If \alpha eq a, \beta eq b, \gamma eq c and |\beginarraylll\alpha & b & c a & \beta & c a & b & \gamma\endarray|=0 , then \fraca\alpha-a+\fracb\beta-b+\frac\gamma\gamma-c is equal to :

Matrices and Determinants PYQ — JEE Main 2024 · 8 Apr · Shift 2 · Q37 with Solution

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Correct answer: D

Let $x=\alpha-a,\quad y=\beta-b,\quad z=\gamma-c.$ Given $\alpha\ne a,\beta\ne b,\gamma\ne c$ , we have $x,y,z\ne 0$ .

Then the determinant becomes

\alpha & b & c\\ a & \beta & c\\ a & b & \gamma \end{vmatrix} = \begin{vmatrix} a+x & b & c\\ a & b+y & c\\ a & b & c+z \end{vmatrix}=0.$$ 2. Apply row operations: $$R_1\to R_1-R_2,\qquad R_2\to R_2-R_3.$$ Then $$\begin{vmatrix} a+x & b & c\\ a & b+y & c\\ a & b & c+z \end{vmatrix} = \begin{vmatrix} x & -y & 0\\ 0 & y & -z\\ a & b & c+z \end{vmatrix}.$$ Now expand along the first row: \begin{align*} \Delta &= x\begin{vmatrix} y & -z \\ b & c+z \end{vmatrix}-(-y)\begin{vmatrix} 0 & -z \\ a & c+z \end{vmatrix} \\ &= x\big(y(c+z)+bz\big)+y(az) \\ &= xyc+xyz+xbz+ayz. \end{align*} Since $\Delta=0$, $$xyc+xyz+xbz+ayz=0.$$ 3. A cleaner way is to expand the original determinant directly: \begin{align*} 0&=\begin{vmatrix} a+x & b & c\\ a & b+y & c\\ a & b & c+z \end{vmatrix}\\ &=(a+x)(b+y)(c+z)+abc+abc-ab(c+z)-ac(b+y)-bc(a+x)-a(b+y)c-(a+x)b c\\ &=xyz+ayz+bxz+cxy. \end{align*} So, $$xyz+ayz+bxz+cxy=0.$$ 4. Divide by $xyz$ (allowed since $x,y,z\ne 0$): $$1+\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=0.$$ Hence $$\frac{a}{\alpha-a}+\frac{b}{\beta-b}+\frac{c}{\gamma-c}=-1.$$ 5. The expression asked in the question is $$\frac{a}{\alpha-a}+\frac{b}{\beta-b}+\frac{\gamma}{\gamma-c}.$$ Since $\gamma=c+z$, $$\frac{\gamma}{\gamma-c}=\frac{c+z}{z}=\frac{c}{z}+1.$$ Therefore, \begin{align*} \frac{a}{\alpha-a}+\frac{b}{\beta-b}+\frac{\gamma}{\gamma-c} &=\left(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}\right)+1 \\ &=(-1)+1=0. \end{align*} 6. Hence the correct option is $$\boxed{0}.$$

View written solutionFree

Correct answer: D

Let $x=\alpha-a,\quad y=\beta-b,\quad z=\gamma-c.$ Given $\alpha\ne a,\beta\ne b,\gamma\ne c$ , we have $x,y,z\ne 0$ .

Then the determinant becomes

\alpha & b & c\\ a & \beta & c\\ a & b & \gamma \end{vmatrix} = \begin{vmatrix} a+x & b & c\\ a & b+y & c\\ a & b & c+z \end{vmatrix}=0.$$ 2. Apply row operations: $$R_1\to R_1-R_2,\qquad R_2\to R_2-R_3.$$ Then $$\begin{vmatrix} a+x & b & c\\ a & b+y & c\\ a & b & c+z \end{vmatrix} = \begin{vmatrix} x & -y & 0\\ 0 & y & -z\\ a & b & c+z \end{vmatrix}.$$ Now expand along the first row: \begin{align*} \Delta &= x\begin{vmatrix} y & -z \\ b & c+z \end{vmatrix}-(-y)\begin{vmatrix} 0 & -z \\ a & c+z \end{vmatrix} \\ &= x\big(y(c+z)+bz\big)+y(az) \\ &= xyc+xyz+xbz+ayz. \end{align*} Since $\Delta=0$, $$xyc+xyz+xbz+ayz=0.$$ 3. A cleaner way is to expand the original determinant directly: \begin{align*} 0&=\begin{vmatrix} a+x & b & c\\ a & b+y & c\\ a & b & c+z \end{vmatrix}\\ &=(a+x)(b+y)(c+z)+abc+abc-ab(c+z)-ac(b+y)-bc(a+x)-a(b+y)c-(a+x)b c\\ &=xyz+ayz+bxz+cxy. \end{align*} So, $$xyz+ayz+bxz+cxy=0.$$ 4. Divide by $xyz$ (allowed since $x,y,z\ne 0$): $$1+\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=0.$$ Hence $$\frac{a}{\alpha-a}+\frac{b}{\beta-b}+\frac{c}{\gamma-c}=-1.$$ 5. The expression asked in the question is $$\frac{a}{\alpha-a}+\frac{b}{\beta-b}+\frac{\gamma}{\gamma-c}.$$ Since $\gamma=c+z$, $$\frac{\gamma}{\gamma-c}=\frac{c+z}{z}=\frac{c}{z}+1.$$ Therefore, \begin{align*} \frac{a}{\alpha-a}+\frac{b}{\beta-b}+\frac{\gamma}{\gamma-c} &=\left(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}\right)+1 \\ &=(-1)+1=0. \end{align*} 6. Hence the correct option is $$\boxed{0}.$$