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# Direct Methods for Solving Linear Systems

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Tsung-Ming Huang

Department of Mathematics National Taiwan Normal University, Taiwan

E-mail: min@math.ntnu.edu.tw

November 30, 2014

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### Outline

1 Linear systems of equations

2 Pivoting Strategies

3 Matrix factorization

4 Special types of matrices

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### Outline

1 Linear systems of equations

2 Pivoting Strategies

3 Matrix factorization

4 Special types of matrices

3 / 255

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### Outline

1 Linear systems of equations

2 Pivoting Strategies

3 Matrix factorization

4 Special types of matrices

4 / 255

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### Outline

1 Linear systems of equations

2 Pivoting Strategies

3 Matrix factorization

4 Special types of matrices

5 / 255

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### Linear systems of equations

Three operations to simplify the linear system:

1 (λEi) → (Ei): Equation Ei can be multiplied by λ 6= 0 with the resulting equation used in place of Ei.

2 (Ei+ λEj) → (Ei): Equation Ej can be multiplied by λ 6= 0 and added to equation Eiwith the resulting equation used in place of Ei.

3 (Ei) ↔ (Ej): Equation Ei and Ej can be transposed in order.

Example 1

E1 : x1 + x2 + 3x4 = 4,

E2 : 2x1 + x2 − x3 + x4 = 1, E3 : 3x1 − x2 − x3 + 2x4 = −3, E4 : −x1 + 2x2 + 3x3 − x4 = 4.

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### Linear systems of equations

Three operations to simplify the linear system:

1 (λEi) → (Ei): Equation Ei can be multiplied by λ 6= 0 with the resulting equation used in place of Ei.

2 (Ei+ λEj) → (Ei): Equation Ej can be multiplied by λ 6= 0 and added to equation Eiwith the resulting equation used in place of Ei.

3 (Ei) ↔ (Ej): Equation Ei and Ej can be transposed in order.

Example 1

E1 : x1 + x2 + 3x4 = 4,

E2 : 2x1 + x2 − x3 + x4 = 1, E3 : 3x1 − x2 − x3 + 2x4 = −3, E4 : −x1 + 2x2 + 3x3 − x4 = 4.

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### Linear systems of equations

Three operations to simplify the linear system:

1 (λEi) → (Ei): Equation Ei can be multiplied by λ 6= 0 with the resulting equation used in place of Ei.

2 (Ei+ λEj) → (Ei): Equation Ej can be multiplied by λ 6= 0 and added to equation Eiwith the resulting equation used in place of Ei.

3 (Ei) ↔ (Ej): Equation Ei and Ej can be transposed in order.

Example 1

E1 : x1 + x2 + 3x4 = 4,

E2 : 2x1 + x2 − x3 + x4 = 1, E3 : 3x1 − x2 − x3 + 2x4 = −3, E4 : −x1 + 2x2 + 3x3 − x4 = 4.

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### Linear systems of equations

Three operations to simplify the linear system:

1 (λEi) → (Ei): Equation Ei can be multiplied by λ 6= 0 with the resulting equation used in place of Ei.

2 (Ei+ λEj) → (Ei): Equation Ej can be multiplied by λ 6= 0 and added to equation Eiwith the resulting equation used in place of Ei.

3 (Ei) ↔ (Ej): Equation Ei and Ej can be transposed in order.

Example 1

E1 : x1 + x2 + 3x4 = 4,

E2 : 2x1 + x2 − x3 + x4 = 1, E3 : 3x1 − x2 − x3 + 2x4 = −3, E4 : −x1 + 2x2 + 3x3 − x4 = 4.

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### Solution:

(E2− 2E1) → (E2), (E3− 3E1) → (E3)and (E4+ E1) → (E4):

E1: x1 + x2 + 3x4 = 4,

E2: − x2 − x3 − 5x4 = −7, E3: − 4x2 − x3 − 7x4 = −15,

E4: 3x2 + 3x3 + 2x4 = 8.

(E3− 4E2) → (E3)and (E4+ 3E2) → (E4):

E1: x1 + x2 + 3x4 = 4,

E2: − x2 − x3 − 5x4 = −7,

E3: 3x3 + 13x4 = 13,

E4: − 13x4 = −13.

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### Solution:

(E2− 2E1) → (E2), (E3− 3E1) → (E3)and (E4+ E1) → (E4):

E1: x1 + x2 + 3x4 = 4,

E2: − x2 − x3 − 5x4 = −7, E3: − 4x2 − x3 − 7x4 = −15,

E4: 3x2 + 3x3 + 2x4 = 8.

(E3− 4E2) → (E3)and (E4+ 3E2) → (E4):

E1: x1 + x2 + 3x4 = 4,

E2: − x2 − x3 − 5x4 = −7,

E3: 3x3 + 13x4 = 13,

E4: − 13x4 = −13.

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Backward-substitution process:

1 E4 ⇒ x4= 1

2 Solve E3for x3: x3=1

3(13 − 13x4) =1

3(13 − 13) = 0.

3 E2gives

x2= −(−7 + 5x4+ x3) = −(−7 + 5 + 0) = 2.

4 E1gives

x1= 4 − 3x4− x2= 4 − 3 − 2 = −1.

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Backward-substitution process:

1 E4 ⇒ x4= 1

2 Solve E3for x3: x3=1

3(13 − 13x4) =1

3(13 − 13) = 0.

3 E2gives

x2= −(−7 + 5x4+ x3) = −(−7 + 5 + 0) = 2.

4 E1gives

x1= 4 − 3x4− x2= 4 − 3 − 2 = −1.

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Backward-substitution process:

1 E4 ⇒ x4= 1

2 Solve E3for x3: x3=1

3(13 − 13x4) =1

3(13 − 13) = 0.

3 E2gives

x2= −(−7 + 5x4+ x3) = −(−7 + 5 + 0) = 2.

4 E1gives

x1= 4 − 3x4− x2= 4 − 3 − 2 = −1.

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Backward-substitution process:

1 E4 ⇒ x4= 1

2 Solve E3for x3: x3=1

3(13 − 13x4) =1

3(13 − 13) = 0.

3 E2gives

x2= −(−7 + 5x4+ x3) = −(−7 + 5 + 0) = 2.

4 E1gives

x1= 4 − 3x4− x2= 4 − 3 − 2 = −1.

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Backward-substitution process:

1 E4 ⇒ x4= 1

2 Solve E3for x3: x3=1

3(13 − 13x4) =1

3(13 − 13) = 0.

3 E2gives

x2= −(−7 + 5x4+ x3) = −(−7 + 5 + 0) = 2.

4 E1gives

x1= 4 − 3x4− x2= 4 − 3 − 2 = −1.

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Solve linear systems of equations









a11x1+ a12x2+ · · · + a1nxn = b1

a21x1+ a22x2+ · · · + a2nxn = b2

...

an1x1+ an2x2+ · · · + annxn = bn

Rewrite in the matrix form

Ax = b, (1)

where

A =

a11 a12 · · · a1n a21 a22 · · · a2n

... ... . .. ... an1 an2 · · · ann

 , b =

 b1 b2

... bn

, x =

 x1 x2

... xn

 and [A, b] is called the augmented matrix.

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Solve linear systems of equations









a11x1+ a12x2+ · · · + a1nxn = b1

a21x1+ a22x2+ · · · + a2nxn = b2

...

an1x1+ an2x2+ · · · + annxn = bn

Rewrite in the matrix form

Ax = b, (1)

where

A =

a11 a12 · · · a1n a21 a22 · · · a2n

... ... . .. ... an1 an2 · · · ann

 , b =

 b1 b2

... bn

, x =

 x1 x2

... xn

 and [A, b] is called the augmented matrix.

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Solve linear systems of equations









a11x1+ a12x2+ · · · + a1nxn = b1

a21x1+ a22x2+ · · · + a2nxn = b2

...

an1x1+ an2x2+ · · · + annxn = bn

Rewrite in the matrix form

Ax = b, (1)

where

A =

a11 a12 · · · a1n a21 a22 · · · a2n

... ... . .. ... an1 an2 · · · ann

 , b =

 b1 b2

... bn

, x =

 x1 x2

... xn

 and [A, b] is called the augmented matrix.

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### Gaussian elimination with backward substitution

The augmented matrix in previous example is

1 1 0 3 4

2 1 −1 1 1

3 −1 −1 2 −3

−1 2 3 −1 4

.

(E2− 2E1) → (E2), (E3− 3E1) → (E3)and (E4+ E1) → (E4):

1 1 0 3 4

0 −1 −1 −5 −7

0 −4 −1 −7 −15

0 3 3 2 8

.

(E3− 4E2) → (E3)and (E4+ 3E2) → (E4):

1 1 0 3 4

0 −1 −1 −5 −7

0 0 3 13 13

0 0 0 −13 −13

.

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### Gaussian elimination with backward substitution

The augmented matrix in previous example is

1 1 0 3 4

2 1 −1 1 1

3 −1 −1 2 −3

−1 2 3 −1 4

.

(E2− 2E1) → (E2), (E3− 3E1) → (E3)and (E4+ E1) → (E4):

1 1 0 3 4

0 −1 −1 −5 −7

0 −4 −1 −7 −15

0 3 3 2 8

.

(E3− 4E2) → (E3)and (E4+ 3E2) → (E4):

1 1 0 3 4

0 −1 −1 −5 −7

0 0 3 13 13

0 0 0 −13 −13

.

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### Gaussian elimination with backward substitution

The augmented matrix in previous example is

1 1 0 3 4

2 1 −1 1 1

3 −1 −1 2 −3

−1 2 3 −1 4

.

(E2− 2E1) → (E2), (E3− 3E1) → (E3)and (E4+ E1) → (E4):

1 1 0 3 4

0 −1 −1 −5 −7

0 −4 −1 −7 −15

0 3 3 2 8

.

(E3− 4E2) → (E3)and (E4+ 3E2) → (E4):

1 1 0 3 4

0 −1 −1 −5 −7

0 0 3 13 13

0 0 0 −13 −13

.

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### The general Gaussian elimination procedure

Provided a116= 0, for each i = 2, 3, . . . , n,



Ei ai1 a11

E1



→ (Ei).

Transform all the entries in the first col. below the diagonal are zero. Denote the new entry in the ith row and jth col. by aij. For i = 2, 3 . . . , n − 1, providedaii 6= 0,



Ejaji

aiiEi



→ (Ej), ∀ j = i + 1, i + 2, . . . , n.

Transform all the entries in the ith column below the diagonal are zero.

Result anupper triangularmatrix:

a11 a12 · · · a1n b1

0 a22 · · · a2n b2

... . .. ... ... ... 0 · · · 0 ann bn

.

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### The general Gaussian elimination procedure

Provided a116= 0, for each i = 2, 3, . . . , n,



Ei ai1 a11

E1



→ (Ei).

Transform all the entries in the first col. below the diagonal are zero. Denote the new entry in the ith row and jth col. by aij. For i = 2, 3 . . . , n − 1, providedaii 6= 0,



Ejaji

aiiEi



→ (Ej), ∀ j = i + 1, i + 2, . . . , n.

Transform all the entries in the ith column below the diagonal are zero.

Result anupper triangularmatrix:

a11 a12 · · · a1n b1

0 a22 · · · a2n b2

... . .. ... ... ... 0 · · · 0 ann bn

.

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### The general Gaussian elimination procedure

Provided a116= 0, for each i = 2, 3, . . . , n,



Ei ai1 a11

E1



→ (Ei).

Transform all the entries in the first col. below the diagonal are zero. Denote the new entry in the ith row and jth col. by aij. For i = 2, 3 . . . , n − 1, providedaii 6= 0,



Ejaji

aiiEi



→ (Ej), ∀ j = i + 1, i + 2, . . . , n.

Transform all the entries in the ith column below the diagonal are zero.

Result anupper triangularmatrix:

a11 a12 · · · a1n b1

0 a22 · · · a2n b2

... . .. ... ... ... 0 · · · 0 ann bn

.

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### The general Gaussian elimination procedure

Provided a116= 0, for each i = 2, 3, . . . , n,



Ei ai1 a11

E1



→ (Ei).

Transform all the entries in the first col. below the diagonal are zero. Denote the new entry in the ith row and jth col. by aij. For i = 2, 3 . . . , n − 1, providedaii 6= 0,



Ejaji

aiiEi



→ (Ej), ∀ j = i + 1, i + 2, . . . , n.

Transform all the entries in the ith column below the diagonal are zero.

Result anupper triangularmatrix:

a11 a12 · · · a1n b1

0 a22 · · · a2n b2

... . .. ... ... ... 0 · · · 0 ann bn

.

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### The general Gaussian elimination procedure

Provided a116= 0, for each i = 2, 3, . . . , n,



Ei ai1 a11

E1



→ (Ei).

Transform all the entries in the first col. below the diagonal are zero. Denote the new entry in the ith row and jth col. by aij. For i = 2, 3 . . . , n − 1, providedaii 6= 0,



Ejaji

aiiEi



→ (Ej), ∀ j = i + 1, i + 2, . . . , n.

Transform all the entries in the ith column below the diagonal are zero.

Result anupper triangularmatrix:

a11 a12 · · · a1n b1

0 a22 · · · a2n b2

... . .. ... ... ... 0 · · · 0 ann bn

.

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### The general Gaussian elimination procedure

Provided a116= 0, for each i = 2, 3, . . . , n,



Ei ai1 a11

E1



→ (Ei).

Transform all the entries in the first col. below the diagonal are zero. Denote the new entry in the ith row and jth col. by aij. For i = 2, 3 . . . , n − 1, providedaii 6= 0,



Ejaji

aiiEi



→ (Ej), ∀ j = i + 1, i + 2, . . . , n.

Transform all the entries in the ith column below the diagonal are zero.

Result anupper triangularmatrix:

a11 a12 · · · a1n b1

0 a22 · · · a2n b2

... . .. ... ... ... 0 · · · 0 ann bn

.

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The process of Gaussian elimination result in a sequence of matrices as follows:

A = A(1) → A(2)→ · · · → A(n)=upper triangular matrix The matrix A(k)has the following form:

A(k)=

a(1)11 · · · a(1)1,k−1 a(1)1k · · · a(1)1j · · · a(1)1n

... . .. ... ... ... ...

0 · · · a(k−1)k−1,k−1 a(k−1)k−1,k · · · a(k−1)k−1,j · · · a(k−1)k−1,n 0 · · · 0 a(k)kk · · · a(k)kj · · · a(k)kn

... ... ... ... ...

0 · · · 0 a(k)ik · · · a(k)ij · · · a(k)in

... ... ... ... ...

0 · · · 0 a(k)nk · · · a(k)nj · · · a(k)nn

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The process of Gaussian elimination result in a sequence of matrices as follows:

A = A(1) → A(2)→ · · · → A(n)=upper triangular matrix The matrix A(k)has the following form:

A(k)=

a(1)11 · · · a(1)1,k−1 a(1)1k · · · a(1)1j · · · a(1)1n

... . .. ... ... ... ...

0 · · · a(k−1)k−1,k−1 a(k−1)k−1,k · · · a(k−1)k−1,j · · · a(k−1)k−1,n 0 · · · 0 a(k)kk · · · a(k)kj · · · a(k)kn

... ... ... ... ...

0 · · · 0 a(k)ik · · · a(k)ij · · · a(k)in

... ... ... ... ...

0 · · · 0 a(k)nk · · · a(k)nj · · · a(k)nn

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The entries of A(k)are produced by the formula

a(k)ij =





a(k−1)ij , for i = 1, . . . , k − 1, j = 1, . . . , n;

0, for i = k, . . . , n, j = 1, . . . , k − 1;

a(k−1)ija

(k−1) i,k−1

a(k−1)k−1,k−1 × a(k−1)k−1,j, for i = k, . . . , n, j = k, . . . , n.

The procedure will fail if one of the elements a(1)11, a(2)22, . . . , a(n)nn is zero.

a(i)ii is called the pivot element.

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The entries of A(k)are produced by the formula

a(k)ij =





a(k−1)ij , for i = 1, . . . , k − 1, j = 1, . . . , n;

0, for i = k, . . . , n, j = 1, . . . , k − 1;

a(k−1)ija

(k−1) i,k−1

a(k−1)k−1,k−1 × a(k−1)k−1,j, for i = k, . . . , n, j = k, . . . , n.

The procedure will fail if one of the elements a(1)11, a(2)22, . . . , a(n)nn is zero.

a(i)ii is called the pivot element.

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The entries of A(k)are produced by the formula

a(k)ij =





a(k−1)ij , for i = 1, . . . , k − 1, j = 1, . . . , n;

0, for i = k, . . . , n, j = 1, . . . , k − 1;

a(k−1)ija

(k−1) i,k−1

a(k−1)k−1,k−1 × a(k−1)k−1,j, for i = k, . . . , n, j = k, . . . , n.

The procedure will fail if one of the elements a(1)11, a(2)22, . . . , a(n)nn is zero.

a(i)ii is called the pivot element.

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### Backward substitution

The new linear system is triangular:

a11x1 + a12x2 + · · · + a1nxn = b1, a22x2 + · · · + a2nxn = b2,

...

annxn = bn

Solving the nth equation for xngives xn= bn

ann.

Solving the (n − 1)th equation for xn−1and using the value for xnyields

xn−1= bn−1− an−1,nxn

an−1,n−1

. In general,

xi= biPn

j=i+1aijxj

aii , ∀ i = n − 1, n − 2, . . . , 1.

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### Backward substitution

The new linear system is triangular:

a11x1 + a12x2 + · · · + a1nxn = b1, a22x2 + · · · + a2nxn = b2,

...

annxn = bn

Solving the nth equation for xngives xn= bn

ann.

Solving the (n − 1)th equation for xn−1and using the value for xnyields

xn−1= bn−1− an−1,nxn

an−1,n−1

. In general,

xi= biPn

j=i+1aijxj

aii , ∀ i = n − 1, n − 2, . . . , 1.

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### Backward substitution

The new linear system is triangular:

a11x1 + a12x2 + · · · + a1nxn = b1, a22x2 + · · · + a2nxn = b2,

...

annxn = bn

Solving the nth equation for xngives xn= bn

ann.

Solving the (n − 1)th equation for xn−1and using the value for xnyields

xn−1= bn−1− an−1,nxn

an−1,n−1

. In general,

xi= biPn

j=i+1aijxj

aii , ∀ i = n − 1, n − 2, . . . , 1.

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### Backward substitution

The new linear system is triangular:

a11x1 + a12x2 + · · · + a1nxn = b1, a22x2 + · · · + a2nxn = b2,

...

annxn = bn

Solving the nth equation for xngives xn= bn

ann.

Solving the (n − 1)th equation for xn−1and using the value for xnyields

xn−1= bn−1− an−1,nxn

an−1,n−1

. In general,

xi= biPn

j=i+1aijxj

aii , ∀ i = n − 1, n − 2, . . . , 1.

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Algorithm 1 (Backward Substitution)

Suppose thatU ∈ Rn×n isnonsingular upper triangularand b ∈ Rn. This algorithm computes the solution ofU x = b.

For i = n, . . . , 1 tmp = 0

For j = i + 1, . . . , n

tmp = tmp + U (i, j) ∗ x(j) End for

x(i) = (b(i) − tmp)/U (i, i) End for

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Example 2

Solve system of linear equations.

6 −2 2 4

12 −8 6 10

3 −13 9 3

−6 4 1 −18

 x1 x2

x3

x4

=

 12 34 27

−38

Solution:

1st step Use 6 as pivot element, the first row as pivot row, and multipliers 2,12, −1are produced to reduce the system to

6 −2 2 4

0 −4 2 2

0 −12 8 1

0 2 3 −14

 x1 x2

x3

x4

=

 12 10 21

−26

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Example 2

Solve system of linear equations.

6 −2 2 4

12 −8 6 10

3 −13 9 3

−6 4 1 −18

 x1 x2

x3

x4

=

 12 34 27

−38

Solution:

1st step Use 6 as pivot element, the first row as pivot row, and multipliers 2,12, −1are produced to reduce the system to

6 −2 2 4

0 −4 2 2

0 −12 8 1

0 2 3 −14

 x1 x2

x3

x4

=

 12 10 21

−26

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2nd step Use −4 as pivot element, the second row as pivot row, and multipliers 3, −12 are computed to reduce the system to

6 −2 2 4

0 −4 2 2

0 0 2 −5

0 0 4 −13

 x1

x2

x3 x4

=

 12 10

−9

−21

3rdstep Use 2 as pivot element, the third row as pivot row, and multipliers 2 is found to reduce the system to

6 −2 2 4

0 −4 2 2

0 0 2 −5

0 0 0 −3

 x1 x2

x3 x4

=

 12 10

−9

−3

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2nd step Use −4 as pivot element, the second row as pivot row, and multipliers 3, −12 are computed to reduce the system to

6 −2 2 4

0 −4 2 2

0 0 2 −5

0 0 4 −13

 x1

x2

x3 x4

=

 12 10

−9

−21

3rdstep Use 2 as pivot element, the third row as pivot row, and multipliers 2 is found to reduce the system to

6 −2 2 4

0 −4 2 2

0 0 2 −5

0 0 0 −3

 x1 x2

x3 x4

=

 12 10

−9

−3

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4thstep The backward substitution is applied:

x4 = −3

−3 = 1, x3 = −9 + 5x4

2 = −9 + 5 2 = −2, x2 = 10 − 2x4− 2x3

−4 = 10 − 2 + 4

−4 = −3, x1 = 12 − 4x4− 2x3+ 2x2

6 = 12 − 4 + 4 − 6

6 = 1.

This example is done since a(k)kk 6= 0 for all k = 1, 2, 3, 4.

How to do if a(k)kk = 0for some k?

43 / 255

(44)

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4thstep The backward substitution is applied:

x4 = −3

−3 = 1, x3 = −9 + 5x4

2 = −9 + 5 2 = −2, x2 = 10 − 2x4− 2x3

−4 = 10 − 2 + 4

−4 = −3, x1 = 12 − 4x4− 2x3+ 2x2

6 = 12 − 4 + 4 − 6

6 = 1.

This example is done since a(k)kk 6= 0 for all k = 1, 2, 3, 4.

How to do if a(k)kk = 0for some k?

44 / 255

(45)

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4thstep The backward substitution is applied:

x4 = −3

−3 = 1, x3 = −9 + 5x4

2 = −9 + 5 2 = −2, x2 = 10 − 2x4− 2x3

−4 = 10 − 2 + 4

−4 = −3, x1 = 12 − 4x4− 2x3+ 2x2

6 = 12 − 4 + 4 − 6

6 = 1.

This example is done since a(k)kk 6= 0 for all k = 1, 2, 3, 4.

How to do if a(k)kk = 0for some k?

45 / 255

(46)

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Example 3

Solve system of linear equations.

1 −1 2 −1 2 −2 3 −3

1 1 1 0

1 −1 4 3

 x1 x2

x3

x4

=

−8

−20

−2 4

Solution:

1st step Use 1 as pivot element, the first row as pivot row, and multipliers 2, 1, 1 are produced to reduce the system to

1 −1 2 −1

0 0 −1 −1

0 2 −1 1

0 0 2 4

 x1 x2

x3

x4

=

−8

−4 6 12

46 / 255

(47)

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Example 3

Solve system of linear equations.

1 −1 2 −1 2 −2 3 −3

1 1 1 0

1 −1 4 3

 x1 x2

x3

x4

=

−8

−20

−2 4

Solution:

1st step Use 1 as pivot element, the first row as pivot row, and multipliers 2, 1, 1 are produced to reduce the system to

1 −1 2 −1

0 0 −1 −1

0 2 −1 1

0 0 2 4

 x1 x2

x3

x4

=

−8

−4 6 12

47 / 255

(48)

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2nd step Since a(2)22 = 0and a(2)32 6= 0, the operation

(E2) ↔ (E3)is performed to obtain a new system

1 −1 2 −1

0 2 −1 1

0 0 −1 −1

0 0 2 4

 x1

x2 x3 x4

=

−8 6

−4 12

3rdstep Use −1 as pivot element, the third row as pivot row, and multipliers −2 is found to reduce the system to

1 −1 2 −1

0 2 −1 1

0 0 −1 −1

0 0 0 2

 x1

x2 x3 x4

=

−8 6

−4 4

48 / 255

(49)

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2nd step Since a(2)22 = 0and a(2)32 6= 0, the operation

(E2) ↔ (E3)is performed to obtain a new system

1 −1 2 −1

0 2 −1 1

0 0 −1 −1

0 0 2 4

 x1

x2 x3 x4

=

−8 6

−4 12

3rdstep Use −1 as pivot element, the third row as pivot row, and multipliers −2 is found to reduce the system to

1 −1 2 −1

0 2 −1 1

0 0 −1 −1

0 0 0 2

 x1

x2 x3 x4

=

−8 6

−4 4

49 / 255

(50)

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4thstep The backward substitution is applied:

x4 = 4 2 = 2, x3 = −4 + x4

−1 = 2, x2 = 6 − x4+ x3

2 = 3,

x1 = −8 + x4− 2x3+ x2

1 = −7.

This example illustrates what is done if a(k)kk = 0for some k.

If a(k)pk 6= 0 for some p with k + 1 ≤ p ≤ n, then the operation (Ek) ↔ (Ep)is performed to obtain new matrix.

If a(k)pk = 0for each p, then the linear system does not have a unique solution and the procedure stops.

50 / 255

(51)

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4thstep The backward substitution is applied:

x4 = 4 2 = 2, x3 = −4 + x4

−1 = 2, x2 = 6 − x4+ x3

2 = 3,

x1 = −8 + x4− 2x3+ x2

1 = −7.

This example illustrates what is done if a(k)kk = 0for some k.

If a(k)pk 6= 0 for some p with k + 1 ≤ p ≤ n, then the operation (Ek) ↔ (Ep)is performed to obtain new matrix.

If a(k)pk = 0for each p, then the linear system does not have a unique solution and the procedure stops.

51 / 255

An additional senior teacher post, to be offset by a post in the rank of Certificated Master/Mistress or Assistant Primary School Master/ Mistress as appropriate, is provided

An additional senior teacher post, to be offset by a post in the rank of CM or Assistant Primary School Master/Mistress (APSM) as appropriate, is provided to each primary

An additional senior teacher post, to be offset by a post in the rank of CM or Assistant Primary School Master/Mistress (APSM) as appropriate, is provided to each primary

An additional senior teacher post, to be offset by a post in the rank of Certificated Master/Mistress or Assistant Primary School Master/Mistress as appropriate, is provided to

An additional senior teacher post, to be offset by a post in the rank of APSM, is provided to each primary special school/special school with primary section that operates six or

An additional senior teacher post, to be offset by a post in the rank of APSM, is provided to each primary special school/special school with primary section that operates six or

An additional senior teacher post, to be offset by a post in the rank of CM or APSM as appropriate, is provided to each primary special school/special school with

An additional senior teacher post, to be offset by a post in the rank of CM or Assistant Primary School Master/Mistress (APSM) as appropriate, is provided to each primary