You are here
Ex 3.2, 5 - Chapter 3 Class 12 Matrices
Last updated at Dec. 16, 2024 by Teachoo
Chapter 3 Class 12 Matrices
Concept wise
- Formation and order of matrix
- Types of matrices
- Equal matrices
-
Addition/ subtraction of matrices
- Statement questions - Addition/Subtraction of matrices
- Multiplication of matrices
- Statement questions - Multiplication of matrices
- Solving Equation
- Finding unknown - Element
- Finding unknown - Matrice
- Transpose of a matrix
- Symmetric and skew symmetric matrices
- Proof using property of transpose
- Inverse of matrix using elementary transformation
- Proof using mathematical induction
Transcript
Ex 3.2, 5 If A = [■8(2/3&1&5/3@1/3&2/3&4/3@7/3&2&2/3)] and B = [■8(2/5&3/5&1@1/5&2/5&4/5@7/5&6/5&2/5)] ,then compute 3A – 5B 3A – 5B = 3 [■8(2/3&1&5/3@1/3&2/3&4/3@7/3&2&2/3)] - 5 [■8(2/5&3/5&1@1/5&2/5&4/5@7/5&6/5&2/5)] = [■8(2/3 ×3&1×3&5/3×3@1/3×3&2/3×3&4/3×3@7/3×3&2×3&2/3×3)] – [■8(2/5×5&3/5×5&1×5@1/5×5&2/5×5&4/5×5@7/5×5&6/5×5&2/5×5 )] = [■8(𝟐&𝟑&𝟓@𝟏&𝟐&𝟒@𝟕&𝟔&𝟐)] – [■8(𝟐&𝟑&𝟓@𝟏&𝟐&𝟒@𝟕&𝟔&𝟐)] = [■8(2−2&3−3&5−5@1−1&2−2&4−4@7−7&6−6&2−2)] = [■8(𝟎&𝟎&𝟎@𝟎&𝟎&𝟎@𝟎&𝟎&𝟎)] Hence, 3A – 5B = [■8(0&0&0@0&0&0@0&0&0)]
