Exercise 1

Calculate the matrix inverse by the gaussian elimination method.

A = \begin{pmatrix} 1 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix}

The best Maths tutors available
Intasar
4.9
4.9 (36 reviews)
Intasar
£48
/h
Gift icon
1st lesson free!
Paolo
4.9
4.9 (28 reviews)
Paolo
£30
/h
Gift icon
1st lesson free!
Shane
4.9
4.9 (23 reviews)
Shane
£25
/h
Gift icon
1st lesson free!
Jamie
5
5 (16 reviews)
Jamie
£25
/h
Gift icon
1st lesson free!
Matthew
5
5 (17 reviews)
Matthew
£30
/h
Gift icon
1st lesson free!
Petar
4.9
4.9 (12 reviews)
Petar
£40
/h
Gift icon
1st lesson free!
Harinder
5
5 (14 reviews)
Harinder
£15
/h
Gift icon
1st lesson free!
Farooq
4.9
4.9 (17 reviews)
Farooq
£40
/h
Gift icon
1st lesson free!
Intasar
4.9
4.9 (36 reviews)
Intasar
£48
/h
Gift icon
1st lesson free!
Paolo
4.9
4.9 (28 reviews)
Paolo
£30
/h
Gift icon
1st lesson free!
Shane
4.9
4.9 (23 reviews)
Shane
£25
/h
Gift icon
1st lesson free!
Jamie
5
5 (16 reviews)
Jamie
£25
/h
Gift icon
1st lesson free!
Matthew
5
5 (17 reviews)
Matthew
£30
/h
Gift icon
1st lesson free!
Petar
4.9
4.9 (12 reviews)
Petar
£40
/h
Gift icon
1st lesson free!
Harinder
5
5 (14 reviews)
Harinder
£15
/h
Gift icon
1st lesson free!
Farooq
4.9
4.9 (17 reviews)
Farooq
£40
/h
Gift icon
1st lesson free!
Let's go

Exercise 2

Calculate the matrix inverse by the gaussian elimination method.

A = \begin{pmatrix} 1 & -1 & 0 \\ 0 & 1 & 1 \\ 2 & 0 & 1 \end{pmatrix}

Exercise 3

Calculate the matrix inverse by determinants.

A = \begin{pmatrix} 2 & 0 & 1 \\ 1 & 1 & -4 \\ 3 & 7 & -3 \end{pmatrix}

Exercise 4

For what values of m in the matrix A = \begin{pmatrix} 1 & 1 & m \\ m & 0 & -1 \\ 6 & -1 & 0 \end{pmatrix} does not support an inverse?

 

Exercise 5

For what values of x in the matrix A = \begin{pmatrix} 3 & x & x \\ 1 & -1 & 0 \\ 3 & -2 & 0 \end{pmatrix} does the matrix inverse not support?

 

Solution of exercise 1

Calculate the matrix inverse by the gaussian elimination method.

A = \begin{pmatrix} 1 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix}

 

\begin{pmatrix} 1 & 1 & 0 & : & 1 & 0 & 0 \\ 1 & 0 & 1 & : & 0 & 1 & 0 \\ 0 & 1 & 0 & : & 0 & 0 & 1 \end{pmatrix}

r2 - r1 r3 + r2
\begin{pmatrix} 1 & 1 & 0 & : & 1 & 0 & 0 \\ 0 & -1 & 1 & : & -1 & 1 & 0 \\ 0 & 1 & 0 & : & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 1 & 0 & : & 1 & 0 & 0 \\ 0 & -1 & 1 & : & -1 & 1 & 0 \\ 0 & 0 & 1 & : & -1 & 1 & 1 \end{pmatrix}
r2 - r3 r1 + r2
\begin{pmatrix} 1 & 1 & 0 & : & 1 & 0 & 0 \\ 0 & -1 & 0 & : & 0 & 0 & -1 \\ 0 & 0 & 1 & : & -1 & 1 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 & : & 1 & 0 & -1 \\ 0 & -1 & 0 & : & 0 & 0 & -1 \\ 0 & 0 & 1 & : & -1 & 1 & 1 \end{pmatrix}
(-1) r2 Matrix Inverse
\begin{pmatrix} 1 & 0 & 0 & : & 1 & 0 & -1 \\ 0 & 1 & 0 & : & 0 & 0 & 1 \\ 0 & 0 & 1 & : & -1 & 1 & 1 \end{pmatrix} { A }^{ -1 } = \begin{pmatrix} 1 & 0 & -1 \\ 0 & 0 & 1 \\ -1 & 1 & 1 \end{pmatrix}

 

Solution of exercise 2

Calculate the matrix inverse by the gaussian elimination method.

A = \begin{pmatrix} 1 & -1 & 0 \\ 0 & 1 & 1 \\ 2 & 0 & 1 \end{pmatrix}

 

\begin{pmatrix} 1 & -1 & 0 & : & 1 & 0 & 0 \\ 0 & 1 & 0 & : & 0 & 1 & 0 \\ 2 & 0 & 1 & : & 0 & 0 & 1 \end{pmatrix}

 

r3 - 2r1 r3 - 2r2
\begin{pmatrix} 1 & -1 & 0 & : & 1 & 0 & 0 \\ 0 & 1 & 0 & : & 0 & 1 & 0 \\ 0 & 2 & 1 & : & -2 & 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & -1 & 0 & : & 1 & 0 & 0 \\ 0 & 1 & 0 & : & 0 & 1 & 0 \\ 0 & 0 & 1 & : & -2 & -2 & 1 \end{pmatrix}
r1 + r2 Matrix Inverse
\begin{pmatrix} 1 & 0 & 0 & : & 1 & 1 & 0 \\ 0 & 1 & 0 & : & 0 & 1 & 0 \\ 0 & 0 & 1 & : & -2 & -2 & 1 \end{pmatrix} { A }^{ -1 } = \begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ -2 & -2 & 1 \end{pmatrix}

 

Solution of exercise 3

Calculate the matrix inverse by determinants.

A = \begin{pmatrix} 2 & 0 & 1 \\ 1 & 1 & -4 \\ 3 & 7 & -3 \end{pmatrix}

 

A = \begin{vmatrix} 2 & 0 & 1 \\ 1 & 1 & -4 \\ 3 & 7 & -3 \end{vmatrix} = 54

A* = \begin{pmatrix} 25 & -9 & 4 \\ 7 & -9 & -14 \\ -1 & 9 & 2 \end{pmatrix}

{ (A*) }^{ t } = \begin{pmatrix} 25 & 7 & -1 \\ -9 & -9 & 9 \\ 4 & -14 & 2 \end{pmatrix}

{ A }^{ -1 } = \begin{pmatrix} \frac { 25 }{ 54 } & \frac { 7 }{ 54 } & \frac { -1 }{ 54 } \\ \frac { -9 }{ 54 } & \frac { -9 }{ 54 } & \frac { 9 }{ 54 } \\ \frac { 4 }{ 54 } & \frac { -14 }{ 54 } & \frac { 2 }{ 54 } \end{pmatrix}

{ A }^{ -1 } = \begin{pmatrix} \frac { 25 }{ 54 } & \frac { 7 }{ 54 } & \frac { -1 }{ 54 } \\ \frac { -1 }{ 6 } & \frac { -1 }{ 6 } & \frac { 1 }{ 6 } \\ \frac { 2 }{ 27 } & \frac { -7 }{ 27 } & \frac { 1 }{ 27 } \end{pmatrix}

 

Solution of exercise 4

For what values of m in the matrix A = \begin{pmatrix} 1 & 1 & m \\ m & 0 & -1 \\ 6 & -1 & 0 \end{pmatrix} does not support an inverse?

 

\left| A \right| = \begin{vmatrix} 1 & 1 & m \\ m & 0 & -1 \\ 6 & -1 & 0 \end{vmatrix} = 0 - { m }^{ 2 } - 6 - 0 - 1 - 0 = -{ m }^{ 2 } - 7

-{ m }^{ 2 } - 7 = 0

m = \pm \sqrt { -7 } \notin R

For any real value of m, there is the inverse { A }^{ -1 }.

 

Solution of exercise 5

For what values of x in the matrix A = \begin{pmatrix} 3 & x & x \\ 1 & -1 & 0 \\ 3 & -2 & 0 \end{pmatrix} does the matrix inverse not support?

 

A = \begin{vmatrix} 3 & x & x \\ 1 & -1 & 0 \\ 3 & -2 & 0 \end{vmatrix} = x \begin{vmatrix} 1 & -1 \\ 3 & -2 \end{vmatrix} = x

For x = 0  , Matrix A has no inverse.

 

>

The platform that connects tutors and students

Did you like this article? Rate it!

1 Star2 Stars3 Stars4 Stars5 Stars 4.00 (2 rating(s))
Loading...

Emma

I am passionate about travelling and currently live and work in Paris. I like to spend my time reading, gardening, running, learning languages and exploring new places.