Exercise 1

Given the matrices:

A = \begin{pmatrix} 2 & 3 \\ 1 & 2 \end{pmatrix} \qquad B = \begin{pmatrix} 3 & 1 \\ 2 & -5 \end{pmatrix}

Solve the matrix equation:

A . X = B

The best Maths tutors available
1st lesson free!
Intasar
4.9
4.9 (23 reviews)
Intasar
£42
/h
1st lesson free!
Matthew
5
5 (17 reviews)
Matthew
£25
/h
1st lesson free!
Dr. Kritaphat
4.9
4.9 (6 reviews)
Dr. Kritaphat
£49
/h
1st lesson free!
Paolo
4.9
4.9 (11 reviews)
Paolo
£25
/h
1st lesson free!
Petar
4.9
4.9 (9 reviews)
Petar
£27
/h
1st lesson free!
Rajan
4.9
4.9 (11 reviews)
Rajan
£15
/h
1st lesson free!
Farooq
5
5 (13 reviews)
Farooq
£35
/h
1st lesson free!
Myriam
5
5 (15 reviews)
Myriam
£20
/h
1st lesson free!
Intasar
4.9
4.9 (23 reviews)
Intasar
£42
/h
1st lesson free!
Matthew
5
5 (17 reviews)
Matthew
£25
/h
1st lesson free!
Dr. Kritaphat
4.9
4.9 (6 reviews)
Dr. Kritaphat
£49
/h
1st lesson free!
Paolo
4.9
4.9 (11 reviews)
Paolo
£25
/h
1st lesson free!
Petar
4.9
4.9 (9 reviews)
Petar
£27
/h
1st lesson free!
Rajan
4.9
4.9 (11 reviews)
Rajan
£15
/h
1st lesson free!
Farooq
5
5 (13 reviews)
Farooq
£35
/h
1st lesson free!
Myriam
5
5 (15 reviews)
Myriam
£20
/h
First Lesson Free>

Exercise 2

Given the matrices:

A = \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix} \qquad B = \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix} \qquad C = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}

Solve the matrix equation:

X . A + B = C

Exercise 3

Given the matrices:

A = \begin{pmatrix} 1 & 1 \\ 3 & 4 \end{pmatrix} \qquad B = \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix} \qquad C = \begin{pmatrix} 1 & 2 \\ 1 & 3 \end{pmatrix}

Solve the matrix equations:

  1. X A = B + I
  2. AX + B = C
  3. XA + B = 2C
  4. AX + BX = C
  5. XAB - XC = 2C

Exercise 4

Given the matrices:

A = \begin{pmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 1 & 1 \end{pmatrix} \qquad B = \begin{pmatrix} 0 & 1 & 1 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix} \qquad C = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{pmatrix}

Solve the matrix equation:

A X + 2B = 3C

Exercise 5

Solve the matrix equation:

A . X + 2 . B = 3 . C

A = \begin{pmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 1 & 1 \end{pmatrix} \qquad B = \begin{pmatrix} 0 & 1 & 1 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix} \qquad C = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{pmatrix}

Exercise 6

Solve the system using a matrix equation.

\begin{cases} x + y + z \\ x + 2y + 5z \\ x + 4y + 25z \end{cases} \begin{matrix} = 6 \\ = 12 \\ =36 \end{matrix}

Exercise 7

Calculate A and B:

\begin{cases} 2A + B = \begin{pmatrix} 1 & 2 & 2 \\ -2 & 1 & 0 \end{pmatrix} \\ A - 3B = \begin{pmatrix} -4 & -3 & -2 \\ -1 & 0 & -1 \end{pmatrix} \end{cases}

Exercise 8

Solve the following equations without developing the determinants.

1   \begin{vmatrix} 1 & 1 & 1 \\ 1 & x & 1 \\ 1 & 1 & { x }^{ 2 } \end{vmatrix} = 0

2  \begin{vmatrix} a & b & c \\ a & x & c \\ a & b & x \end{vmatrix} = 0

 

Solution of exercise 1

1Given the matrices:

A = \begin{pmatrix} 2 & 3 \\ 1 & 2 \end{pmatrix} \qquad B = \begin{pmatrix} 3 & 1 \\ 2 & -5 \end{pmatrix}

 

Solve the matrix equation:

A . X = B

\left| A \right| = 1 \neq 0, there is the inverse { A }^{ -1 }

{ A }^{ -1 } (A . X) = { A }^{ -1 } . B

I . X = { A }^{ -1 } . B

X = { A }^{ -1 } . B

{ A }^{ -1 } = \begin{pmatrix} 2 & -3 \\ -1 & 2 \end{pmatrix}

X = \begin{pmatrix} 2 & -3 \\ -1 & 2 \end{pmatrix} \begin{pmatrix} 3 & 1 \\ 2 & -5 \end{pmatrix} = \begin{pmatrix} 0 & 17 \\ 1 & -11 \end{pmatrix}

 

Solution of exercise 2

Given the matrices:

A = \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix} \qquad B = \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix} \qquad C = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}

Solve the matrix equation:

X . A + B = C

\left| A \right| = 1 \neq 0, there is the inverse { A }^{ -1 }

(X . A + B) - B = C - B

X . A + (B - B) = C - B

X . A + 0 = C - B

X . A = C - B

X . A . { A }^{ -1 } = (C - B) . { A }^{ -1 }

X (A . { A }^{ -1 }) = (C - B) . { A }^{ -1 }

X . I = (C - B) . { A }^{ -1 }

X = (C - B) . { A }^{ -1 }

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

X = [ \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} - \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix}] \begin{pmatrix} 2 & -1 \\ -1 & 1 \end{pmatrix} =

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

 

Solution of exercise 3

Given the matrices:

A = \begin{pmatrix} 1 & 1 \\ 3 & 4 \end{pmatrix} \qquad B = \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix} \qquad C = \begin{pmatrix} 1 & 2 \\ 1 & 3 \end{pmatrix}

 

Solve the matrix equations:

1. XA = B +I

X A { A }^{ -1 } = (B + I) { A }^{ -1 }

X I = (B + I) { A }^{ -1 }

X = (B + I) { A }^{ -1 }

X = \begin{pmatrix} 9 & -2 \\ -2 & 1 \end{pmatrix}

 

2. AX + B = C

AX = C - B

{ A }^{ -1 } A X = { A }^{ -1 } (C - B)

I X = { A }^{ -1 } (C - B)

X = { A }^{ -1 } (C - B)

X = \begin{pmatrix} -4 & 2 \\ 3 & -1 \end{pmatrix}

 

3. XA + B = 2C

X A { A }^{ -1 } = (2C - B) { A }^{ -1 }

X I = (2C - B) { A }^{ -1 }

X = (2C - B) { A }^{ -1 }

X = \begin{pmatrix} -9 & 3 \\ -11 & 4 \end{pmatrix}

 

4. AX + BX = C

(A + B) X = C

{ (A + B) }^{ -1 } (A + B) X = { (A+B) }^{ -1 } C

I X = { (A + B) }^{ -1 } C

X = { (A + B) }^{ -1 } C

X = \begin{pmatrix} \frac{ 3 }{ 7 } & \frac { 4 }{ 7 } \\ \frac { -1 }{ 7 } & \frac { 1 }{ 7 } \end{pmatrix}

 

5. XAB - XC = 2C

X(AB - C) = 2C

X (AB - C){ (AB - C) }^{ -1 } = 2C { (AB - C) }^{ -1 }

XI = 2C { (AB - C) }^{ -1 }

X = 2C { (AB - C) }^{ -1 }

X = \begin{pmatrix} \frac{ -7 }{ 2 } & 1 \\ \frac { -23 }{ 4 } & \frac { 3 }{ 2 } \end{pmatrix}

 

Solution of exercise 4

Given the matrices:

A = \begin{pmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 1 & 1 \end{pmatrix} \qquad B = \begin{pmatrix} 0 & 1 & 1 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix} \qquad C = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{pmatrix}

Solve the matrix equation:

A X + 2B = 3C

AX = 3C - 2B

IX = { A }^{ -1 } (3C - 2B)

AX + 2B = 3C

X = { A }^{ -1 } (3C - 2B)

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

X = { A }^{ -1 } (3C - 2B)

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

X = \begin{pmatrix} 1 & 0 & 0 \\ -1 & 1 & 0 \\ 0 & -1 & 1 \end{pmatrix} [ 3 \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{pmatrix} - 2 \begin{pmatrix} 0 & 1 & 1 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}] =

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

= \begin{pmatrix} 3 & -2 & -2 \\ -5 & 5 & 2 \\ 5 & -3 & 1 \end{pmatrix}

 

Solution of exercise 5

Solve the matrix equation:

A . X + 2 . B = 3 . C

A = \begin{pmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 1 & 1 \end{pmatrix} \qquad B = \begin{pmatrix} 0 & 1 & 1 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix} \qquad C = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{pmatrix}

 

\left| A \right| = 1 \neq 0,

(A . X + 2 . B) - 2 . B = 3 . C - 2B

A . X + (2 . B - 2 . B) = 3 . C - 2B

A . X + 0 = 3 . C - 2B

A . X = 3 . C - 2B

({ A }^{ -1 } . A) . X = { A }^{ -1 } . (3 . C - 2B)

I . X = { A }^{ -1 } . (3 . C - 2B)

I . X = { A }^{ -1 } . (3 . C - 2B)

X = { A }^{ -1 } . (3 . C - 2B)

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

X = \begin{pmatrix} 1 & 0 & 0 \\ -1 & 1 & 0 \\ 0 & -1 & 1 \end{pmatrix} [ 3 \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{pmatrix} - 2 \begin{pmatrix} 0 & 1 & 1 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}] =

= \begin{pmatrix} 1 & 0 & 0 \\ -1 & 1 & 0 \\ 0 & -1 & 1 \end{pmatrix} \begin{pmatrix} 3 & -2 & -2 \\ -2 & 3 & 0 \\ 3 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 3 & -2 & -2 \\ -5 & 5 & 2 \\ 5 & -3 & 1 \end{pmatrix}

 

Solution of exercise 6

Solve the system using a matrix equation.

\begin{cases} x + y + z \\ x + 2y + 5z \\ x + 4y + 25z \end{cases} \begin{matrix} = 6 \\ = 12 \\ =36 \end{matrix}

\begin{pmatrix} 1 & 1 & 1 \\ 1 & 2 & 5 \\ 1 & 4 & 25 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 6 \\ 12 \\ 36 \end{pmatrix}

A . X = C

{ A }^{ -1 } A . X = { A }^{ -1 } C

I . X = { A }^{ -1 } C

X = { A }^{ -1 } C

{ A }^{ -1 } = \frac { 1 }{ 12 } \begin{pmatrix} 30 & -21 & 3 \\ -20 & 24 & -4 \\ 2 & -3 & 1 \end{pmatrix}

\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \frac { 1 }{ 12 } \begin{pmatrix} 30 & -21 & 3 \\ -20 & 24 & -4 \\ 2 & -3 & 1 \end{pmatrix} \begin{pmatrix} 6 \\ 12 \\ 36 \end{pmatrix} = \frac { 1 }{ 12 } \begin{pmatrix} 36 \\ 24 \\ 12 \end{pmatrix}

\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 3 \\ 2 \\ 1 \end{pmatrix}

x = 3 \qquad y = 2 \qquad z = 1

 

Solution of exercise 7

Calculate A and B:

\begin{cases} 2A + B = \begin{pmatrix} 1 & 2 & 2 \\ -2 & 1 & 0 \end{pmatrix} \\ A - 3B = \begin{pmatrix} -4 & -3 & -2 \\ -1 & 0 & -1 \end{pmatrix} \end{cases}

Multiply the second equation by −2.

\begin{cases} 2A + B = \begin{pmatrix} 1 & 2 & 2 \\ -2 & 1 & 0 \end{pmatrix} \\ -2A + 6B = \begin{pmatrix} 8 & 6 & 4 \\ 2 & 0 & 2 \end{pmatrix} \end{cases}

Add the equations.

7B = \begin{pmatrix} 9 & 8 & 6 \\ 0 & 1 & 2 \end{pmatrix} \qquad B = \begin{pmatrix} \frac { 9 }{ 7 } & \frac { 8 }{ 7 } & \frac { 6 }{ 7 }  \\ 0 & \frac { 1 }{ 7 } & \frac { 2 }{ 7 }  \end{pmatrix}

Multiply the first equation by 3 and add the equations:

7A = \begin{pmatrix} -1 & 3 & 4 \\ -7 & 3 & -1 \end{pmatrix} \qquad A = \begin{pmatrix} \frac { -1 }{ 7 } & \frac { 3 }{ 7 } & \frac { 4 }{ 7 }  \\ -1 & \frac { 3 }{ 7 } & \frac { -1 }{ 7 } \end{pmatrix}

 

Solution of exercise 8

Solve the following equations without developing the determinants.

1   \begin{vmatrix} 1 & 1 & 1 \\ 1 & x & 1 \\ 1 & 1 & { x }^{ 2 } \end{vmatrix} = 0

\begin{vmatrix} 1 & 1 & 1 \\ 1 & x & 1 \\ 1 & 1 & { x }^{ 2 } \end{vmatrix}

{ r }_{ 2 } \Rightarrow { r }_{ 2 } - { r }_{ 1 } \qquad { r }_{ 3 } \Rightarrow { r }_{ 3 } - { r }_{ 1 }

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

(x - 1)( { x }^{ 2 } - 1) = 0 \begin{matrix} \qquad x = 1 \\ x = -1 \end{matrix}

2  \begin{vmatrix} a & b & c \\ a & x & c \\ a & b & x \end{vmatrix} = 0

\begin{vmatrix} a & b & c \\ a & x & c \\ a & b & x \end{vmatrix} = a \begin{vmatrix} 1 & b & c \\ 1 & x & c \\ 1 & b & x \end{vmatrix} = a \begin{vmatrix} 1 & b & c \\ 0 & b - x & 0 \\ 0 & 0 & c - x \end{vmatrix}

= a (b - x)(c - x) a(b - x)(c - x) = 0 \qquad \begin{matrix} x = b \\ x = c \end{matrix}

 

Need a Maths teacher?

Did you like the article?

1 Star2 Stars3 Stars4 Stars5 Stars 3.33/5 - 3 vote(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.