Find out the answer to "Solve the system by inverting the coefficient matrix..."

Anika Klein

Answered question

2022-01-23

Solve the system by inverting the coefficient matrix

x_{1} + 3 x_{2} + x_{3} = 4

2 x_{1} + 2 x_{2} + x_{3} = - 1

2 x_{1} + 3 x_{2} + x_{3} = 3

sineurosi0f

Beginner2022-01-24Added 14 answers

Step 1
Represent the linear system by matrices:

A = [\begin{array}{ccc} 1 & 3 & 1 \\ 2 & 2 & 1 \\ 2 & 3 & 1 \end{array}]

x = [\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \end{matrix}]

b = [\begin{matrix} 4 \\ - 1 \\ 3 \end{matrix}]

As by theorem 1.6.2

A x = b

x = A^{- 1} b

Step 2
Now we need to get

A^{- 1}

So we reduce A to the identity matrix by row operations and simultaneously apply these operations to I to produce

A^{- 1}

. To accomplish this we will adjoin the identity matrix to the right side of A, there by producing a partitoned matrix of the form:

[A ∣ I]

Adjoin A with the identity matrix:

[\begin{array}{cccccc} 1 & 3 & 1 & 1 & 0 & 0 \\ 2 & 2 & 1 & 0 & 1 & 0 \\ 2 & 3 & 1 & 0 & 0 & 1 \end{array}]

Apply row operations:
1)

R_{2} = R_{3} - R_{2}

[\begin{array}{cccccc} 1 & 3 & 1 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 & - 1 & 0 \\ 2 & 3 & 1 & 0 & 0 & 1 \end{array}]

Apply row operations:
2)

R_{3} = 2 R_{1} - R_{3}

Step 3

[\begin{array}{cccccc} 1 & 3 & 1 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 & - 1 & 1 \\ 0 & 3 & 1 & 2 & 0 & - 1 \end{array}]

Apply row operations:
3)

R_{3} = - 3 R_{2} + R_{3}

[\begin{array}{cccccc} 1 & 3 & 1 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 & - 1 & 1 \\ 0 & 0 & 1 & 2 & 3 & - 4 \end{array}]

Apply row operations:
4)

R_{1} = - 3 R_{2} + R_{1}

Step 4

[\begin{array}{cccccc} 1 & 0 & 1 & 1 & 3 & - 3 \\ 0 & 1 & 0 & 0 & - 1 & 1 \\ 0 & 0 & 1 & 2 & 3 & - 4 \end{array}]

Apply row operations:
5)

R_{1} = - R_{3} + R_{1}

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