A square systems of linear equations is a linear system of equations that contains as many equations as unkown variables. For square systems of linear equations the determinant of the coefficient matrix plays a central role for the solutions of the systems. We discuss Cramer’s rule, by which certain square systems of linear equations can be solved in terms of determinants.
Theoretical Remarks
2.5.Consider the square system of linear equations
A x = b, (2.5.1)
where A is an n× n matrix and b ∈ Rn.
1. System (2.5.1) can be solved by the use of the Gauss elimination method. See Theoretical Remark 2.4 for a detailed description of this method.
2. System (2.5.1) admits a unique solution x∈ Rn for every b∈ Rnif and only if A is invertible. The unique solution of (2.5.1) is then
x = A−1b.
3. System (2.5.1) admits a unique solution if and only if det A = 0. Therefore, if det A = 0 then system (2.5.1) may admit infinitely many solutions or the system may be inconsistent.
4. If system (2.5.1) is consistent, then its unique solution can be calculated by the use of Cramer’s rule, which states the following:
Cramer’s rule:
If det A = 0 then the unique solution x = (x1, x2, . . . xn) of (2.5.1) is given by the formula
xj = det Aj(b)
det A , j = 1, 2, . . . , n,
where Aj(b) is the matrix that has been obtained from matrix A by replacing the jth column in A by the vector b. In the case where det A = 0, Cramer’s rule states the following:
If det A = 0 and det Aj(b) = 0 for at least one j, then the system (2.5.1) is incon-sistent. If det A = 0 and det Aj(b) = 0 for every j = 1, 2, . . . , n, then the system (2.5.1) admits infinitely many solutions.
Problem
2.5.1.Consider the square system of linear equations Ax = b with
A =
k 1 2 2 1 k k 0 1
, b =
1
−7 3
, x =
x1 x2 x3
,
where k is an unspecified real parameter.
a) Find all values of k, such that the given system has a unique solution. For which values of k is the matrix A invertible?
b) Find all values of k, such that the given system admits infinitely many solutions and give all values of k for which the system is inconsistent.
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Solution
2.5.1.a) We recall that a linear square system Ax = b has a unique solution if and only if det A = 0 and then A is invertible. Calculating the determinant for A, we obtain
det A =
k 1 2 2 1 k k 0 1
= (k + 1)(k− 2).
Thus the linear system has a unique solution for all k∈ R\{−1, 2} and this unique solution is
x = A−1b for all k∈ R\{−1, 2}.
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b) To find the values of k for which the linear system Ax = b might admit infinitely many solutions, we have to investigate the two cases k = −1 and k = 2, since the determinant of A is zero for those two values of k.
For k =−1, the augmented matrix of the system is
By the third row of the previous matrix it is clear that the system is inconsistent in this case, i.e. for k =−1.
For k = 2, the augmented matrix of the system is
By the second row of the previous matrix it is clear that the system is also inconsis-tent in this case (k = 2).
We recall that the given system has a unique solution for all k ∈ R\{−1, 2} and that the system is inconsistent for k =−1 as well as for k = 2. Therefore there exist no real value of k for which the system may admit infinitely many solutions.
Problem
2.5.2.Consider the following linear system:
where k is an unspecified real parameter.
a) Find all values of k, such that the given linear system admits a unique solution and find then this solution by the use of Cramer’s rule.
b) Find all values of k, such that the given linear system is inconsistent and all values of k for which it admits infinitely many solutions. Find all solutions.
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Solution
2.5.2.The determinant of A is
det A = k2− 3k + 2 = (k − 1)(k − 2)
so that A is invertible if and only if k∈ R\{1, 2}. The system has therefore a unique solution for all real k, except for k = 1 or k = 2. To find this unique solution we use Cramer’s rule and calculate
xj = det Aj(b)
Therefore, by Cramer’s rule, the system admits infinitely many solutions for k = 2.
For k = 1, we have
det A = 0 and det Aj(b)= 0 for j = 1, 2, 3.
Therefore, by Cramer’s rule, the system is inconsistent for k = 1.
b) The augmented matrix for the given system is
We multiply the first row by −1 and add this to the second row, to obtain the row equivalent matrix the above augmented matrix has the following reduced echelon form:
The 3rd column is not a pivot column and we can therefore choose x3 arbitrary. We let x3= t. Hence the solutions of the given system for k = 2 are
Consider the following 4× 4 matrix:
A =
where k is an unspecified real parameter.
a) Consider the homogeneous linear system Ax = 0,
where x ∈ R4. Find all values of k, such that this system admits only the trivial (zero) solution, as well as all values of k for which the system admits infinitely many solutions.
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b) Consider the non-homogeneous linear system
Ax = b, b =
1 0 1 0
.
Find all values of k, such that this system admits infinitely many solutions x∈ R4 and give all those solutions. Find also all values of k for which the system admits a unique solution, as well as all k for which the system is inconsistent.
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© Deloitte & Touche LLP and affiliated entities.
360° thinking .
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Solution
2.5.3.a) We recall that the square system Ax = 0 admits only the zero solution, x = 0, if and only if A is an invertible matrix. Moreover, A is invertible if and only if det A = 0.
We therefore calculate the determinant of A:
det A =
From the above we conclude that Ax = 0 admits only the zero solution for all k∈ R\{−1, 1/2} and that the system admits infinitely many solutions for k = −1 as well as for k = 1/2.
b) Since A is invertible for all k ∈ R\{−1, 1/2}, the system has a unique solution for all those values of k and the unique solution of the system is
x = A−1b.
For k =−1 the augmented matrix and its reduced echelon form are as follows:
Since the 4th column of the augmented matrix is not pivot, we set x4 = t, where t is an arbitrary parameter. Then the solutions are
x = t
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For k = 1/2 the augmented matrix and its reduced echelon form are as follows:
Since the 2nd column of the augmented matrix is not pivot, we set x2 = t, where t is an arbitrary parameter. Then the solutions are
x = t
We conclude that the system has infinitely many solutions for both k = 1 and for k = 1/2 and it has a unique solution for all other values of k, so that there exist no values of k for which the system is inconsistent.