Transpose of a matrix

Chapter 3 Class 12 Matrices
Concept wise

In transpose of a matrix,

• Rows become columns and,
• Column become rows

For matrix

It’s transpose is

We denote it by A’

Similarly for

Let’s look at some properties of transpose

## Properties of transpose of a matrix

• (A’)’ = A
• (kA)’ = kA’
• (A + B)’ = A’ + B’
• (AB)’ = B’ A’

Let’s try to prove them one by one

Let

(A’)’ = A

Therefore,

(A’)’ = A

Therefore,

(4A)’ = 4A’

## (AB)’ = B’ A’

Therefore,

(AB)’ = B’ A’

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### Transcript

For matrix A = [■8(3&[email protected]&4)] It’s transpose is A’ = [■8(3&[email protected]&4)] B = [■8(3&[email protected]&[email protected]&3)] B’ = [■8(3&1&[email protected]&4&3)] A = [■8(3&[email protected]&4)] B = [■8(−8&[email protected]−4&0)] A = [■8(3&[email protected]&4)] A’ = [■8(3&[email protected]&4)] (A’)’ = [■8(3&[email protected]&4)]^′ = [■8(3&[email protected]&4)] = A A = [■8(3&[email protected]&4)] Let k = 4 (4A)’ 4A = 4 [■8(3&[email protected]&4)] = [■8(4×3&4×[email protected]×1&4×4)] = [■8(12&[email protected]&16)] 4A’ A = [■8(3&[email protected]&4)] A’ = [■8(3&[email protected]&4)] (4A)’ = [■8(12&[email protected]&16)]^′ = [■8(12&[email protected]&16)] 4A’ = 4[■8(3&[email protected]&4)] = [■8(4×3&4×[email protected]×2&4×4)] = [■8(12&[email protected]&16)] Let A = [■8(3&[email protected]&4)], B = [■8(−8&[email protected]−4&0)] (A + B)’ A + B = [■8(3&[email protected]&4)] + [■8(−8&[email protected]−4&0)] = [■8(3+(−8)&[email protected]+(−4)&4+0)] = [■8(−5&[email protected]−3&4)] A’ + B’ A’ = [■8(3&[email protected]&4)] B’ = [■8(−8&[email protected]−4&0)] (A + B)’ = [■8(−5&−[email protected]&4)] A’ + B’ = [■8(3&[email protected]&4)] + [■8(−8&−[email protected]&0)] = [■8(3+(−8)&1+(−4)@2+2&4+0)] = [■8(−5&−[email protected]&4)] Let A = [■8(3&[email protected]&4)], B = [■8(−8&[email protected]−4&0)] (AB)’ AB = [■8(3&[email protected]&4)] [■8(−8&[email protected]−4&0)] = [■8(3×(−8)+2×(−4)&3×2+2×[email protected]×(−8)+4×(−4)&1×2+4×8)] = [■8(−24−8&[email protected]−8−16&2+0)] = [■8(−32&[email protected]−24&2)] (AB)’ = [■8(−32&−[email protected]&2)] B’A’ A = [■8(3&[email protected]&4)] A’ = [■8(3&[email protected]&4)] B = [■8(−8&[email protected]−4&0)] B’ = [■8(−8&−[email protected]&0)] B’A’ =[■8(−8&−[email protected]&0)][■8(3&[email protected]&4)] = [■8((−8)×3+(−4)×2&(−8)×1+(−4)×[email protected]×3+0×2&2×1+0×4)] = [■8(−24−8&−8−[email protected]+0&2+0)] = [■8(−32&−[email protected]&2)]