Elmer NoconAngelo Bernabe

Mark Kenneth Hitosis

3

2,

1

0,

1

1cba

3

2

1

0

1

1yx

can also be written as

3

2

11

01

y

x

Given:First we write this from the given

3

2

11

01

y

x

now we call this matrix A

using what we learned from the previous groups we’ll try to find the inverse of A

1A

first we get the determinant of A

11

01

which is (1 * 1) – (0 * -1) = 1

then divide 1 by the determinant

1

1

now we multiply it with the adjoint of this matrix, which is:

11

01

then we multiply the inverse to both sides of the equation

3

2

11

01

11

01

11

01

y

x

10

01

y

x

3

2

11

01

y

x

now we multiply

5

2

y

x

final answer:

We multiply matrix Awith the inverse of A to get an identity matrix

Multiplying x and yto an identity matrix will still give x and y

now what is this?

5

2

y

x

then we multiply this to the given

1

0,

1

1ba

5

05,

2

22 ba

first we graph the given

then graph

with the graph we draw a parallelogram we draw a diagonal line

through the parallelogram

you’ll notice that the diagonal line ends at the given

3

2c

if you got this right then then you already found x and y such that the vector (2, 3) is a linear combination of the form (2, 3) = x(1, -1) + y(0, 1)

7

5,

0

1,

2

1cba

7

5

0

1

2

1yx

can also be written as

7

5

02

11

y

x

Given:First we write this from the given

7

5

02

11

y

x

now we call this matrix B

using what we learned from the previous groups we’ll try to find the inverse of B

1B

first we get the determinant of B

02

11

which is (1 * 0) – (-1 * 2) = 2

then divide 1 by the determinant

2

1

now we multiply it with the adjoint of this matrix, which is:

12

10

then we multiply the inverse to both sides of the equation

7

5

12

10

2

1

02

11

12

10

2

1

y

x

10

01

y

x

7

5

12

10

2

1

y

x

now we multiply

2

32

7

y

x

final answer:

We multiply matrix Bwith the inverse of B to get an identity matrix

Multiplying x and yto an identity matrix will still give x and y

2

32

7

y

x

then we multiply this to the given

0

1,

2

1ba

02

3

2

3,

72

7

2

7ba

first we graph the given

then graph

with the graph we draw a parallelogram we draw a diagonal line

through the parallelogram

you’ll notice that the diagonal line ends at the given

if you got this right then then you already found x and y such that the vector (5, 7) is a linear combination of the form (5, 7) = x(1, 2) + y(-1, 0)

7

5c