lecture 4: lu decomposition and matrix inversethanghuynh.org/teaching/math102_lecture_4.pdfgaussian...

Lecture 4: 𝐿𝑈 Decomposition and MatrixInverse

Thang Huynh, UC San Diego1/17/2018

Gaussian elimination revisited

▶ Example. Keeping track of the elementary matrices duringGaussian elimination on 𝐴:

𝐴 = ⎡⎢⎣2 14 −6

⎤⎥⎦

𝐸𝐴 = ⎡⎢⎣

1 0−2 1

⎤⎥⎦

⎡⎢⎣2 14 −6

⎤⎥⎦

= ⎡⎢⎣2 10 −8

⎤⎥⎦

.

Note that𝐴 = 𝐸−1 ⎡⎢

⎣2 10 −8

⎤⎥⎦

= ⎡⎢⎣1 02 1

⎤⎥⎦

⎡⎢⎣2 10 −8

⎤⎥⎦

We factored 𝐴 as the product of a lower and upper triangularmatrix! We say that 𝐴 has triangular factorization.

1

Gaussian elimination revisited

▶ Example. Keeping track of the elementary matrices duringGaussian elimination on 𝐴:

𝐴 = ⎡⎢⎣2 14 −6

⎤⎥⎦

𝐸𝐴 = ⎡⎢⎣

1 0−2 1

⎤⎥⎦

⎡⎢⎣2 14 −6

⎤⎥⎦

= ⎡⎢⎣2 10 −8

⎤⎥⎦

.

Note that𝐴 = 𝐸−1 ⎡⎢

⎣2 10 −8

⎤⎥⎦

= ⎡⎢⎣1 02 1

⎤⎥⎦

⎡⎢⎣2 10 −8

⎤⎥⎦

We factored 𝐴 as the product of a lower and upper triangularmatrix! We say that 𝐴 has triangular factorization.

1

Gaussian elimination revisited

▶ Example. Keeping track of the elementary matrices duringGaussian elimination on 𝐴:

𝐴 = ⎡⎢⎣2 14 −6

⎤⎥⎦

𝐸𝐴 = ⎡⎢⎣

1 0−2 1

⎤⎥⎦

⎡⎢⎣2 14 −6

⎤⎥⎦

= ⎡⎢⎣2 10 −8

⎤⎥⎦

.

Note that𝐴 = 𝐸−1 ⎡⎢

⎣2 10 −8

⎤⎥⎦

= ⎡⎢⎣1 02 1

⎤⎥⎦

⎡⎢⎣2 10 −8

⎤⎥⎦

We factored 𝐴 as the product of a lower and upper triangularmatrix! We say that 𝐴 has triangular factorization.

1

Gaussian elimination revisited

▶ Example. Keeping track of the elementary matrices duringGaussian elimination on 𝐴:

𝐴 = ⎡⎢⎣2 14 −6

⎤⎥⎦

𝐸𝐴 = ⎡⎢⎣

1 0−2 1

⎤⎥⎦

⎡⎢⎣2 14 −6

⎤⎥⎦

= ⎡⎢⎣2 10 −8

⎤⎥⎦

.

Note that𝐴 = 𝐸−1 ⎡⎢

⎣2 10 −8

⎤⎥⎦

= ⎡⎢⎣1 02 1

⎤⎥⎦

⎡⎢⎣2 10 −8

⎤⎥⎦

We factored 𝐴 as the product of a lower and upper triangularmatrix! We say that 𝐴 has triangular factorization.

1

Gaussian elimination revisited

𝐴 = 𝐿𝑈 is known as the LU decomposition of 𝐴.

▶ Definition.lower triangular

⎡⎢⎢⎢⎢⎢⎢⎣

∗ 0 0 0 0∗ ∗ 0 0 0∗ ∗ ∗ 0 0∗ ∗ ∗ ∗ 0∗ ∗ ∗ ∗ ∗

⎤⎥⎥⎥⎥⎥⎥⎦

upper triangular

⎡⎢⎢⎢⎢⎢⎢⎣

∗ ∗ ∗ ∗ ∗0 ∗ ∗ ∗ ∗0 0 ∗ ∗ ∗0 0 0 ∗ ∗0 0 0 0 ∗

⎤⎥⎥⎥⎥⎥⎥⎦

2

Gaussian elimination revisited

𝐴 = 𝐿𝑈 is known as the LU decomposition of 𝐴.

▶ Definition.lower triangular

⎡⎢⎢⎢⎢⎢⎢⎣

∗ 0 0 0 0∗ ∗ 0 0 0∗ ∗ ∗ 0 0∗ ∗ ∗ ∗ 0∗ ∗ ∗ ∗ ∗

⎤⎥⎥⎥⎥⎥⎥⎦

upper triangular

⎡⎢⎢⎢⎢⎢⎢⎣

∗ ∗ ∗ ∗ ∗0 ∗ ∗ ∗ ∗0 0 ∗ ∗ ∗0 0 0 ∗ ∗0 0 0 0 ∗

⎤⎥⎥⎥⎥⎥⎥⎦

2

𝐿𝑈 decompostion

▶ Example. Factor 𝐴 =⎡⎢⎢⎣

2 1 14 −6 0

−2 7 2

⎤⎥⎥⎦

as 𝐴 = 𝐿𝑈.

▶ Solution.

𝐸1𝐴 =⎡⎢⎢⎣

1 0 0−2 1 00 0 1

⎤⎥⎥⎦

⎡⎢⎢⎣

2 1 14 −6 0

−2 7 2

⎤⎥⎥⎦

=⎡⎢⎢⎣

2 1 10 −8 −2

−2 7 2

⎤⎥⎥⎦

𝐸2(𝐸1𝐴) =⎡⎢⎢⎣

1 0 00 1 01 0 1

⎤⎥⎥⎦

⎡⎢⎢⎣

2 1 10 −8 −2

−2 7 2

⎤⎥⎥⎦

=⎡⎢⎢⎣

2 1 10 −8 −20 8 3

⎤⎥⎥⎦

𝐸3𝐸2𝐸1𝐴 =⎡⎢⎢⎣

1 0 00 1 00 1 1

⎤⎥⎥⎦

⎡⎢⎢⎣

2 1 10 −8 −20 8 3

⎤⎥⎥⎦

=⎡⎢⎢⎣

2 1 10 −8 −20 0 1

⎤⎥⎥⎦

= 𝑈

3

𝐿𝑈 decompostion

▶ Example. Factor 𝐴 =⎡⎢⎢⎣

2 1 14 −6 0

−2 7 2

⎤⎥⎥⎦

as 𝐴 = 𝐿𝑈.

▶ Solution.

𝐸1𝐴 =⎡⎢⎢⎣

1 0 0−2 1 00 0 1

⎤⎥⎥⎦

⎡⎢⎢⎣

2 1 14 −6 0

−2 7 2

⎤⎥⎥⎦

=⎡⎢⎢⎣

2 1 10 −8 −2

−2 7 2

⎤⎥⎥⎦

𝐸2(𝐸1𝐴) =⎡⎢⎢⎣

1 0 00 1 01 0 1

⎤⎥⎥⎦

⎡⎢⎢⎣

2 1 10 −8 −2

−2 7 2

⎤⎥⎥⎦

=⎡⎢⎢⎣

2 1 10 −8 −20 8 3

⎤⎥⎥⎦

𝐸3𝐸2𝐸1𝐴 =⎡⎢⎢⎣

1 0 00 1 00 1 1

⎤⎥⎥⎦

⎡⎢⎢⎣

2 1 10 −8 −20 8 3

⎤⎥⎥⎦

=⎡⎢⎢⎣

2 1 10 −8 −20 0 1

⎤⎥⎥⎦

= 𝑈

3

𝐿𝑈 decompostion

▶ Example. Factor 𝐴 =⎡⎢⎢⎣

2 1 14 −6 0

−2 7 2

⎤⎥⎥⎦

as 𝐴 = 𝐿𝑈.

▶ Solution.

𝐸1𝐴 =⎡⎢⎢⎣

1 0 0−2 1 00 0 1

⎤⎥⎥⎦

⎡⎢⎢⎣

2 1 14 −6 0

−2 7 2

⎤⎥⎥⎦

=⎡⎢⎢⎣

2 1 10 −8 −2

−2 7 2

⎤⎥⎥⎦

𝐸2(𝐸1𝐴) =⎡⎢⎢⎣

1 0 00 1 01 0 1

⎤⎥⎥⎦

⎡⎢⎢⎣

2 1 10 −8 −2

−2 7 2

⎤⎥⎥⎦

=⎡⎢⎢⎣

2 1 10 −8 −20 8 3

⎤⎥⎥⎦

𝐸3𝐸2𝐸1𝐴 =⎡⎢⎢⎣

1 0 00 1 00 1 1

⎤⎥⎥⎦

⎡⎢⎢⎣

2 1 10 −8 −20 8 3

⎤⎥⎥⎦

=⎡⎢⎢⎣

2 1 10 −8 −20 0 1

⎤⎥⎥⎦

= 𝑈

3

𝐿𝑈 decompostion

▶ Example. Factor 𝐴 =⎡⎢⎢⎣

2 1 14 −6 0

−2 7 2

⎤⎥⎥⎦

as 𝐴 = 𝐿𝑈.

▶ Solution.

𝐸1𝐴 =⎡⎢⎢⎣

1 0 0−2 1 00 0 1

⎤⎥⎥⎦

⎡⎢⎢⎣

2 1 14 −6 0

−2 7 2

⎤⎥⎥⎦

=⎡⎢⎢⎣

2 1 10 −8 −2

−2 7 2

⎤⎥⎥⎦

𝐸2(𝐸1𝐴) =⎡⎢⎢⎣

1 0 00 1 01 0 1

⎤⎥⎥⎦

⎡⎢⎢⎣

2 1 10 −8 −2

−2 7 2

⎤⎥⎥⎦

=⎡⎢⎢⎣

2 1 10 −8 −20 8 3

⎤⎥⎥⎦

𝐸3𝐸2𝐸1𝐴 =⎡⎢⎢⎣

1 0 00 1 00 1 1

⎤⎥⎥⎦

⎡⎢⎢⎣

2 1 10 −8 −20 8 3

⎤⎥⎥⎦

=⎡⎢⎢⎣

2 1 10 −8 −20 0 1

⎤⎥⎥⎦

= 𝑈

3

𝐿𝑈 decompostion

𝐸3𝐸2𝐸1𝐴 = 𝑈

⟹ 𝐴 = 𝐸−11 𝐸−1

2 𝐸−13 𝑈

The factor 𝐿 is given by

𝐿 = 𝐸−11 𝐸−1

2 𝐸−13

=⎡⎢⎢⎣

1 0 02 1 00 0 1

⎤⎥⎥⎦

⎡⎢⎢⎣

1 0 00 1 0

−1 0 1

⎤⎥⎥⎦

⎡⎢⎢⎣

1 0 00 1 00 −1 1

⎤⎥⎥⎦

=⎡⎢⎢⎣

1 0 02 1 0

−1 −1 1

⎤⎥⎥⎦

4

𝐿𝑈 decompostion

𝐸3𝐸2𝐸1𝐴 = 𝑈 ⟹ 𝐴 = 𝐸−11 𝐸−1

2 𝐸−13 𝑈

The factor 𝐿 is given by

𝐿 = 𝐸−11 𝐸−1

2 𝐸−13

=⎡⎢⎢⎣

1 0 02 1 00 0 1

⎤⎥⎥⎦

⎡⎢⎢⎣

1 0 00 1 0

−1 0 1

⎤⎥⎥⎦

⎡⎢⎢⎣

1 0 00 1 00 −1 1

⎤⎥⎥⎦

=⎡⎢⎢⎣

1 0 02 1 0

−1 −1 1

⎤⎥⎥⎦

4

𝐿𝑈 decompostion

𝐸3𝐸2𝐸1𝐴 = 𝑈 ⟹ 𝐴 = 𝐸−11 𝐸−1

2 𝐸−13 𝑈

The factor 𝐿 is given by

𝐿 = 𝐸−11 𝐸−1

2 𝐸−13

=⎡⎢⎢⎣

1 0 02 1 00 0 1

⎤⎥⎥⎦

⎡⎢⎢⎣

1 0 00 1 0

−1 0 1

⎤⎥⎥⎦

⎡⎢⎢⎣

1 0 00 1 00 −1 1

⎤⎥⎥⎦

=⎡⎢⎢⎣

1 0 02 1 0

−1 −1 1

⎤⎥⎥⎦

4

𝐿𝑈 decompostion

We found the following 𝐿𝑈 decomposition of 𝐴:

𝐴 =⎡⎢⎢⎣

2 1 14 −6 0

−2 7 2

⎤⎥⎥⎦

= 𝐿𝑈 =⎡⎢⎢⎣

1 0 02 1 0

−1 −1 1

⎤⎥⎥⎦

⎡⎢⎢⎣

2 1 10 −8 −20 0 1

⎤⎥⎥⎦

.

5

Why 𝐿𝑈 decomposition?Once we have 𝐴 = 𝐿𝑈, it is simple to solve 𝐴𝑥𝑥𝑥 = 𝑏𝑏𝑏.

𝐴𝑥𝑥𝑥 = 𝑏𝑏𝑏𝐿(𝑈𝑥𝑥𝑥) = 𝑏𝑏𝑏

𝐿𝑐𝑐𝑐 = 𝑏𝑏𝑏 and 𝑈𝑥𝑥𝑥 = 𝑐𝑐𝑐.Both of the final systems are triangular and hence easily solved:

• 𝐿𝑐𝑐𝑐 = 𝑏𝑏𝑏 by forward substitution to find 𝑐𝑐𝑐, and then• 𝑈𝑥𝑥𝑥 = 𝑐𝑐𝑐 by backward substitution to find 𝑥𝑥𝑥.

▶ Example. Solve

⎡⎢⎢⎣

2 1 14 −6 0

−2 7 2

⎤⎥⎥⎦

𝑥𝑥𝑥 =⎡⎢⎢⎣

410−3

⎤⎥⎥⎦

.

6

Why 𝐿𝑈 decomposition?Once we have 𝐴 = 𝐿𝑈, it is simple to solve 𝐴𝑥𝑥𝑥 = 𝑏𝑏𝑏.

𝐴𝑥𝑥𝑥 = 𝑏𝑏𝑏𝐿(𝑈𝑥𝑥𝑥) = 𝑏𝑏𝑏

𝐿𝑐𝑐𝑐 = 𝑏𝑏𝑏 and 𝑈𝑥𝑥𝑥 = 𝑐𝑐𝑐.Both of the final systems are triangular and hence easily solved:

• 𝐿𝑐𝑐𝑐 = 𝑏𝑏𝑏 by forward substitution to find 𝑐𝑐𝑐, and then• 𝑈𝑥𝑥𝑥 = 𝑐𝑐𝑐 by backward substitution to find 𝑥𝑥𝑥.

▶ Example. Solve

⎡⎢⎢⎣

2 1 14 −6 0

−2 7 2

⎤⎥⎥⎦

𝑥𝑥𝑥 =⎡⎢⎢⎣

410−3

⎤⎥⎥⎦

.

6

Why 𝐿𝑈 decomposition?Once we have 𝐴 = 𝐿𝑈, it is simple to solve 𝐴𝑥𝑥𝑥 = 𝑏𝑏𝑏.

𝐴𝑥𝑥𝑥 = 𝑏𝑏𝑏𝐿(𝑈𝑥𝑥𝑥) = 𝑏𝑏𝑏

𝐿𝑐𝑐𝑐 = 𝑏𝑏𝑏 and 𝑈𝑥𝑥𝑥 = 𝑐𝑐𝑐.Both of the final systems are triangular and hence easily solved:

• 𝐿𝑐𝑐𝑐 = 𝑏𝑏𝑏 by forward substitution to find 𝑐𝑐𝑐, and then• 𝑈𝑥𝑥𝑥 = 𝑐𝑐𝑐 by backward substitution to find 𝑥𝑥𝑥.

▶ Example. Solve

⎡⎢⎢⎣

2 1 14 −6 0

−2 7 2

⎤⎥⎥⎦

𝑥𝑥𝑥 =⎡⎢⎢⎣

410−3

⎤⎥⎥⎦

.

6

The inverse of a matrix

▶ Definition. An 𝑛 × 𝑛 matrix 𝐴 is invertible if there is a matrix 𝐵such that

𝐴𝐵 = 𝐵𝐴 = 𝐼𝑛×𝑛.

In that case, 𝐵 is the inverse of 𝐴 and is denoted by 𝐴−1.

▶ Remark.

• The inverse of a matrix is unique. (Why?)• Do not write 𝐴

𝐵 .

7

The inverse of a matrix

▶ Definition. An 𝑛 × 𝑛 matrix 𝐴 is invertible if there is a matrix 𝐵such that

𝐴𝐵 = 𝐵𝐴 = 𝐼𝑛×𝑛.

In that case, 𝐵 is the inverse of 𝐴 and is denoted by 𝐴−1.▶ Remark.

• The inverse of a matrix is unique. (Why?)

• Do not write 𝐴𝐵 .

7

The inverse of a matrix

▶ Definition. An 𝑛 × 𝑛 matrix 𝐴 is invertible if there is a matrix 𝐵such that

𝐴𝐵 = 𝐵𝐴 = 𝐼𝑛×𝑛.

In that case, 𝐵 is the inverse of 𝐴 and is denoted by 𝐴−1.▶ Remark.

• The inverse of a matrix is unique. (Why?)• Do not write 𝐴

𝐵 .

7

The inverse of a matrix

▶ Example. Let 𝐴 = ⎡⎢⎣𝑎 𝑏𝑐 𝑑

⎤⎥⎦

. If 𝑎𝑑 − 𝑏𝑐 ≠ 0, then

𝐴−1 = 1𝑎𝑑 − 𝑏𝑐

⎡⎢⎣

𝑑 −𝑏−𝑐 𝑎

⎤⎥⎦

.

▶ Example. The matrix 𝐴 = ⎡⎢⎣0 10 0

⎤⎥⎦

is not invertible.

▶ Example. A 2 × 2 matrix ⎡⎢⎣𝑎 𝑏𝑐 𝑑

⎤⎥⎦

is invertible if and only if

𝑎𝑑 − 𝑏𝑐 ≠ 0.

8

The inverse of a matrix

▶ Example. Let 𝐴 = ⎡⎢⎣𝑎 𝑏𝑐 𝑑

⎤⎥⎦

. If 𝑎𝑑 − 𝑏𝑐 ≠ 0, then

𝐴−1 = 1𝑎𝑑 − 𝑏𝑐

⎡⎢⎣

𝑑 −𝑏−𝑐 𝑎

⎤⎥⎦

.

▶ Example. The matrix 𝐴 = ⎡⎢⎣0 10 0

⎤⎥⎦

is not invertible.

▶ Example. A 2 × 2 matrix ⎡⎢⎣𝑎 𝑏𝑐 𝑑

⎤⎥⎦

is invertible if and only if

𝑎𝑑 − 𝑏𝑐 ≠ 0.

8

The inverse of a matrix

▶ Example. Let 𝐴 = ⎡⎢⎣𝑎 𝑏𝑐 𝑑

⎤⎥⎦

. If 𝑎𝑑 − 𝑏𝑐 ≠ 0, then

𝐴−1 = 1𝑎𝑑 − 𝑏𝑐

⎡⎢⎣

𝑑 −𝑏−𝑐 𝑎

⎤⎥⎦

.

▶ Example. The matrix 𝐴 = ⎡⎢⎣0 10 0

⎤⎥⎦

is not invertible.

▶ Example. A 2 × 2 matrix ⎡⎢⎣𝑎 𝑏𝑐 𝑑

⎤⎥⎦

is invertible if and only if

𝑎𝑑 − 𝑏𝑐 ≠ 0.

8

The inverse of a matrix

Suppose 𝐴 and 𝐵 are invertible. Then

• 𝐴−1 is invertible and (𝐴−1)−1 = 𝐴.

• 𝐴𝑇 is invertible and (𝐴𝑇 )−1 = (𝐴−1)𝑇 .• 𝐴𝐵 is invertible and (𝐴𝐵)−1 = 𝐵−1𝐴−1. (Why?)

9

The inverse of a matrix

Suppose 𝐴 and 𝐵 are invertible. Then

• 𝐴−1 is invertible and (𝐴−1)−1 = 𝐴.• 𝐴𝑇 is invertible and (𝐴𝑇 )−1 = (𝐴−1)𝑇 .

• 𝐴𝐵 is invertible and (𝐴𝐵)−1 = 𝐵−1𝐴−1. (Why?)

9

The inverse of a matrix

Suppose 𝐴 and 𝐵 are invertible. Then

• 𝐴−1 is invertible and (𝐴−1)−1 = 𝐴.• 𝐴𝑇 is invertible and (𝐴𝑇 )−1 = (𝐴−1)𝑇 .• 𝐴𝐵 is invertible and (𝐴𝐵)−1 = 𝐵−1𝐴−1. (Why?)

9

Solving systems using matrix inverse

Theorem. Let 𝐴 be invertible. Then the system 𝐴𝑥𝑥𝑥 = 𝑏𝑏𝑏 has theunique solution 𝑥𝑥𝑥 = 𝐴−1𝑏𝑏𝑏.

10

Computing the inverse

▶ To solve 𝐴𝑥𝑥𝑥 = 𝑏𝑏𝑏, we do row reduction on [ 𝐴 ∣ 𝑏 ].

▶ To solve 𝐴𝑋 = 𝐼, we do row reduction on [ 𝐴 ∣ 𝐼 ].▶ To compute 𝐴−1 (The Gauss-Jordan Method):

• Form the augmented matrix [ 𝐴 ∣ 𝐼 ].• Compute the reduced echelon form.• If 𝐴 is invertible, the result is of the form [ 𝐼 ∣ 𝐴−1 ].

11

Computing the inverse

▶ To solve 𝐴𝑥𝑥𝑥 = 𝑏𝑏𝑏, we do row reduction on [ 𝐴 ∣ 𝑏 ].▶ To solve 𝐴𝑋 = 𝐼, we do row reduction on [ 𝐴 ∣ 𝐼 ].

▶ To compute 𝐴−1 (The Gauss-Jordan Method):

• Form the augmented matrix [ 𝐴 ∣ 𝐼 ].• Compute the reduced echelon form.• If 𝐴 is invertible, the result is of the form [ 𝐼 ∣ 𝐴−1 ].

11

Computing the inverse

▶ To solve 𝐴𝑥𝑥𝑥 = 𝑏𝑏𝑏, we do row reduction on [ 𝐴 ∣ 𝑏 ].▶ To solve 𝐴𝑋 = 𝐼, we do row reduction on [ 𝐴 ∣ 𝐼 ].▶ To compute 𝐴−1 (The Gauss-Jordan Method):

• Form the augmented matrix [ 𝐴 ∣ 𝐼 ].• Compute the reduced echelon form.• If 𝐴 is invertible, the result is of the form [ 𝐼 ∣ 𝐴−1 ].

11

Computing the inverse

▶ Example. Find the inverse of 𝐴 =⎡⎢⎢⎣

2 1 14 −6 0

−2 7 2

⎤⎥⎥⎦, if it exists.

▶ Solution.

⎡⎢⎢⎣

2 1 1 1 0 04 −6 0 0 1 0

−2 7 2 0 0 1

⎤⎥⎥⎦

𝑅2⟶𝑅2−2𝑅1−−−−−−−−−−→𝑅3⟶𝑅3+𝑅1

⎡⎢⎢⎣

2 1 1 1 0 00 −8 −2 −2 1 00 8 3 1 0 1

⎤⎥⎥⎦

𝑅3⟶𝑅3+𝑅2−−−−−−−−−→⎡⎢⎢⎣

2 1 1 1 0 00 −8 −2 −2 1 00 0 1 −1 1 1

⎤⎥⎥⎦

−−−−−−→ …

12

Computing the inverse

▶ Example. Find the inverse of 𝐴 =⎡⎢⎢⎣

2 1 14 −6 0

−2 7 2

⎤⎥⎥⎦, if it exists.

▶ Solution.

⎡⎢⎢⎣

2 1 1 1 0 04 −6 0 0 1 0

−2 7 2 0 0 1

⎤⎥⎥⎦

𝑅2⟶𝑅2−2𝑅1−−−−−−−−−−→𝑅3⟶𝑅3+𝑅1

⎡⎢⎢⎣

2 1 1 1 0 00 −8 −2 −2 1 00 8 3 1 0 1

⎤⎥⎥⎦

𝑅3⟶𝑅3+𝑅2−−−−−−−−−→⎡⎢⎢⎣

2 1 1 1 0 00 −8 −2 −2 1 00 0 1 −1 1 1

⎤⎥⎥⎦

−−−−−−→ …

12

Computing the inverse

▶ Example. Find the inverse of 𝐴 =⎡⎢⎢⎣

2 1 14 −6 0

−2 7 2

⎤⎥⎥⎦, if it exists.

▶ Solution.

⎡⎢⎢⎣

2 1 1 1 0 04 −6 0 0 1 0

−2 7 2 0 0 1

⎤⎥⎥⎦

𝑅2⟶𝑅2−2𝑅1−−−−−−−−−−→𝑅3⟶𝑅3+𝑅1

⎡⎢⎢⎣

2 1 1 1 0 00 −8 −2 −2 1 00 8 3 1 0 1

⎤⎥⎥⎦

𝑅3⟶𝑅3+𝑅2−−−−−−−−−→⎡⎢⎢⎣

2 1 1 1 0 00 −8 −2 −2 1 00 0 1 −1 1 1

⎤⎥⎥⎦

−−−−−−→ …

12

Computing the inverse

▶ Example. Find the inverse of 𝐴 =⎡⎢⎢⎣

2 1 14 −6 0

−2 7 2

⎤⎥⎥⎦, if it exists.

▶ Solution.

⎡⎢⎢⎣

2 1 1 1 0 04 −6 0 0 1 0

−2 7 2 0 0 1

⎤⎥⎥⎦

𝑅2⟶𝑅2−2𝑅1−−−−−−−−−−→𝑅3⟶𝑅3+𝑅1

⎡⎢⎢⎣

2 1 1 1 0 00 −8 −2 −2 1 00 8 3 1 0 1

⎤⎥⎥⎦

𝑅3⟶𝑅3+𝑅2−−−−−−−−−→⎡⎢⎢⎣

2 1 1 1 0 00 −8 −2 −2 1 00 0 1 −1 1 1

⎤⎥⎥⎦

−−−−−−→ …

12

Computing the inverse

▶ Example. Find the inverse of 𝐴 =⎡⎢⎢⎣

2 1 14 −6 0

−2 7 2

⎤⎥⎥⎦, if it exists.

▶ Solution.

⎡⎢⎢⎣

2 1 1 1 0 04 −6 0 0 1 0

−2 7 2 0 0 1

⎤⎥⎥⎦

𝑅2⟶𝑅2−2𝑅1−−−−−−−−−−→𝑅3⟶𝑅3+𝑅1

⎡⎢⎢⎣

2 1 1 1 0 00 −8 −2 −2 1 00 8 3 1 0 1

⎤⎥⎥⎦

𝑅3⟶𝑅3+𝑅2−−−−−−−−−→⎡⎢⎢⎣

2 1 1 1 0 00 −8 −2 −2 1 00 0 1 −1 1 1

⎤⎥⎥⎦

−−−−−−→ …

12

Computing the inverse

−−−−−−→⎡⎢⎢⎣

1 0 0 1216

−516

−616

0 1 0 48

−38

−28

0 0 1 −1 1 1

⎤⎥⎥⎦

𝐴−1 =⎡⎢⎢⎣

1216

−516

−616

48

−38

−28

−1 1 1

⎤⎥⎥⎦

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Why does it work?

• Each row reduction corresponds to multiplying with anelementary matrix 𝐸:

[ 𝐴∣ 𝐼 ] → [ 𝐸1𝐴∣ 𝐸1𝐼 ] → [ 𝐸2𝐸1𝐴∣ 𝐸2𝐸1 ] → …

… → [ 𝐹𝐴 ∣ 𝐹] where 𝐹 = 𝐸𝑟 … 𝐸2𝐸1.

• If we manage to reduce [ 𝐴∣ 𝐼 ] to [ 𝐼∣ 𝐹 ], this means

𝐹𝐴 = 𝐼 ⟹ 𝐴−1 = 𝐹.

14

Why does it work?

• Each row reduction corresponds to multiplying with anelementary matrix 𝐸:

[ 𝐴∣ 𝐼 ] → [ 𝐸1𝐴∣ 𝐸1𝐼 ] → [ 𝐸2𝐸1𝐴∣ 𝐸2𝐸1 ] → …

… → [ 𝐹𝐴 ∣ 𝐹] where 𝐹 = 𝐸𝑟 … 𝐸2𝐸1.

• If we manage to reduce [ 𝐴∣ 𝐼 ] to [ 𝐼∣ 𝐹 ], this means

𝐹𝐴 = 𝐼 ⟹ 𝐴−1 = 𝐹.

14

Why does it work?

• Each row reduction corresponds to multiplying with anelementary matrix 𝐸:

[ 𝐴∣ 𝐼 ] → [ 𝐸1𝐴∣ 𝐸1𝐼 ] → [ 𝐸2𝐸1𝐴∣ 𝐸2𝐸1 ] → …

… → [ 𝐹𝐴 ∣ 𝐹] where 𝐹 = 𝐸𝑟 … 𝐸2𝐸1.

• If we manage to reduce [ 𝐴∣ 𝐼 ] to [ 𝐼∣ 𝐹 ], this means

𝐹𝐴 = 𝐼 ⟹ 𝐴−1 = 𝐹.

14