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Published byΞΡοΟΟΟΞ»Ξ·ΞΌΞΏΟ ΞΞ½Ξ±ΟΟΞ±ΟΞΉΞ¬Ξ΄Ξ·Ο Modified over 6 years ago
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FP1 Matrices Introduction
BAT multiply matrices
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Matrix multiplication
These numbers will give the dimensions of the answer π¨ Γ π© = πͺ πΓπ πΓπ πΓπ These numbers must be the same for the multiplication to be possible You multiply each row in the first matrix, by each column in the second matrix The product will have the same number of rows as the first matrix, and the same number of columns as the second
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WB 7 π¨= 1 β2 3 4 π©= β3 2 Find AB and BA 1 β2 3 4 β3 2 =
1 β β3 2 = Dimensions for AB (2 x 2) x (2 x 1) = (2 x 1) To multiply matrices you first multiply corresponding elements of the rows and columns, then add them up (youβll get it with practice!) Then once you have done all the columns, do the same thing, but using the second rowβ¦ After this, work out each part, and you then have the final matrix answer! 1Γβ3 +(β2Γ2) 3Γβ3 +(4Γ2) β3 +(β4) β9 +(8) π΄π΅= β7 β1 Dimensions for BA (2 x 1) x (2 x 2) = not possible
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WB 8a π¨= β1 0 2 3 π©= 4 1 0 β2 Find AB and BA Dimensions for AB
Calculating AB Dimensions for AB (2 x 2) x (2 x 2) = (2 x 2) β β2 = Multiply the first row by each column as in the previous example You always fill in the top row of the answer first A quick check β you have probably done this correctly if the highlighted (green) numbers are the same! β1Γ4 +(0Γ0) β1Γ1 +(0Γβ2) 2Γ4 +(3Γ0) 2Γ1 +(3Γβ2) β4 +(0) β1 +(0) 8 +(0) 2 +(β6) π¨π©= β4 β1 8 β4
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WB 8b π¨= β1 0 2 3 π©= 4 1 0 β2 Find AB and BA Dimensions for BA
Calculating BA Dimensions for BA (2 x 2) x (2 x 2) = (2 x 2) 4 1 0 β2 β = Multiply the first row by each column as in the previous example You always fill in the top row of the answer first A quick check β you have probably done this correctly if the highlighted (green) numbers are the same! 4Γβ1 +(1Γ2) 4Γ0 +(1Γ3) 0Γβ1 +(β2Γ2) 0Γ0 +(β2Γ3) β4 +(2) 0 +(3) 0 +(β4) 0 +(β6) π©π¨= β2 3 β4 β6 ABβ π΅π΄ Multiplication using matrices is NOT commutative This means the order of multiplication does matter as a different order gives different answers!
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AB = dimensions (1x3) x (1x2) not possible
WB 9abc π¨= 1 β1 2 π©= 3 β πͺ= 4 5 Find a) AB b) BC c) CA d) BCA AB = dimensions (1x3) x (1x2) not possible As the central numbers are not equal, these matrices cannot be combined BC = 3 β = 12+(β10) = 2 CA = β1 2 = 4Γ1 4Γβ1 4Γ2 5Γ1 5Γβ1 5Γ2 = 4 β β5 10
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WB 9d π¨= 1 β1 2 π©= 3 β2 πͺ= 4 5 Find a) AB b) BC c) CA d) BCA
π¨= 1 β1 2 π©= 3 β πͺ= 4 5 Find a) AB b) BC c) CA d) BCA Calculating BCA This can be done in one of two ways (1) π©πͺ = 2 1) (BC)A Multiply B by C, and the answer to that by A (in that order) π©πͺπ¨= β1 2 π©πͺπ¨= 2 β2 4 2) B(CA) Multiply C by A, and multiply B by the answer to that (in that order) Remember you cannot change the order, so for method 2, do not do CA and then x B after, the B must go at the front! (2) πͺπ¨ = 4 β β5 10 π©πͺπ¨ = 3 β β β5 10 π©πͺπ¨ = 2 β2 4
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Matrices - multiplying
WB 10abcd Use each row of the first matrix with each column of the second matrix. The result gives you the element in row a column b of the product matrix. π¨= 2 3 β π©= β πͺ= β β π«= 1 β Find (where possible): AB ii) BA iii) BC iv) CB v) DA vi) AD vii) CD viii) DC AB = β β2 = β18 BA = β β1 5 = β β7 BC = β2 β β2 = β1 38 β16 β5 β6 4 CB = β β β2 = not possible To multiply matrices what must be true?
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Matrices - multiplying
WB 10ef Use each row of the first matrix with each column of the second matrix. The result gives you the element in row a column b of the product matrix. π¨= 2 3 β π©= β πͺ= β β π«= 1 β Find (where possible): v) DA vi) AD vii) CD viii) DC DA = 1 β β2 = β6 10 8 AD = β β = not possible BC = β2 β β2 = β1 38 β16 β5 β6 4 To multiply matrices what must be true?
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Matrices - multiplying
WB 10gh Use each row of the first matrix with each column of the second matrix. The result gives you the element in row a column b of the product matrix. π¨= 2 3 β π©= β πͺ= β β π«= 1 β Find (where possible): v) DA vi) AD vii) CD viii) DC CD = β β β = β3 9 β3 3 DC = 1 β β β2 = β4 β β6 β2 20 β8 To multiply matrices what must be true?
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Given that BA = (0), show that AB = β2π β2 2 π 2 2π
WB π¨= β1 π π©= π 2 Given that BA = (0), show that AB = β2π β2 2 π 2 2π π©π¨= π β1 π = βπ+2π As BA = 0, the implication is that βb + 2a = 0 βπ+2π=0 2π=π π¨π©= β1 π π 2 = βπ β2 ππ 2π We know from before that 2a = b so we can replace the b termsβ¦ π¨π©= β2π β2 2 π 2 2π
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Crucial points: Summary
Make sure that you can do matrix multiplication confidently Remember that matrix multiplication is NOT commutative AB οΉ BA
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