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Matrix
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REVIEW LAST LECTURE
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Keyword
• Parametric form• Augmented Matrix• Elementary Operation• Gaussian Elimination• Row Echelon form• Reduced Row Echelon form• Leading 1’s
• Rank• Homogeneous System
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Goal of Elementary Operation
• To arrive at an easy system
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Theorem 3
• Suppose a system of m equation on n variables has a solution, if the rank of the augmented matrix is r • the set of the solution involve exactly n-r
parameters
The number of leading
1’s
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Homogeneous Equation
When b = 0What is the solution?
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MATRIX REVIEW
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Matrix Review
column matrixOr
column vector
A has 2 rows 3 columns
A is a 2 x 3 matrix
a22
a13 c21
Square matrix (number of row
equals number of column
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Matrix Review
• Scalar multiplication• kA = [kaij]
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Matrix Addition Rules
• A + B = B + A (commutative)• A + (B + C) = (A + B) + C (associative)• There is an m x n matrix 0, such that 0 + A = A for
each A (additive identity)• There is an m x n matrix, -A, such that A + (-A) =
0 for each A (additive inverse)• k(A + B) = kA + kB• (k+p)A = kA + pA• (kp)A = k(pA)
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Transpose
• Swap the index of rows and columns• A = [aij]
• AT = [aji]
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Transpose Rule
• If A is an m x n matrix, then AT is n x m matrix
• (AT)T = A• (kA)T = kAT
• (A + B)T = AT + BT
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Main Diagonal & Symmetric
• Main diagonal, the members Aii
• If A = AT, A is called a symmetric
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Example
• A = 2AT
• Solve for A
A = 2AT = 2[2AT]T = 2[a(AT)T] = 4A0 = 3AHence A = 0
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Dot Product
• Step in multiplication
• We need to compute• 3*6 + -1 * 3 + 2 * 5• The multiplication of (3 -1 2) and (6 3 5) is
called a dot product of row 1 and column 3
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Identify Matrix
• A matrix whose main diagonal are 1’s and 0’s are elsewhere
• In most case, we assume that the identity matrix is a square matrix
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Multiplication Rules
• IA = A, BI = B (identity)• A(BC) = (AB)C (associative)• A(B + C) = AB + AC; (distributive)• A(B – C) = AB – AC• (B + C)A = BA + CA; • (B – C)A = BA – CA;
• k(AB) = (kA)B = A(kB)• (AB)T = BTAT
In most caseAB != BA
(no commutative!!)
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Example
• When AB = BA? (when will they commutes?)
• (A – B)(A + B) = A2 – B2
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MATRIX AND LINEAR EQUATION
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Matrix and Linear Equation
factoring
Matrix equation
Linear equation
2 x 1 matrix
2 x 3 and 3 x 1
matrix
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Matrix Equation
A X B
AX = B
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Matrix Equation
AX = B
Coefficient matrix
Constance matrixSolution
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Associated homogeneous system
• Given a particular system AX = B• There is a related system AX = 0• Called associated homogeneous system
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Solution of a linear system
• Let • X1 be a solution to AX = B
• X0 be a solution to AX = 0
• X1 + X0 is also a solution of AX = B
• Why?• A(X1 + X0) = AX1 + AX0 = B + 0 = B
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Theorem 2
• Suppose X1 is a particular solution to the system AX = B of linear equations.
• Then every solution X2 to AX = B has the form• X2 = X1 + X0
• For some solution X0 of the associated homogeneous system AX = 0
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Proof
• Suppose that X* is any solution to AX = B• So, AX* = B• We write Xz = X* – X1
• Then AXz = A(X* + X1) = AX* + AX1 = B – B = 0
• Xz is the solution of AX = 0
• Hence, X* = Xz + X1 is the solution of AX = B
X1 is our particular
solution to AX = B
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Implication of Theorem 2
• Given a particular system AX = B• We can find all solutions by• Find a particular solution to AX = B• Reduce the problem into finding all solution to
AX = 0
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Example
• Find all solution to
• Gaussian Elimination gives parametric form• x = 4 + 2t• y = 2 + t• z = t
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Basic Solution
1 2 3 2
3 6 1 0
2 4 4 2
A
Solve the homogeneous system AX = 0
1 2 3 2 0 1 2 0 1/ 5 0
3 6 1 0 0 0 0 1 3/ 5 0
2 4 4 2 0 0 0 0 0 0
Do the elimination
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Basic Solution
1
21 2
3
4
12
2 1/ 551 0
3 0 3/ 5
5 0 1
s tx
x sX s t sX tX
xt
xt
x1 = 2s + (1/5), x2 = s, x3 = (3/5)t, x4 = t
1 2 3 2 0 1 2 0 1/ 5 0
3 6 1 0 0 0 0 1 3/ 5 0
2 4 4 2 0 0 0 0 0 0
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Basic Solution
• A basic Solution is a solution to the homogeneous system
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Linear Combination
• The solution to the previous system issX1 + tX2
• Solutions in this form are called a linear combination of X1 and X2
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Linear Combination
• Consider the previous solution2 1/ 5
1 0
0 3/ 5
0 1
X s t
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Linear Combination
• Consider the previous solution2 1/ 5 2 1
1 0 1 0/ 5
0 3/ 5 0 3
0 1 0 5
X s t s t
We can let r =t / 5… Hence, it is also another parametric form
but [1 0 3 5]T is a solution as well!!
Hence, a scalar multiple of a basic solution is a basic solution as well
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Relation to Rank
• A system AX = 0• Having n variable and m equation (A is m x n
matrix)
• Suppose the rank of A is r• Then there are n – r parameter (from theorem 3
of the last slide)
• We will have exactly n – r basic solutions• Every solution is a linear combination of
these basic solutions
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BLOCK MULTIPLICATION
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Multiplication by Block
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Block Multiplication
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Compatibility
• Block multiplication is possible when partition is compatible.• i.e., size of partitioning allows multiplication of
the block
Can we divide here?
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MATRIX INVERSE
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Solving equation
• How to solve a scalar equation• ax = b
• Multiply both side by 1/aax/a = b/ax = b/a We need multiplicative
inverse
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Matrix Inverse
• A matrix B is an inverse of a matrix A• If and only if AB = I and BA = I• B is written as A-1
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Example
• Find the inverse of
• Let
• If B is the inverse, we have AB = I Cannot
be I
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Existence of an Inverse
• From the previous example• There is a matrix having no inverse!!!• Zero matrix cannot have an inverse
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Non-Square matrix
• What should be an inverse of non-square?• Let A is m x n matrix• What should be A-1?• We can have B = n x m such that• AB = Im and BA = In
• But this gives m = n• If m < n, there exists a basic solution X (a 1 x n
matrix) for AX = 0 • So X = InX = (BA)X = B(AX) = B(0) = 0 contradict1 2 3 2
3 6 1 0
2 4 4 2
A
Non square matrix has no inverse
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Theorem 2.3.1
• If B and C are both inverse of A, then B = C
• If we have inverse, it must be unique.
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Proof
• Since B and C are inverses• CA = I = AB• Hence• B = IB = (CA)B = C(AB) = CI = C
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Inverse
• For A• A-1 is unique• A-1 is square
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First introduction to Det of 2 x 2 matrix
• Det determinant
• Det of
• is (ad – bc)
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Adjugate of 2 x 2
Adjugate of B
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Det and Inverse
• Letdet
adj
e B
So, if e != 0, we multiply it by 1/egives A(1/e)B = I =(1/e)BA
So, the inverse of a is (1/e)B
AB = eI = BA
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Determinant
• Det exists before matrix• Det is used to determine whether a linear
system has a solution
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Inverse and Linear System
• We have AX = B
• We can solve by
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Inversion Method
• A method to determine the inverse of A based on solving linear equation system
• We have A = 2 x 2 matrix• We need to find A-1
• We write the inverse as
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Inversion Method
• We have AA-1 = I• Gives
• Each are a system• A is the coefficiency matrix
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Solving A
• Find the equivalent systems in a reduced row echelon form• Gives
• This can be done by elementary operation• In fact, we do this at the same time for both
equation
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Inversion Method
• A short hand form [A I] [I A-1]
Double matrix
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Matrix Inversion Algorithm
• If A is a square matrix• There exists a sequence of elementary row
operation that carry A to the identity matrix of the same size.
• This same series carries I to A-1
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Conclusion
• Matrix• Det• Inverse