Using the property of determinants and without expanding, prove that:
$\left|\begin{array}{lll}b+c & q+r & y+z \\ c+a & r+p & z+x \\ a+b & p+q & x+y\end{array}\right|=2\left|\begin{array}{lll}a & p & x \\ b & q & y \\ c & r & z\end{array}\right|$
$\Delta=\left|\begin{array}{ccc}b+c & q+r & y+z \\ c+a & r+p & z+x \\ a+b & p+q & x+y\end{array}\right|$
$=\left|\begin{array}{lll}b+c & q+r & y+z \\ c+a & r+p & z+x \\ a & p & x\end{array}\right|+\left|\begin{array}{lll}b+c & q+r & y+z \\ c+a & r+p & z+x \\ b & q & y\end{array}\right|$
$=\Delta_{1}+\Delta_{2}$ (say) ....(1)
Now, $\Delta_{\mathrm{I}}=\left|\begin{array}{lll}b+c & q+r & y+z \\ c+a & r+p & z+x \\ a & p & x\end{array}\right|$
Applying $R_{2} \rightarrow R_{2}-R_{3}$, we have:
$\Delta_{1}=\left|\begin{array}{lll}b+c & q+r & y+z \\ c & r & z \\ a & p & x\end{array}\right|$
Applying $R_{1} \rightarrow R_{1}-R_{2}$, we have:
$\Delta_{1}=\left|\begin{array}{lll}b & q & y \\ c & r & z \\ a & p & x\end{array}\right|$
Applying $R_{1} \leftrightarrow R_{3}$ and $R_{2} \leftrightarrow R_{3}$, we have:
$\Delta_{1}=(-1)^{2}\left|\begin{array}{lll}a & p & x \\ b & q & y \\ c & r & z\end{array}\right|=\left|\begin{array}{lll}a & p & x \\ b & q & y \\ c & r & z\end{array}\right|$ ....(2)
$\Delta_{2}=\left|\begin{array}{lll}b+c & q+r & y+z \\ c+a & r+p & z+x \\ b & q & y\end{array}\right|$
Applying $R_{1} \rightarrow R_{1}-R_{3}$, we have:
$\Delta_{2}=\left|\begin{array}{lll}c & r & z \\ c+a & r+p & z+x \\ b & q & y\end{array}\right|$
Applying $R_{2} \rightarrow R_{2}-R_{1}$, we have:
$\Delta_{2}=\left|\begin{array}{lll}c & r & z \\ a & p & x \\ b & q & y\end{array}\right|$
Applying $\mathrm{R}_{1} \leftrightarrow \mathrm{R}_{2}$ and $\mathrm{R}_{2} \leftrightarrow \mathrm{R}_{3}$, we have:
$\Delta_{2}=(-1)^{2}\left|\begin{array}{lll}a & p & x \\ b & q & y \\ c & r & z\end{array}\right|=\left|\begin{array}{lll}a & p & x \\ b & q & y \\ c & r & z\end{array}\right|$ ...(3)
From (1), (2), and (3), we have:
$\Delta=2\left|\begin{array}{lll}a & p & x \\ b & q & y \\ c & r & z\end{array}\right|$
Hence, the given result is proved.
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- JEE Main
- Exam Pattern
- Previous Year Papers
- PYQ Chapterwise
- Physics
- Kinematics 1D
- Kinemetics 2D
- Friction
- Work, Power, Energy
- Centre of Mass and Collision
- Rotational Dynamics
- Gravitation
- Calorimetry
- Elasticity
- Thermal Expansion
- Heat Transfer
- Kinetic Theory of Gases
- Thermodynamics
- Simple Harmonic Motion
- Wave on String
- Sound waves
- Fluid Mechanics
- Electrostatics
- Current Electricity
- Capacitor
- Magnetism and Matter
- Electromagnetic Induction
- Atomic Structure
- Dual Nature of Matter
- Nuclear Physics
- Radioactivity
- Semiconductors
- Communication System
- Error in Measurement & instruments
- Alternating Current
- Electromagnetic Waves
- Wave Optics
- X-Rays
- All Subjects
- Physics
- Motion in a Plane
- Law of Motion
- Work, Energy and Power
- Systems of Particles and Rotational Motion
- Gravitation
- Mechanical Properties of Solids
- Mechanical Properties of Fluids
- Thermal Properties of matter
- Thermodynamics
- Kinetic Theory
- Oscillations
- Waves
- Electric Charge and Fields
- Electrostatic Potential and Capacitance
- Current Electricity
- Thermoelectric Effects of Electric Current
- Heating Effects of Electric Current
- Moving Charges and Magnetism
- Magnetism and Matter
- Electromagnetic Induction
- Alternating Current
- Electromagnetic Wave
- Ray Optics and Optical Instruments
- Wave Optics
- Dual Nature of Radiation and Matter
- Atoms
- Nuclei
- Semiconductor Electronics: Materials, Devices and Simple Circuits.
- Chemical Effects of Electric Current,