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3. Find the reduced form \( A_{R} \) of the matrix \( A \), and produce a matrix \( \Omega \) such that \( \Omega A=A_{R} \) \[ A=\left(\begin{array}{ccc} 3 & 2 & -12 \\ 1 & -1 & 1 \\ 4 & 1 & -11 \end

User QiAlex
by
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1 Answer

2 votes

To find the reduced row-echelon form

A

R

of matrix

A, we can use row operations. Here are the steps:

Step 1: Perform row operations to create zeros below the leading coefficient of the first row:

R2 = R2 - (1/3)R1

R3 = R3 - (4/3)R1

The matrix after Step 1 becomes:

(

3

2

12

0

5

3

5

0

5

3

5

)

3

0

0

2

3

5

3

5

−12

5

5

Step 2: Perform row operations to create a leading coefficient of 1 in the second row:

R2 = -\frac{3}{5}R2

The matrix after Step 2 becomes:

(

3

2

12

0

1

1

0

5

3

5

)

3

0

0

2

1

3

5

−12

−1

5

Step 3: Perform row operations to create zeros above and below the leading coefficient of the second row:

R1 = R1 - 2R2

R3 = R3 + \frac{5}{3}R2

The matrix after Step 3 becomes:

(

3

0

10

0

1

1

0

0

0

)

3

0

0

0

1

0

−10

−1

0

Step 4: Perform row operations to create a leading coefficient of 1 in the first row:

R1 = \frac{1}{3}R1

The matrix after Step 4 becomes:

(

1

0

10

3

0

1

1

0

0

0

)

1

0

0

0

1

0

3

10

−1

0

The reduced row-echelon form

A

R

of matrix

A is:

=

(

1

0

10

3

0

1

1

0

0

0

)

A

R

=

1

0

0

0

1

0

3

10

−1

0

To find the matrix

Ω

Ω such that

Ω

=

ΩA=A

R

, we perform the same row operations on the identity matrix:

Ω

=

(

1

3

0

0

0

3

5

0

0

5

3

1

)

Ω=

3

1

0

0

0

5

3

3

5

0

0

1

Therefore,

Ω

=

ΩA=A

R

User Jeprubio
by
8.2k points