Exercise 1

Solve:

\begin{cases} 2x+y+az \\ x+z \\ x+y+z \end{cases}\begin{matrix} =4 \\ =2 \\ =2 \end{matrix}

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Exercise 2

Solve:

\begin{cases} x+y+z \\ x+(a+1)y+z \\ x+y+(1+a)z \end{cases}\begin{matrix} =a \\ =2a \\ =0 \end{matrix}

Exercise 3

Solve:

 

\begin{cases} ax+z+t \\ ay+z-t \\ ay+z-2t \\ az-t \end{cases}\begin{matrix} =1 \\ =1 \\ =2 \\ =0 \end{matrix}

Exercise 4

Consider the following system for different values of a and b:

 

\begin{cases} (a+1)x+y+z \\ x+(a+1)y+z \\ x+y+(1+a)z \end{cases}\begin{matrix} =1 \\ =b \\ = { b }^{ 2 } \end{matrix}

Exercise 5

Solve:

 

\begin{cases} x+y+2t \\ 3x-y+z-t \\ 5x-3y+2z-4t \\ 2x+y+z+t \end{cases}\begin{matrix} =3 \\ =1 \\ = a \\ =2 \end{matrix}

Exercise 6

Consider the following system for different values of a and b:

 

\begin{cases} x+y+z \\ x-y \\ 3x+y+bz \end{cases}\begin{matrix} =a \\ =0 \\ = 0 \end{matrix}

Exercise 7

Determine for what values of k, the following system has infinite solutions:

 

\begin{cases} x+y+z \\ x-y+z \\ kx+z \end{cases}\begin{matrix} =0 \\ =0 \\ = 0 \end{matrix}

Exercise 8

Solve:

 

\begin{cases} x-y+z \\ 2x+my-4z \\ x+y-z \\ -x+y-z \end{cases}\begin{matrix} =7 \\ =m \\ = 1 \\ =3 \end{matrix}

 

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Solution of exercise 1

Solve:

 

\begin{cases} 2x+y+az \\ x+z \\ x+y+z \end{cases}\begin{matrix} =4 \\ =2 \\ =2 \end{matrix}

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

{ A }^{ ' }=\begin{pmatrix} 2 & 1 & a & 4 \\ 1 & 0 & 1 & 2 \\ 1 & 1 & 1 & 2 \end{pmatrix}

\left| A \right| = \begin{vmatrix} 2 & 1 & a \\ 1 & 0 & 1 \\ 1 & 1 & 1 \end{vmatrix}

\left| A \right| = 2 \begin{vmatrix} 0 & 1 \\ 1 & 1 \end{vmatrix} - 1 \begin{vmatrix} 1 & 1 \\ 1 & 1 \end{vmatrix} + \begin{vmatrix} 1 & 0 \\ 1 & 1 \end{vmatrix}

\left| A \right| = 2(0-1)-1(1-1)+a(1-0)

\left| A \right| = -2+a

a-2 = 0

a=2

 

{ A }^{ ' }=\begin{pmatrix} 2 & 1 & a & 4 \\ 1 & 0 & 1 & 2 \\ 1 & 1 & 1 & 2 \end{pmatrix}

{ C }_{ 4 } \rightarrow 2{ C }_{ 1 }

{ A }^{ ' }=\begin{pmatrix} 2 & 1 & a  \\ 1 & 0 & 1  \\ 1 & 1 & 1 \end{pmatrix}

If a=2, r(A) = r(A')=2, n=3

\begin{cases} 2x+y \\ x \end{cases}\begin{matrix} =4-2\lambda \\ =2-\lambda \end{matrix}

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

If a \neq 2, r(A) = r(A')2 = n=3, this means that the system is a consistent independent system!

 

\left| A \right| = \begin{vmatrix} 2 & 1 & a \\ 1 & 0 & 1 \\ 1 & 1 & 1 \end{vmatrix}=2a-4

\left| A \right| = \begin{vmatrix} 2 & 4 & a \\ 1 & 2 & 1 \\ 1 & 2 & 1 \end{vmatrix} = 0

\left| A \right| = \begin{vmatrix} 2 & 1 & 4 \\ 1 & 0 & 2 \\ 1 & 1 & 2 \end{vmatrix} = 0

x=\frac { 2a -4 }{ a-2 } =2, \qquad y=0 \qquad z=0

 

Solution of exercise 2

Solve:

\begin{cases} x+y+z \\ x+(a+1)y+z \\ x+y+(1+a)z \end{cases}\begin{matrix} =a \\ =2a \\ =0 \end{matrix}

A=\begin{pmatrix} 1 & 1 & 1 \\ 1 & a+1 & 1 \\ 1 & 1 & a+1 \end{pmatrix}

A'=\begin{pmatrix} 1 & 1 & 1 & a \\ 1 & a+1 & 1 & 2a \\ 1 & 1 & a+1 & 0 \end{pmatrix}

\left| A \right| = \begin{vmatrix} 1 & 1 & 1 \\ 1 & a+1 & 1 \\ 1 & 1 & a+1 \end{vmatrix} = { a }^{ 2 }

Supposing a = 0:

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

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

r(A) = r(A') = 1 \qquad n = 3, this means that the system is a consistent dependent system!

 

x + y + z = 0, supposing y = \lambda, \qquad z = \mu

x + \lambda+ \mu = 0

x = - \lambda - \mu

In case, a \neq 0, \qquad r(A) = r(A') = n = 3, this means that the system is a consistent independent system!

\left| A \right| = \begin{vmatrix} a & 1 & 1 \\ 2a & a+1 & 1 \\ 0 & 1 & a+1 \end{vmatrix} = { a }^{ 3 }

\left| A \right| = \begin{vmatrix} 1 & a & 1 \\ 1 & 2a & 1 \\ 1 & 0 & a+1 \end{vmatrix} = { a }^{ 2 }

\left| A \right| = \begin{vmatrix} 1 & 1 & a \\ 1 & a+1 & 2a \\ 1 & 1 & 0 \end{vmatrix} = -{ a }^{ 2 }

x=\frac { { a }^{ 3 } }{ { a }^{ 2 } } =a, \qquad y=\frac { { a }^{ 2 } }{ { a }^{ 2 } } = 1 \qquad z=\frac { { -a }^{ 2 } }{ { a }^{ 2 } } = -1

 

Solution of exercise 3

Solve:

\begin{cases} ax+z+t \\ ay+z-t \\ ay+z-2t \\ az-t \end{cases}\begin{matrix} =1 \\ =1 \\ =2 \\ =0 \end{matrix}

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

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

\left| A \right| = \begin{vmatrix} a & 0 & 1 & 1 \\ 0 & a & 1 & -1 \\ 0 & a & 1 & -2 \\ 0 & 0 & a & -1 \end{vmatrix} = { a }^{ 3 } \qquad a=0

If a = 0

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

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

 

\begin{vmatrix} 1 & 1 \\ 1 & -1 \end{vmatrix} \neq 0

\begin{vmatrix} 1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & -2 & 2\end{vmatrix} \neq 0

r(A) = 2, \quad r(A') = 3, this means that the system is an inconsistent system!

If a \neq 0, \qquad r(A)= r(A') = n = 4 this means that the system is a consistent independent system!

\begin{cases} ax+z+t \\ ay+z-t \\ ay+z-2t \\ az-t \end{cases}\begin{matrix} =1 \\ =1 \\ =2 \\ =0 \end{matrix}

Applying row operation { f }_{ 3 } \rightarrow { f }_{ 3 } - { f }_{ 2 }:

\begin{cases} ax+z+t \\ ay+z-t \\ -t \\ az-t \end{cases}\begin{matrix} =1 \\ =1 \\ =2 \\ =0 \end{matrix}

x=\frac { 2a + 1 }{ { a }^{ 2 } } , \qquad y=\frac { 1 }{ { a }^{ 2 } }, \qquad z=\frac { -1 }{ a },  \qquad t=-1

 

Solution of exercise 4

Consider the following system for different values of a and b:

\begin{cases} (a+1)x+y+z \\ x+(a+1)y+z \\ x+y+(1+a)z \end{cases}\begin{matrix} =1 \\ =b \\ = { b }^{ 2 } \end{matrix}

A=\begin{pmatrix} a+1 & 1 & 1 \\ 1 & a+1 & 1  \\ 1 & 1 & a+1 \end{pmatrix}

A'=\begin{pmatrix} a+1 & 1 & 1 & 1 \\ 1 & a+1 & 1 & b \\ 1 & 1 & a+1 & { b }^{ 2 } \end{pmatrix}

\left| A \right| = \begin{vmatrix} a+1 & 1 & 1 \\ 1 & a+1 & 1  \\ 1 & 1 & a+1 \end{vmatrix} = { a }^{ 2 }(a+3) \qquad a=0 \qquad a=-3

If \begin{cases} a\neq 0 \\ \forall b \end{cases} \qquad r(A)=r(A')=n=3, this means that the system is a consistent independent system!

 

If  a = 0:

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

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

\left| A \right| = \begin{vmatrix} 1 & 1 \\ 1 & b \end{vmatrix} = b - 1 \qquad b=1

If \begin{cases} a\neq 0 \\  b \neq 1 \end{cases} \qquad r(A)= 1, \quad r(A')=2, this means that the system is an Inconsistent system!

Otherwiser(A) = 2, \quad r(A')=3, this means that the system is a consistent dependent system!

 

If  a = -3:

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

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

\left| A \right| = \begin{vmatrix} -2 & 1 \\ 1 & -2 \end{vmatrix} \neq 0

\left| A' \right| = \begin{vmatrix} -2 & 1 & 1 \\ 1 & -2 & b \\ 1 & 1 & { b }^{ 2 } \end{vmatrix} = 3({ b }^{ 2 }+b+1)

r(A) = 2, \quad r(A')=3, this means that the system is an Inconsistent system!

 

Solution of exercise 5

Solve:

 

\begin{cases} x+y+2t \\ 3x-y+z-t \\ 5x-3y+2z-4t \\ 2x+y+z+t \end{cases}\begin{matrix} =3 \\ =1 \\ = a \\ =2 \end{matrix}

A=\begin{pmatrix} 1 & 1 & 0 & 2 & 3 \\ 3 & -1 & 1 & -1 & 1 \\ 5 & -3 & 2 & -4 & a \\ 2 & 1 & 1 & 1 & 2 \end{pmatrix}

We will be using row operations:

{ r }_{ 2 } - 3{ r }_{ 1 } \rightarrow { r }_{ 2 }

{ r }_{ 3 } - 5{ r }_{ 1 } \rightarrow { r }_{ 3 }

{ r }_{ 4 } - 2{ r }_{ 1 } \rightarrow { r }_{ 4 }

A=\begin{pmatrix} 1 & 1 & 0 & 2 & 3 \\ 0 & -4 & 1 & -7 & -8 \\ 0 & -8 & 2 & -14 & a-15 \\ 0 & -1 & 1 & -3 & -4 \end{pmatrix}

 

{ r }_{ 3 } - 2{ r }_{ 2 } \rightarrow { r }_{ 3 }

{ r }_{ 4 } - { r }_{ 2 } \rightarrow { r }_{ 4 }

A=\begin{pmatrix} 1 & 1 & 0 & 2 & 3 \\ 0 & -4 & 1 & -7 & -8 \\ 0 & 0 & 0 & 0 & a+1 \\ 0 & 3 & 0 & 4 & 4 \end{pmatrix}

Finding the value of a from the 3rd row:

a+1 = 0

a = -1

In the third row, 0=0 means that the system is a consistent dependent system.

\begin{cases} x+y+2t \\ -4y-7t \\ 3y+4t \end{cases} \begin{matrix} =3 \\ =-8-z \\ =4 \end{matrix}

a \neq -1

0 = k which means inconsistent system.

 

Solution of exercise 6

Consider the following system for different values of a and b:

 

\begin{cases} x+y+z \\ x-y \\ 3x+y+bz \end{cases}\begin{matrix} =a \\ =0 \\ = 0 \end{matrix}

\begin{pmatrix} 1 & 1 & 1 & a \\ 1 & -1 & 0 & 0\\ 3 & 1 & b & 0 \end{pmatrix}

Applying these row operations:

{ r }_{ 1 } + { r }_{ 2 } \rightarrow { r }_{ 1 }

{ r }_{ 3 } + { r }_{ 2 } \rightarrow { r }_{ 3 }

\begin{pmatrix} 2 & 0 & 1 & a \\ 1 & -1 & 0 & 0\\ 4 & 0 & b & 0 \end{pmatrix}

{ r }_{ 3 } - 2{ r }_{ 2 } \rightarrow { r }_{ 3 }

\begin{pmatrix} 2 & 0 & 1 & a \\ 1 & -1 & 0 & 0\\ 0 & 0 & b-2 & -2a \end{pmatrix}

b \neq 2 \quad & \quad kz=-2a, hence it is a consistent indendent system

z = \frac { -2a }{ b-2 } \qquad x = \frac { a }{ b-2 } \qquad y = \frac { a }{ b-2 }

b = 2 \quad a \neq 0 \qquad -2a = 0 which indicates that the system is an inconsistent system

b = 2 \quad a = 0 \qquad hence, 0 = 0 which indicates that the system is a consistent dependent system

\begin{cases} 2x+z \\ x-y \end{cases} \begin{matrix} =0 \\ =0 \end{matrix} \qquad \begin{cases} z \\ -y \end{cases} \begin{matrix} =-2 \lambda \\ =-x \end{matrix}

x = \lambda \qquad y = \lambda \qquad z=-2 \lambda

 

Solution of exercise 7

Determine for what values of k, the following system has infinite solutions:

 

\begin{cases} x+y+z \\ x-y+z \\ kx+z \end{cases}\begin{matrix} =0 \\ =0 \\ = 0 \end{matrix}

\begin{pmatrix} 1 & 1 & 1 & 0 \\ 1 & -1 & 1 & 0\\ k & 0 & 1 & 0 \end{pmatrix}

Applying row operations:

{ r }_{ 2 } + { r }_{ 1 } \rightarrow  { r }_{ 2 }

\begin{pmatrix} 1 & 1 & 1 & 0 \\ 2 & 0 & 2 & 0\\ k & 0 & 1 & 0 \end{pmatrix} \rightarrow \begin{pmatrix} 1 & 1 & 1 & 0 \\ 1 & 0 & 1 & 0\\ k & 0 & 1 & 0 \end{pmatrix} * 2 was taken as common and divided by 0

{ r }_{ 3 } - { r }_{ 1 } \rightarrow  { r }_{ 3 }

\begin{pmatrix} 1 & 1 & 1 & 0 \\ 1 & 0 & 21 & 0\\ k-1 & 0 & 0 & 0 \end{pmatrix}

k - 1 = 0

k = 1

\begin{cases} x+y+z \\ x+z  \end{cases} \qquad x = \lambda, \quad y=0 \quad z=- \lambda

 

Solution of exercise 8

Solve:

 

\begin{cases} x-y+z \\ 2x+my-4z \\ x+y-z \\ -x+y-z \end{cases}\begin{matrix} =7 \\ =m \\ = 1 \\ =3 \end{matrix}

\begin{pmatrix} 1 & -1 & 1 & 7 \\ 2 & m & -4 & m\\ 1 & 1 & -1 & 1 \\ -1 & 1 & -1 & 3 \end{pmatrix}

Applying these row operations:

{ r }_{ 4 } + { r }_{ 1 } \rightarrow { r }_{ 4 }

\begin{pmatrix} 1 & -1 & 1 & 7 \\ 2 & m & -4 & m\\ 1 & 1 & -1 & 1 \\ 0 & 0 & 0 & 10 \end{pmatrix}

0 \neq 10, this means that the system is an inconsistent system (for any value of m)

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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.