CALCU- Solve the given system of linear equations

Solve the given system of linear equations over Z7 and Z11, respectively. 3𝑥 + 2𝑦 = 1 𝑥 + 4𝑦 = 1

Solution by Steps

step 1

To solve the system of equations over

\mathbb{Z}_7

, we first write down the equations:

3x + 2y = 1

and

x + 4y = 1

step 2

We multiply the second equation by 3 to make the coefficient of

x

in both equations the same:

3x + 12y = 3

. Since we are in

\mathbb{Z}_7

,

12y

is equivalent to

5y

because

12 \equiv 5 \mod 7

. So the equation becomes

3x + 5y = 3

step 3

Subtract the first equation from the modified second equation to eliminate

x

:

(3x + 5y) - (3x + 2y) = 3 - 1

. This simplifies to

3y = 2

or

y = 2 \cdot 3^{-1}

. In

\mathbb{Z}_7

,

3^{-1} = 5

because

3 \cdot 5 = 15 \equiv 1 \mod 7

. Thus,

y = 2 \cdot 5 = 10 \equiv 3 \mod 7

step 4

Substitute

y = 3

into the first equation:

3x + 2 \cdot 3 = 1

. This simplifies to

3x + 6 = 1

, and since

6 \equiv -1 \mod 7

, we have

3x - 1 = 1

. Adding 1 to both sides gives

3x = 2

, and multiplying by

3^{-1} = 5

gives

x = 2 \cdot 5 = 10 \equiv 3 \mod 7

Answer

Over

\mathbb{Z}_7

, the solution is

x = 3

and

y = 3

.

Key Concept

Solving systems of linear equations in modular arithmetic

Explanation

We use modular arithmetic properties to solve the system of equations, including finding multiplicative inverses and simplifying coefficients within the given modulus.

---

Solution by Steps

step 1

To solve the system of equations over

\mathbb{Z}_{11}

, we start with the equations:

3x + 2y = 1

and

x + 4y = 1

step 2

Multiply the second equation by 3 to align the

x

coefficients:

3x + 12y = 3

. In

\mathbb{Z}_{11}

,

12y

is equivalent to

1y

because

12 \equiv 1 \mod 11

. The equation becomes

3x + y = 3

step 3

Subtract the first equation from the modified second equation to eliminate

x

:

(3x + y) - (3x + 2y) = 3 - 1

. This simplifies to

-y = 2

, and since

-1 \equiv 10 \mod 11

, we have

10y = 2

. Multiplying by

10^{-1} = 10

gives

y = 2 \cdot 10 = 20 \equiv 9 \mod 11

step 4

Substitute

y = 9

into the first equation:

3x + 2 \cdot 9 = 1

. This simplifies to

3x + 18 = 1

, and since

18 \equiv 7 \mod 11

, we have

3x + 7 = 1

. Subtracting 7 from both sides gives

3x = -6

, and since

-6 \equiv 5 \mod 11

, we have

3x = 5

. Multiplying by

3^{-1} = 4

gives

x = 5 \cdot 4 = 20 \equiv 9 \mod 11

Answer

Over

\mathbb{Z}_{11}

, the solution is

x = 9

and

y = 9

.

Key Concept

Solving systems of linear equations in modular arithmetic

Explanation

The process involves aligning coefficients, eliminating variables, and finding multiplicative inverses in the context of the specified modulus.

Calculate B^2023. 𝐵=(1 1）（0 1)

Solution by Steps

step 1

To calculate

B^{2023}

for

B = \begin{pmatrix} 1 &amp; 1 \\ 0 &amp; 1 \end{pmatrix}

, we recognize that B is an upper triangular matrix

step 2

The powers of an upper triangular matrix can be calculated by taking the powers of the diagonal elements and applying binomial expansion to the off-diagonal elements

step 3

Since the diagonal elements are both 1, their powers remain 1. The off-diagonal element in the first row and second column is 1, and it will be multiplied by 2023 when raising B to the power of 2023

step 4

Therefore,

B^{2023} = \begin{pmatrix} 1 &amp; 2023 \\ 0 &amp; 1 \end{pmatrix}

Answer

B^{2023} = \begin{pmatrix} 1 &amp; 2023 \\ 0 &amp; 1 \end{pmatrix}

Key Concept

Powers of Triangular Matrices

Explanation

The powers of an upper triangular matrix can be calculated by raising the diagonal elements to the power and multiplying the off-diagonal elements by the power, due to the binomial expansion. Since the diagonal elements are 1, they remain unchanged, and the off-diagonal element is multiplied by the power, resulting in the matrix

\begin{pmatrix} 1 &amp; 2023 \\ 0 &amp; 1 \end{pmatrix}

for

B^{2023}

.

What is the maximum rank of a 5 × 3 matrix?

Solution by Steps

step 1

Consider a general 5x3 matrix with elements a(i, j) where i represents the row number and j represents the column number

step 2

The rank of a matrix is the maximum number of linearly independent rows or columns

step 3

Since the matrix has 3 columns, the maximum number of linearly independent columns is 3

step 4

Therefore, the maximum rank of a 5x3 matrix is 3

Answer

The maximum rank of a 5 × 3 matrix is 3.

Key Concept

Rank of a Matrix

Explanation

The rank of a matrix is determined by the number of linearly independent rows or columns, which is limited by the smaller dimension of the matrix.

Solve the given system of linear equations using the Gaussian elimination: 2𝑥 − 4𝑦 = 2 ( 3𝑥−𝑦+4𝑧=0 −𝑥 + 2𝑦 + 2𝑧 = −5

Solution by Steps

step 1

To solve the system of equations using Gaussian elimination, we first write the augmented matrix for the system

step 2

The augmented matrix is

\begin{pmatrix} 2 &amp; -4 &amp; 0 &amp; | &amp; 2 \\ 3 &amp; -1 &amp; 4 &amp; | &amp; 0 \\ -1 &amp; 2 &amp; 2 &amp; | &amp; -5 \end{pmatrix}

step 3

We perform row operations to get the matrix in echelon form

step 4

After applying the row operations, we obtain the matrix

\begin{pmatrix} 2 &amp; -4 &amp; 0 &amp; | &amp; 2 \\ 0 &amp; 5 &amp; 4 &amp; | &amp; 6 \\ 0 &amp; 0 &amp; 1 &amp; | &amp; -2 \end{pmatrix}

step 5

From the last row, we get

z = -2

step 6

Substituting

z = -2

into the second row, we solve for

y

step 7

We find

y = 1

step 8

Substituting

y = 1

and

z = -2

into the first row, we solve for

x

step 9

We find

x = 3

Answer

The solution to the system is

x = 3

,

y = 1

, and

z = -2

.

Key Concept

Gaussian Elimination

Explanation

Gaussian elimination is a method for solving systems of linear equations. It involves performing row operations to transform the system's augmented matrix into echelon form, from which the solutions can be easily obtained.

Show that R2 = 𝑠𝑝𝑎𝑛 ((1 （2 1) 3)).

Solution by Steps

step 1

To show that

R^2

is spanned by the vectors

\begin{pmatrix} 1 \\ 1 \end{pmatrix}

and

\begin{pmatrix} 2 \\ 3 \end{pmatrix}

, we need to prove that any vector in

R^2

can be expressed as a linear combination of these two vectors

step 2

Let

\begin{pmatrix} x \\ y \end{pmatrix}

be an arbitrary vector in

R^2

. We want to find scalars

a

and

b

such that

a\begin{pmatrix} 1 \\ 1 \end{pmatrix} + b\begin{pmatrix} 2 \\ 3 \end{pmatrix} = \begin{pmatrix} x \\ y \end{pmatrix}

step 3

Writing the linear combination explicitly, we have the system of equations:

a + 2b = x

and

a + 3b = y

step 4

To solve for

a

and

b

, we can subtract the first equation from the second to get

b = y - x

step 5

Substituting

b

back into the first equation, we get

a + 2(y - x) = x

, which simplifies to

a = 2x - y

step 6

Since we can find

a

and

b

for any

x

and

y

, this shows that any vector in

R^2

can be written as a linear combination of the two given vectors, and thus they span

R^2

Answer

The set

\left\{ \begin{pmatrix} 1 \\ 1 \end{pmatrix}, \begin{pmatrix} 2 \\ 3 \end{pmatrix} \right\}

spans

R^2

.

Key Concept

Span of a set of vectors

Explanation

A set of vectors spans

R^2

if any vector in

R^2

can be expressed as a linear combination of the vectors in the set. The given vectors

\begin{pmatrix} 1 \\ 1 \end{pmatrix}

and

\begin{pmatrix} 2 \\ 3 \end{pmatrix}

can be used to form any vector in

R^2

through appropriate scalar multiples, hence they span

R^2

.

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