# Matrix 2x2 (Math)

See how to solve 2x2 matrix.

17 examples and their solutions.

## Adding and Subtracting Matrices

### Example

A = 1234, B = 2-101

A + B = ?

Solution A + B = ?

A + B

= 1234 + 2-101

= 1 + 22 + (-1)3 + 04 + 1 - [1]

= 3135

= 1234 + 2-101

= 1 + 22 + (-1)3 + 04 + 1 - [1]

= 3135

[1]

Add the same position elements.

Close

### Example

A = 1234, B = 2-101

A - B = ?

Solution A - B = ?

A - B

= 1234 - 2-101

= 1 - 22 - (-1)3 - 04 - 1

= -1333

= 1234 - 2-101

= 1 - 22 - (-1)3 - 04 - 1

= -1333

Close

### Example

A = 1234, B = 2-101

2A - 5B = ?

Solution 2A - 5B = ?

2A - 5B

= 2 1234 - 5 2-101

= 2⋅12⋅22⋅32⋅4 - 5⋅25⋅(-1)5⋅05⋅1

= 2468 - 10-505

= 2 - 104 - (-5)6 - 08 - 5

= -8963

= 2 1234 - 5 2-101

= 2⋅12⋅22⋅32⋅4 - 5⋅25⋅(-1)5⋅05⋅1

= 2468 - 10-505

= 2 - 104 - (-5)6 - 08 - 5

= -8963

Close

## Multiplying Matrices

### Example

A = 1234, B = 2-101

AB = ?

Solution AB = ?

AB = 1234 2-101

= 1⋅2 + 2⋅01⋅(-1) + 2⋅13⋅2 + 4⋅03⋅(-1) + 4⋅1 - [1]

= 2 + 0-1 + 26 + 0-3 + 4

= 2161

= 1⋅2 + 2⋅01⋅(-1) + 2⋅13⋅2 + 4⋅03⋅(-1) + 4⋅1 - [1]

= 2 + 0-1 + 26 + 0-3 + 4

= 2161

[1]

Row 1, Column 1:

Multiply row 1 [1 2], and column 1 [2 / 0].

1⋅2 + 2⋅0

Row 1, Column 2:

Multiply row 1 [1 2] and column 2 [-1 / 1].

1⋅(-1) + 2⋅1

Row 2, Column 1:

Multiply row 2 [3 4] and column 1 [2 / 0].

3⋅2 + 4⋅0

Row 2, Column 2:

Multiply row 2 [3 4] and column 2 [-1 / 1].

3⋅(-1) + 4⋅1

Multiply row 1 [1 2], and column 1 [2 / 0].

1⋅2 + 2⋅0

Row 1, Column 2:

Multiply row 1 [1 2] and column 2 [-1 / 1].

1⋅(-1) + 2⋅1

Row 2, Column 1:

Multiply row 2 [3 4] and column 1 [2 / 0].

3⋅2 + 4⋅0

Row 2, Column 2:

Multiply row 2 [3 4] and column 2 [-1 / 1].

3⋅(-1) + 4⋅1

Close

### Example

A = 1234, B = 2-101

BA = ?

Solution BA = ?

BA = 2-101 1234

= 2⋅1 + (-1)⋅32⋅2 + (-1)⋅40⋅1 + 1⋅30⋅2 + 1⋅4

= 2 - 34 - 40 + 30 + 4

= -1034 - [1]

= 2⋅1 + (-1)⋅32⋅2 + (-1)⋅40⋅1 + 1⋅30⋅2 + 1⋅4

= 2 - 34 - 40 + 30 + 4

= -1034 - [1]

[1]

In matrix,

AB = BA is not always true.

AB = BA is not always true.

Close

## Zero Matrix

### Definition

I = [0], 0000, 000000000, ...

A zero matrix is a matrixwhose elements are all 0.

So AO = OA = O.

### Property 1

If AB = O,

then A = O or B = O. ( x )

Unlike the numbers,then A = O or B = O. ( x )

AB = O doesn't mean

either A or B is a zero matrix.

(It can be a zero matrix,

but not always a zero matrix.)

### Example

If AB = O, then A = O or B = O.

Counterexample?

Solution Counterexample?

A = 0100, B = 1000 - [1]

AB = 0100 1000

= 0⋅1 + 1⋅00⋅0 + 1⋅00⋅1 + 0⋅00⋅0 + 0⋅0

= 0 + 00 + 00 + 00 + 0

= 0000

= O - [2]

A = 0100, B = 1000

AB = 0100 1000

= 0⋅1 + 1⋅00⋅0 + 1⋅00⋅1 + 0⋅00⋅0 + 0⋅0

= 0 + 00 + 00 + 00 + 0

= 0000

= O - [2]

A = 0100, B = 1000

[1]

Think of two matrices A, B

that are not zero matrices

and that seems to make AB = O.

that are not zero matrices

and that seems to make AB = O.

[2]

A ≠ O, B ≠ O

AB = O.

→ A and B makes the given statement false.

→ A and B are the counterexample.

AB = O.

→ A and B makes the given statement false.

→ A and B are the counterexample.

Close

### Property 2

If AB = O,

then BA = O. ( x )

then BA = O. ( x )

### Example

If AB = O, then BA = O.

Counterexample?

Solution Counterexample?

A = 0010, B = 0001 - [1]

AB = 0010 0001

= 0⋅0 + 0⋅00⋅0 + 0⋅11⋅0 + 0⋅01⋅0 + 0⋅1

= 0 + 00 + 00 + 00 + 0

= 0000

= O

BA = 0001 0010

= 0⋅0 + 0⋅10⋅0 + 0⋅00⋅0 + 1⋅10⋅0 + 1⋅0

= 0 + 00 + 00 + 10 + 0

= 0010

≠ O

A = 0010, B = 0001

AB = 0010 0001

= 0⋅0 + 0⋅00⋅0 + 0⋅11⋅0 + 0⋅01⋅0 + 0⋅1

= 0 + 00 + 00 + 00 + 0

= 0000

= O

BA = 0001 0010

= 0⋅0 + 0⋅10⋅0 + 0⋅00⋅0 + 1⋅10⋅0 + 1⋅0

= 0 + 00 + 00 + 10 + 0

= 0010

≠ O

A = 0010, B = 0001

[1]

Think of two matrices A, B

that are not zero matrices

and that seems to make AB = O.

that are not zero matrices

and that seems to make AB = O.

Close

## Identity Matrix

### Definition

AI = IA = A

The identity matrix is a matrix that satisfiesAI = IA = A.

I = [1], 1001, 100010001, ...

The identity matrix is a square matrix.(number of rows = number of columns)

The diagonal elements are 1.

And the other elements are 0.

### Example

Show that the given statement is true.

(A + I)

Solution (A + I)

^{2}= A^{2}+ 2A + I (A + I)

= A

= A

= A

^{2}= (A + I)(A + I) - [1]= A

^{2}+ AI + IA + I^{2}= A

^{2}+ A + A + I= A

^{2}+ 2A + I[1]

Close

## Cayley-Hamilton Theorem (2x2)

### Theorem

A = abcd

→ A

The Cayley-Hamilton theorem can be used→ A

^{2}- (a + d)A + (ad - bc)I = Oto simplify A

^{n}.

### Example

A = 2310

Show that A

Solution Show that A

^{3}= 7A + 6I is true. A = 2310

A

A

A

A

A

= A(2A + 3I)

= 2A

= 2(2A + 3I) + 3A

= 4A + 6I + 3A

= 7A + 6I

A

A

^{2}- (2 + 0)A + (2⋅0 - 3⋅1)I = OA

^{2}- 2A + (0 - 3)I = OA

^{2}- 2A - 3I = OA

^{2}= 2A + 3IA

^{3}= AA^{2}= A(2A + 3I)

= 2A

^{2}+ 3A= 2(2A + 3I) + 3A

= 4A + 6I + 3A

= 7A + 6I

A

^{3}= 7A + 6IClose

### Example

A = -1111

A

Solution A

^{10}= ? A = -1111

A

A

A

A

A

= (2I)

= 2

= 32I

A

^{2}- (-1 + 1)A + ((-1)⋅1 - 1⋅1)I = OA

^{2}+ (-1 - 1)I = OA

^{2}- 2I = OA

^{2}= 2IA

^{10}= (A^{2})^{5}= (2I)

^{5}= 2

^{5}I^{5}- [1]= 32I

Close

## Determinant (2x2)

### Formula

abcd

→ D = ad - bc

For a 2x2 matrix,→ D = ad - bc

the determinant determines

whether the inverse matrix exists.

The determinant is written as

D, det(A), abcd.

D ≠ 0 → Inverse matrix exists.

D = 0 → Inverse matrix doesn't exist.

D = 0 → Inverse matrix doesn't exist.

### Example

A = 1234Inverse matrix of A exists?

Solution A = 1234

D = 1⋅4 - 2⋅3

= 4 - 6

= -2 ≠ 0

→ The inverse matrix exists.

D = 1⋅4 - 2⋅3

= 4 - 6

= -2 ≠ 0

→ The inverse matrix exists.

Close

## Inverse Matrix (2x2)

### Definition

AA

The inverse matrix A^{-1}= A^{-1}A = I^{-1}is a matrix

that satisfy this condition.

If you multiply A and A

^{-1},

you get the identity matrix I.

### Formula

A = abcd

→ A

(D = ad - bc)

First find the determinant D.→ A

^{-1}= 1D d-b-ca(D = ad - bc)

If D ≠ 0, find A

^{-1}:

Switch a and d.

Change the signs of b and c.

If D = O,

then the inverse matrix A

^{-1}does not exist.

### Example

A = 4131

A

Solution A

^{-1}= ? A = 4131

D = 4⋅1 - 3⋅1

= 4 - 3

= 1 ≠ 0

A

= 1-1-34

D = 4⋅1 - 3⋅1

= 4 - 3

= 1 ≠ 0

A

^{-1}= 11 1-1-34= 1-1-34

Close

### Example

A = 6834

A

Solution A

^{-1}= ? A = 6834

D = 6⋅4 - 3⋅8

= 24 - 24

= 0

→ A

D = 6⋅4 - 3⋅8

= 24 - 24

= 0

→ A

^{-1}doesn't exist.Close

## Matrix Equation (2x2)

### Formula

AX = B

→ X = A

If A→ X = A

^{-1}B^{-1}doesn't exist,

then the matrix equation has either

infinitely many solutions

or no solution.

### Example

5332 X = 8553

X = ?

Solution X = ?

5332 X = 8553

D = 5⋅2 - 3⋅3

= 10 - 9

= 1

X = 5332

= 11 2-3-35 8553 - [1]

= 2-3-35 8553

= 2⋅8 + (-3)⋅52⋅5 + (-3)⋅3-3⋅8 + 5⋅5-3⋅5 + 5⋅3

= 16 - 1510 - 9-24 + 25-15 + 15

= 1110

D = 5⋅2 - 3⋅3

= 10 - 9

= 1

X = 5332

^{-1}8553= 11 2-3-35 8553 - [1]

= 2-3-35 8553

= 2⋅8 + (-3)⋅52⋅5 + (-3)⋅3-3⋅8 + 5⋅5-3⋅5 + 5⋅3

= 16 - 1510 - 9-24 + 25-15 + 15

= 1110

Close

## System of Linear Equations: Using Matrix

### Example

x - y = 4

2x + y = 5

System of Linear Equations 2x + y = 5

Solution

x - y = 4

2x + y = 5

1-121 xy = 45 - [1]

D = 1⋅1 - (-1)⋅2

= 1 - (-2)

= 1 + 2

= 3

xy = 1-121

= 13 11-21 45

= 13 1⋅4 + 1⋅5-2⋅4 + 1⋅5

= 13 4 + 5-8 + 5

= 13 9-3

= 3-1

x = 3, y = -1

2x + y = 5

1-121 xy = 45 - [1]

D = 1⋅1 - (-1)⋅2

= 1 - (-2)

= 1 + 2

= 3

xy = 1-121

^{-1}45= 13 11-21 45

= 13 1⋅4 + 1⋅5-2⋅4 + 1⋅5

= 13 4 + 5-8 + 5

= 13 9-3

= 3-1

x = 3, y = -1

[1]

Close

### Example

x - y = 4

2x - 2y = 8

Solution 2x - 2y = 8

x - y = 4

2x - 2y = 8

1-12-2 xy = 48

D = 1⋅(-2) - (-1)⋅2

= -2 - (-2)

= -2 + 2

= 0

12 = -1-2 = 48 ( o ) - [1]

Infinitely many solutions - [2]

2x - 2y = 8

1-12-2 xy = 48

D = 1⋅(-2) - (-1)⋅2

= -2 - (-2)

= -2 + 2

= 0

12 = -1-2 = 48 ( o ) - [1]

Infinitely many solutions - [2]

[1]

D = 0

Then set a proportion

by using the elements of the matrix equation.

See if the proportion is true.

Then set a proportion

by using the elements of the matrix equation.

See if the proportion is true.

[2]

The proportion is true.

Then the given system has

Infinitely many solutions.

Then the given system has

Infinitely many solutions.

Close

### Example

x - y = 4

x - y = -3

Solution x - y = -3

x - y = 4

x - y = -3

1-11-1 xy = 4-3

D = 1⋅(-1) - (-1)⋅1

= -1 - (-1)

= -1 + 1

= 0

11 = -1-1 = 4-3 ( x )

No solution - [1]

x - y = -3

1-11-1 xy = 4-3

D = 1⋅(-1) - (-1)⋅1

= -1 - (-1)

= -1 + 1

= 0

11 = -1-1 = 4-3 ( x )

No solution - [1]

[1]

The proportion is false.

Then the given system has

no solution.

Then the given system has

no solution.

Close