Practice Start Test

Matrices and Determinants

JEE Mathematics · 271 questions · Page 18 of 28 · Click an option or "Show Solution" to reveal answer

Q171
Let R=(x000y000z)R=\left(\begin{array}{ccc}x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z\end{array}\right) be a non-zero 3×33 \times 3 matrix, where xsin⁡θ=ysin⁡(θ+2π3)=zsin⁡(θ+4π3)≠0,θ∈(0,2π)x \sin \theta=y \sin \left(\theta+\dfrac{2 \pi}{3}\right)=z \sin \left(\theta+\dfrac{4 \pi}{3}\right) \neq 0, \theta \in(0,2 \pi). For a square matrix MM, let trace (M)(M) denote the sum of all the diagonal entries of MM. Then, among the statements: (I) Trace (R)=0(R)=0 (II) If trace (adj⁡(adj⁡(R))=0(\operatorname{adj}(\operatorname{adj}(R))=0, then RR has exactly one non-zero entry.
A Only (I) is true
B Only (II) is true
C Both (I) and (II) are true
D Neither (I) nor (II) is true
Correct Answer
Option D
Solution
xsin⁡θ=ysin⁡(θ+2π3)=zsin⁡(θ+4π3)≠0⇒x,y,z≠0\begin{aligned} & x \sin \theta=y \sin \left(\theta+\frac{2 \pi}{3}\right)=z \sin \left(\theta+\frac{4 \pi}{3}\right) \neq 0 \\ & \Rightarrow x, y, z \neq 0 \end{aligned}

Also,

sin⁡θ+sin⁡(θ+2π3)+sin⁡(θ+4π3)=0∀θ∈R⇒1x+1y+1z=0⇒xy+yz+zx=0\begin{aligned} & \sin \theta+\sin \left(\theta+\frac{2 \pi}{3}\right)+\sin \left(\theta+\frac{4 \pi}{3}\right)=0 \forall \theta \in \mathrm{R} \\ & \Rightarrow \frac{1}{\mathrm{x}}+\frac{1}{\mathrm{y}}+\frac{1}{\mathrm{z}}=0 \\ & \Rightarrow \mathrm{xy}+\mathrm{yz}+\mathrm{zx}=0 \end{aligned}

(i)

Trace⁡(R)=x+y+z\quad \operatorname{Trace}(\mathrm{R})=\mathrm{x}+\mathrm{y}+\mathrm{z}

If

x+y+z=0x+y+z=0

and

xy+yz+zx=0x y+y z+z x=0
⇒x=y=z=0\Rightarrow \mathrm{x}=\mathrm{y}=\mathrm{z}=0

Statement (i) is False (ii)

Adj⁡(Adj⁡(R))=∣R∣R\quad \operatorname{Adj}(\operatorname{Adj}(\mathrm{R}))=|\mathrm{R}| \mathrm{R}

Trace

(Adj⁡(Adj⁡(R)))(\operatorname{Adj}(\operatorname{Adj}(\mathrm{R})))
=xyz(x+y+z)≠0=x y z(x+y+z) \neq 0

Statement (ii) is also False

Q172
Consider the system of linear equations x+y+z=5,x+2y+λ2z=9,x+3y+λz=μx+y+z=5, x+2 y+\lambda^2 z=9, x+3 y+\lambda z=\mu, where λ,μ∈R\lambda, \mu \in \mathbb{R}. Then, which of the following statement is NOT correct?
A System is consistent if λ≠1\lambda \neq 1 and μ=13\mu=13
B System is inconsistent if λ=1\lambda=1 and μ≠13\mu \neq 13
C System has unique solution if λ≠1\lambda \neq 1 and μ≠13\mu \neq 13
D System has infinite number of solutions if λ=1\lambda=1 and μ=13\mu=13
Correct Answer
Option C
Solution
∣11112λ213λ∣=0⇒2λ2−λ−1=0λ=1,−12∣1152λ293λμ∣=0⇒μ=13\begin{aligned} & \left|\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 2 & \lambda^2 \\ 1 & 3 & \lambda \end{array}\right|=0 \\ & \Rightarrow 2 \lambda^2-\lambda-1=0 \\ & \lambda=1,-\frac{1}{2} \\ & \left|\begin{array}{ccc} 1 & 1 & 5 \\ 2 & \lambda^2 & 9 \\ 3 & \lambda & \mu \end{array}\right|=0 \Rightarrow \mu=13 \end{aligned}

Infinite solution

λ=1&μ=13\lambda=1 \& \mu=13

For unique

sol⁡nλ≠1\operatorname{sol}^{\mathrm{n}} \lambda \neq 1

For no

sol⁡nλ=1&μ≠13\operatorname{sol}^{\mathrm{n}} \lambda=1 \& \mu \neq 13

If

λ≠1\lambda \neq 1

and

μ≠13\mu \neq 13

Considering the case when

λ=−12\lambda=-\frac{1}{2}

and

μ≠13\mu \neq 13

this will generate no solution case

Q173
Consider the system of linear equations x+y+z=4μ,x+2y+2λz=10μ,x+3y+4λ2z=μ2+15x+y+z=4 \mu, x+2 y+2 \lambda z=10 \mu, x+3 y+4 \lambda^2 z=\mu^2+15 where λ,μ∈R\lambda, \mu \in \mathbf{R}. Which one of the following statements is NOT correct ?
A The system has unique solution if λ≠12\lambda \neq \dfrac{1}{2} and μ≠1,15\mu \neq 1,15
B The system has infinite number of solutions if λ=12\lambda=\dfrac{1}{2} and μ=15\mu=15
C The system is consistent if λ≠12\lambda \neq \dfrac{1}{2}
D The system is inconsistent if λ=12\lambda=\dfrac{1}{2} and μ≠1\mu \neq 1
Correct Answer
Option D
Solution
x+y+z=4μ,x+2y+2λz=10μ,x+3y+4λ2z=μ2+15x+y+z=4 \mu, x+2 y+2 \lambda z=10 \mu, x+3 y+4 \lambda{ }^2 z=\mu^2+15

,

Δ=∣111122λ134λ2∣=(2λ−1)2\Delta=\left|\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 2 & 2 \lambda \\ 1 & 3 & 4 \lambda^2 \end{array}\right|=(2 \lambda-1)^2

For unique solution

Δ≠0,2λ−1≠0,(λ≠12)\Delta \neq 0,2 \lambda-1 \neq 0,\left(\lambda \neq \frac{1}{2}\right)

Let

Δ=0,λ=12\Delta=0, \lambda=\frac{1}{2}
Δy=0,Δx=Δz=∣4μ1110μ21μ2+1531∣=(μ−15)(μ−1)\begin{aligned} & \Delta_y=0, \Delta_x=\Delta_z=\left|\begin{array}{ccc} 4 \mu & 1 & 1 \\ 10 \mu & 2 & 1 \\ \mu^2+15 & 3 & 1 \end{array}\right| \\ & =(\mu-15)(\mu-1) \end{aligned}

For infinite solution

λ=12,μ=1\lambda=\frac{1}{2}, \mu=1

or 15

Q174
Let B=[1315]B=\left[\begin{array}{ll}1 & 3 \\ 1 & 5\end{array}\right] and AA be a 2×22 \times 2 matrix such that AB−1=A−1A B^{-1}=A^{-1}. If BCB−1=AB C B^{-1}=A and C4+αC2+βI=OC^4+\alpha C^2+\beta I=O, then 2β−α2 \beta-\alpha is equal to
A 16
B 10
C 8
D 2
Correct Answer
Option B
Solution
B=[1315]AB−1=A−1⇒A2=B\begin{aligned} & B=\left[\begin{array}{ll} 1 & 3 \\ 1 & 5 \end{array}\right] \\ & A B^{-1}=A^{-1} \\ & \Rightarrow A^2=B \end{aligned}

Also,

BCB−1=AB C B^{-1}=A
⇒C=B−1AB⇒C4=(B−1AB)(B−1AB)(B−1AB)(B−1AB)=B−1A4B=B−1B2B⇒C4=B2\begin{aligned} \Rightarrow C & =B^{-1} A B \\ \Rightarrow C^4 & =\left(B^{-1} A B\right)\left(B^{-1} A B\right)\left(B^{-1} A B\right)\left(B^{-1} A B\right) \\ & =B^{-1} A^4 B \\ & =B^{-1} B^2 B \\ \Rightarrow \quad C^4 & =B^2 \end{aligned}

Also,

C2=(B−1AB)(B−1AB)C^2=\left(B^{-1} A B\right)\left(B^{-1} A B\right)
=B−1A2B=B−1BB\begin{aligned} & =B^{-1} A^2 B \\ & =B^{-1} B B \end{aligned}
⇒C2=B⇒C4+αC2+βI=0⇒B2+αB+βI=0B2=[1315][1315]=[418628]⇒[418628]+α[1315]+[β00β]=[0000]⇒4+α+β=0\begin{aligned} & \Rightarrow C^2=B \\ & \Rightarrow C^4+\alpha C^2+\beta I=0 \\ & \Rightarrow B^2+\alpha B+\beta I=0 \\ & B^2=\left[\begin{array}{ll} 1 & 3 \\ 1 & 5 \end{array}\right]\left[\begin{array}{ll} 1 & 3 \\ 1 & 5 \end{array}\right]=\left[\begin{array}{ll} 4 & 18 \\ 6 & 28 \end{array}\right] \\ & \Rightarrow\left[\begin{array}{ll} 4 & 18 \\ 6 & 28 \end{array}\right]+\alpha\left[\begin{array}{ll} 1 & 3 \\ 1 & 5 \end{array}\right]+\left[\begin{array}{ll} \beta & 0 \\ 0 & \beta \end{array}\right]=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right] \\ & \Rightarrow 4+\alpha+\beta=0 \end{aligned}

and

18+3α=018+3 \alpha=0
⇒α=−6⇒β=2⇒2β−α=10\begin{aligned} & \Rightarrow \alpha=-6 \\ & \Rightarrow \beta=2 \\ & \Rightarrow 2 \beta-\alpha=10 \end{aligned}
Q175
Let λ,μ∈R\lambda, \mu \in \mathbf{R}. If the system of equations 3x+5y+λz=37x+11y−9z=297x+155y−189z=μ\begin{aligned} & 3 x+5 y+\lambda z=3 \\ & 7 x+11 y-9 z=2 \\ & 97 x+155 y-189 z=\mu \end{aligned} has infinitely many solutions, then μ+2λ\mu+2 \lambda is equal to :
A 24
B 25
C 27
D 22
Correct Answer
Option B
Solution
3x+5y+λz=37x+11y−9z=297x+155y−189z=μ\begin{aligned} & 3 x+5 y+\lambda z=3 \\ & 7 x+11 y-9 z=2 \\ & 97 x+155 y-189 z=\mu \end{aligned}
[35λ711−997155−189][xyz]=[32μ]AX=BX=A−1BX=adj⁡A∣A∣B\begin{aligned} & {\left[\begin{array}{ccc} 3 & 5 & \lambda \\ 7 & 11 & -9 \\ 97 & 155 & -189 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{l} 3 \\ 2 \\ \mu \end{array}\right]} \\ & A X=B \\ & X=A^{-1} B \\ & X=\frac{\operatorname{adj} A}{|A|} B \end{aligned}

For Infinitely many solution

∣A∣=0 and (adj⁡A)B=0∣35λ711−997155−189∣=0−2052+2250+18λ=0⇒λ=−11adj⁡A=[−684−76076450500−501820−2](adj⁡A)B=[684−76076450500−501820−2][32μ]=[000]⇒54+40−2μ=0⇒2μ=94⇒μ=47⇒μ+2λ=25\begin{aligned} & |A|=0 \text{ and }(\operatorname{adj} A) B=0 \\ & \left|\begin{array}{ccc} 3 & 5 & \lambda \\ 7 & 11 & -9 \\ 97 & 155 & -189 \end{array}\right|=0 \\ & -2052+2250+18 \lambda=0 \\ & \Rightarrow \lambda=-11 \\ & \operatorname{adj} A=\left[\begin{array}{ccc} -684 & -760 & 76 \\ 450 & 500 & -50 \\ 18 & 20 & -2 \end{array}\right] \\ & (\operatorname{adj} A) B=\left[\begin{array}{ccc} 684 & -760 & 76 \\ 450 & 500 & -50 \\ 18 & 20 & -2 \end{array}\right]\left[\begin{array}{l} 3 \\ 2 \\ \mu \end{array}\right]=\left[\begin{array}{l} 0 \\ 0 \\ 0 \end{array}\right] \\ & \Rightarrow 54+40-2 \mu=0 \\ & \Rightarrow 2 \mu=94 \\ & \Rightarrow \mu=47 \\ & \Rightarrow \mu+2 \lambda=25 \end{aligned}
Q176
Let α∈(0,∞)\alpha \in(0, \infty) and A=[12α101012]A=\left[\begin{array}{lll}1 & 2 & \alpha \\ 1 & 0 & 1 \\ 0 & 1 & 2\end{array}\right]. If det⁡(adj⁡(2A−AT)⋅adj⁡(A−2AT))=28\operatorname{det}\left(\operatorname{adj}\left(2 A-A^T\right) \cdot \operatorname{adj}\left(A-2 A^T\right)\right)=2^8, then (det⁡(A))2(\operatorname{det}(A))^2 is equal to:
A 16
B 36
C 49
D 1
Correct Answer
Option A
Solution
∣adj⁡(A−2AT)⋅adj⁡(2A−AT)∣=28P=A−2A⊤Q=2AT−A⇒QT=2AT−A=−P∣adj⁡(P)adj⁡(Q)∣⇒∣PQ∣=−24⇒∣P∣(−∣P∣)=−24⇒∣P∣=4 and ∣Q∣=−4∣A−2AT∣=4A−2AT=[12α101012]−2[110201α12]=[−10α−30−1−2α−1−2]⇒∣A−2AT∣=1+3α=4⇒α=1⇒∣A∣=−4⇒∣A∣2=16\begin{aligned} & \left|\operatorname{adj}\left(A-2 A^T\right) \cdot \operatorname{adj}\left(2 A-A^T\right)\right|=2^8 \\ & P=A-2 A^{\top} \\ & Q=2 A^T-A \Rightarrow Q^T=2 A^T-A=-P \\ & |\operatorname{adj}(P) \operatorname{adj}(Q)| \Rightarrow|P Q|=-2^4 \\ & \Rightarrow|P|(-|P|)=-2^4 \Rightarrow|P|=4 \text{ and }|Q|=-4 \\ & \left|A-2 A^T\right|=4 \\ & A-2 A^T=\left[\begin{array}{lll} 1 & 2 & \alpha \\ 1 & 0 & 1 \\ 0 & 1 & 2 \end{array}\right]-2\left[\begin{array}{lll} 1 & 1 & 0 \\ 2 & 0 & 1 \\ \alpha & 1 & 2 \end{array}\right]=\left[\begin{array}{ccc} -1 & 0 & \alpha \\ -3 & 0 & -1 \\ -2 \alpha & -1 & -2 \end{array}\right] \\ & \Rightarrow\left|A-2 A^T\right|=1+3 \alpha=4 \Rightarrow \alpha=1 \Rightarrow|A|=-4 \\ & \Rightarrow|A|^2=16 \end{aligned}
Q177
If the system of equations 11x+y+λz=−52x+3y+5z=38x−19y−39z=μ\begin{array}{r} 11 x+y+\lambda z=-5 \\ 2 x+3 y+5 z=3 \\ 8 x-19 y-39 z=\mu \end{array} has infinitely many solutions, then λ4−μ\lambda^4-\mu is equal to :
A 51
B 45
C 47
D 49
Correct Answer
Option C
Solution
11x+y+λz=−52x+3y+5z=38x−19y−39z=μΔ=0⇒∣111λ2358−19−39∣=011(−39.3+19.5)−1(−39.2−40)+λ(−38−24)=0=11(−117+95)−1(−118)−62λ=0=−242+118=62λ⇒λ=−2Δ2=0⇒∣111−52338−19μ∣=0\begin{aligned} & 11 x+y+\lambda z=-5 \\ & 2 x+3 y+5 z=3 \\ & 8 x-19 y-39 z=\mu \\ & \Delta=0 \Rightarrow\left|\begin{array}{ccc} 11 & 1 & \lambda \\ 2 & 3 & 5 \\ 8 & -19 & -39 \end{array}\right|=0 \\ & 11(-39.3+19.5)-1(-39.2-40)+\lambda(-38-24)=0 \\ & =11(-117+95)-1(-118)-62 \lambda=0 \\ & =-242+118=62 \lambda \\ & \Rightarrow \lambda=-2 \\ & \Delta 2=0 \\ & \Rightarrow\left|\begin{array}{ccc} 11 & 1 & -5 \\ 2 & 3 & 3 \\ 8 & -19 & \mu \end{array}\right|=0 \end{aligned}
11(3μ+57)−1(2μ−24)−5(−38−24)=033μ+627−2μ+24+310=0μ=−31⇒λ4−31=16+31=47\begin{aligned} & 11(3 \mu+57)-1(2 \mu-24)-5(-38-24)=0 \\ & 33 \mu+627-2 \mu+24+310=0 \\ & \mu=-31 \\ & \Rightarrow \lambda^4-31 \\ & =16+31 \\ & =47 \end{aligned}
Q178
If the system of equations x+(2sin⁡α)y+(2cos⁡α)z=0x+(cos⁡α)y+(sin⁡α)z=0x+(sin⁡α)y−(cos⁡α)z=0\begin{aligned} & x+(\sqrt{2} \sin \alpha) y+(\sqrt{2} \cos \alpha) z=0 \\ & x+(\cos \alpha) y+(\sin \alpha) z=0 \\ & x+(\sin \alpha) y-(\cos \alpha) z=0 \end{aligned} has a non-trivial solution, then α∈(0,π2)\alpha \in\left(0, \dfrac{\pi}{2}\right) is equal to :
A 5π24\dfrac{5 \pi}{24}
B 11π24\dfrac{11 \pi}{24}
C 7π24\dfrac{7 \pi}{24}
D 3π4\dfrac{3 \pi}{4}
Correct Answer
Option A
Solution
x+(2sin⁡α)y+(2cos⁡α)z=0x+(cos⁡α)y+(sin⁡α)z=0x+(sin⁡α)y−(cos⁡α)z=0\begin{aligned} & x+(\sqrt{2} \sin \alpha) y+(\sqrt{2} \cos \alpha) z=0 \\ & x+(\cos \alpha) y+(\sin \alpha) z=0 \\ & x+(\sin \alpha) y-(\cos \alpha) z=0 \end{aligned}

∵\because Non-trivial solution

⇒D=0\Rightarrow D=0
∣12sin⁡α2cos⁡1cos⁡αsin⁡α1sin⁡α−cos⁡α∣=01[−cos⁡2α−sin⁡α]−1[−2sin⁡αcos⁡α−2sin⁡αcos⁡α]+1[2sin⁡2α−2cos⁡2α]=0\begin{aligned} & \left|\begin{array}{ccc} 1 & \sqrt{2} \sin \alpha & \sqrt{2} \cos \\ 1 & \cos \alpha & \sin \alpha \\ 1 & \sin \alpha & -\cos \alpha \end{array}\right|=0 \\ & 1{\left[ { - {{\cos }^2}\alpha - \sin \alpha } \right]^{ - 1}}\left[ { - \sqrt 2 \sin \alpha \cos \alpha - \sqrt 2 \sin \alpha \cos \alpha } \right] + 1\left[ {\sqrt 2 {{\sin }^2}\alpha - \sqrt 2 {{\cos }^2}\alpha } \right] = 0 \\ \end{aligned}
−1+22sin⁡αcos⁡α+2(sin⁡2α−cos⁡2α)=02sin⁡2α−2cos⁡2α=1sin⁡2α2−cos⁡2α2=12sin⁡(2α−π4)=sin⁡π6⇒2α−π4=nπ+(−1)nπ6 for n=0⇒α=5π24\begin{aligned} & -1+2 \sqrt{2} \sin \alpha \cos \alpha+\sqrt{2}\left(\sin ^2 \alpha-\cos ^2 \alpha\right)=0 \\ & \sqrt{2} \sin 2 \alpha-\sqrt{2} \cos 2 \alpha=1 \\ & \frac{\sin 2 \alpha}{\sqrt{2}}-\frac{\cos 2 \alpha}{\sqrt{2}}=\frac{1}{2} \\ & \sin \left(2 \alpha-\frac{\pi}{4}\right)=\sin \frac{\pi}{6} \\ & \Rightarrow 2 \alpha-\frac{\pi}{4}=n \pi+(-1)^n \frac{\pi}{6} \text{ for } n=0 \\ & \Rightarrow \alpha=\frac{5 \pi}{24} \end{aligned}
Q179
Let A=[1201]A=\left[\begin{array}{ll}1 & 2 \\ 0 & 1\end{array}\right] and B=I+adj⁡(A)+(adj⁡A)2+…+(adj⁡A)10B=I+\operatorname{adj}(A)+(\operatorname{adj} A)^2+\ldots+(\operatorname{adj} A)^{10}. Then, the sum of all the elements of the matrix BB is:
A −-110
B 22
C −-124
D −-88
Correct Answer
Option D
Solution
adj⁡(A)=[1−201](adj⁡A)2=[1−401](adj⁡A)3=[1−601](adj⁡A)4=[1−801](adj⁡A)r=[1(−2r)01]\begin{aligned} & \operatorname{adj}(A)=\left[\begin{array}{ll} 1 & -2 \\ 0 & 1 \end{array}\right] \\ & (\operatorname{adj} A)^2=\left[\begin{array}{ll} 1 & -4 \\ 0 & 1 \end{array}\right] \\ & (\operatorname{adj} A)^3=\left[\begin{array}{cc} 1 & -6 \\ 0 & 1 \end{array}\right] \\ & (\operatorname{adj} A)^4=\left[\begin{array}{cc} 1 & -8 \\ 0 & 1 \end{array}\right] \\ & (\operatorname{adj} A)^r=\left[\begin{array}{cc} 1 & (-2 r) \\ 0 & 1 \end{array}\right] \end{aligned}
B=∑r=010(adj⁡A)r=[∑r=0101∑r=010(−2r)∑r=010(0)∑r=010(1)]B=[11−110011]\begin{aligned} & B=\sum_{r=0}^{10}(\operatorname{adj} A)^r=\left[\begin{array}{ll} \sum\limits_{r=0}^{10} 1 & \sum\limits_{r=0}^{10}(-2 r) \\ \sum\limits_{r=0}^{10}(0) & \sum\limits_{r=0}^{10}(1) \end{array}\right] \\ & B=\left[\begin{array}{cc} 11 & -110 \\ 0 & 11 \end{array}\right] \end{aligned}
 Sum of elements =−110+11+11=−88\text{ Sum of elements }=-110+11+11=-88
Q180
If α≠a,β≠b,γ≠c\alpha \neq \mathrm{a}, \beta \neq \mathrm{b}, \gamma \neq \mathrm{c} and ∣αbcaβcabγ∣=0\left|\begin{array}{lll}\alpha & \mathrm{b} & \mathrm{c} \\ \mathrm{a} & \beta & \mathrm{c} \\ \mathrm{a} & \mathrm{b} & \gamma\end{array}\right|=0, then aα−a+bβ−b+γγ−c\dfrac{\mathrm{a}}{\alpha-\mathrm{a}}+\dfrac{\mathrm{b}}{\beta-\mathrm{b}}+\dfrac{\gamma}{\gamma-\mathrm{c}} is equal to :
A 2
B 3
C 1
D 0
Correct Answer
Option D
Solution
∣αbcaβcabγ∣=0\left|\begin{array}{lll} \alpha & b & c \\ a & \beta & c \\ a & b & \gamma \end{array}\right|=0
R1→R1−R2,R2→R2−R3⇒∣α−ab−β00β−bc−γabγ∣=0\begin{aligned} & R_1 \rightarrow R_1-R_2, R_2 \rightarrow R_2-R_3 \\ & \Rightarrow\left|\begin{array}{ccc} \alpha-a & b-\beta & 0 \\ 0 & \beta-b & c-\gamma \\ a & b & \gamma \end{array}\right|=0 \end{aligned}

Take α\alpha-a, β\beta-b, γ\gamma-c common from column-1, 2 and 3 respectively

(α−a)(β−b)(γ−c)∣1−1001−1aα−abβ−bγγ−c∣=0⇒γγ−c+bβ−b+aα−a=0\begin{aligned} & (\alpha-a)(\beta-b)(\gamma-c)\left|\begin{array}{ccc} 1 & -1 & 0 \\ 0 & 1 & -1 \\ \frac{a}{\alpha-a} & \frac{b}{\beta-b} & \frac{\gamma}{\gamma-c} \end{array}\right|=0 \\ & \Rightarrow \frac{\gamma}{\gamma-c}+\frac{b}{\beta-b}+\frac{a}{\alpha-a}=0 \end{aligned}
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