \]