If
(i) (AB) C = A (BC)
(ii) A (B + C) = AB + AC.
Given, $A=\left[\begin{array}{cc}1 & 2 \\ -2 & 1\end{array}\right], B=\left[\begin{array}{cc}2 & 3 \\ 3 & -4\end{array}\right]$ and $C=\left[\begin{array}{cc}1 & 0 \\ -1 & 0\end{array}\right]$
(i) $A B=\left[\begin{array}{cc}1 & 2 \\ -2 & 1\end{array}\right]\left[\begin{array}{cc}2 & 3 \\ 3 & -4\end{array}\right]=\left[\begin{array}{cc}2+6 & 3-8 \\ -4+3 & -6-4\end{array}\right]=\left[\begin{array}{cc}8 & -5 \\ -1 & -10\end{array}\right]$
and $(A B) C=\left[\begin{array}{cc}8 & -5 \\ -1 & -10\end{array}\right]\left[\begin{array}{cc}1 & 0 \\ -1 & 0\end{array}\right]=\left[\begin{array}{cc}8+5 & 0 \\ -1+10 & 0\end{array}\right]=\left[\begin{array}{cc}13 & 0 \\ 9 & 0\end{array}\right]$...(i)
Again, $(B C)=\left[\begin{array}{cc}2 & 3 \\ 3 & -4\end{array}\right]\left[\begin{array}{cc}1 & 0 \\ -1 & 0\end{array}\right]=\left[\begin{array}{cc}2-3 & 0 \\ 3+4 & 0\end{array}\right]=\left[\begin{array}{cc}-1 & 0 \\ 7 & 0\end{array}\right]$
And $A(B C)=\left[\begin{array}{cc}1 & 2 \\ -2 & 1\end{array}\right]\left[\begin{array}{cc}-1 & 0 \\ 7 & 0\end{array}\right]=\left[\begin{array}{cc}-1+14 & 0 \\ 2+7 & 0\end{array}\right]=\left[\begin{array}{cc}13 & 0 \\ 9 & 0\end{array}\right]$$\ldots(i i)$
From (i) and (ii), we get
Hence, $(\mathbf{A B}) \mathbf{C}=\mathbf{A}(\mathbf{B C})$
(ii) $B+C=\left[\begin{array}{cc}2 & 3 \\ 3 & -4\end{array}\right]+\left[\begin{array}{cc}1 & 0 \\ -1 & 0\end{array}\right]=\left[\begin{array}{cc}3 & 3 \\ 2 & -4\end{array}\right]$
Now,
$A \cdot(B+C)=\left[\begin{array}{cc}1 & 2 \\ -2 & 1\end{array}\right]\left[\begin{array}{cc}3 & 3 \\ 2 & -4\end{array}\right]$
$=\left[\begin{array}{cc}3+4 & 3-8 \\ -6+2 & -6-4\end{array}\right]$
$=\left[\begin{array}{cc}7 & -5 \\ -4 & -10\end{array}\right]$ ...(iii)
$A B=\left[\begin{array}{cc}1 & 2 \\ -2 & 1\end{array}\right]\left[\begin{array}{cc}2 & 3 \\ 3 & -4\end{array}\right]=\left[\begin{array}{cc}2+6 & 3-8 \\ -4+3 & -6-4\end{array}\right]=\left[\begin{array}{cc}8 & -5 \\ -1 & -10\end{array}\right]$
and, $A C=\left[\begin{array}{cc}1 & 2 \\ -2 & 1\end{array}\right]\left[\begin{array}{cc}1 & 0 \\ -1 & 0\end{array}\right]=\left[\begin{array}{cc}1-2 & 0 \\ -2-1 & 0\end{array}\right]=\left[\begin{array}{cc}-1 & 0 \\ -3 & 0\end{array}\right]$
Thus, $A B+A C=\left[\begin{array}{cc}8 & -5 \\ -1 & -10\end{array}\right]+\left[\begin{array}{cc}-1 & 0 \\ -3 & 0\end{array}\right]=\left[\begin{array}{cc}7 & -5 \\ -4 & -10\end{array}\right]$ $\ldots$ (iv)
Hence from equation (iii) and (iv), we have
$A \cdot(B+C)=A \cdot B+A \cdot C$
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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
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- Semiconductor Electronics: Materials, Devices and Simple Circuits.
- Chemical Effects of Electric Current,