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

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

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

$a-2 = 0$

$a=2$

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

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

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

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!

Otherwise $r(A) = 2, \quad r(A')=3$ , 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.

Linear System of Equations with Parameters Exercises

Exercise 1

Exercise 2

Exercise 3

Exercise 4

Exercise 5

Exercise 6

Exercise 7

Exercise 8

Solution of exercise 1

Solution of exercise 2

Solution of exercise 3

Solution of exercise 4

Solution of exercise 5

Solution of exercise 6

Solution of exercise 7

Solution of exercise 8

Theory

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