Notes on Determinant

ENGG2012B Advanced Engineering Mathematics Notes on Determinant Lecturer: Kenneth Shum Lecture 9-18/02/2013 The determinant of a system of linear equations determines whether the solution is unique, without actually calculating the solution. This note is the preliminary to the calculation of eigenvalues and eigenvectors. We want to determine whether a system of homogeneous linear equations Ax = 0, where A is an n n matrix, has nontrivial solution or not. It is obvious that x = 0 is a solution. Any solution with at least one component non-zero is said to be non-trivial. Corollary 10 at the end of this note gives a necessary and sufficient condition for the existence of non-trivial solution. It is analogous to the discriminant of a quadratic equation ax 2 + bx + c = 0. We can determine whether the roots are distinct or repeated from the discriminant b 2 4ac, without actually calculating the roots. The objectives of this notes are: 1. Define determinant properly. 2. Derive some basic properties of determinant. 1 2 2 determinant The determinant of 2 2 matrix is defined as [ ] a b det := ad bc. (1) c d (The symbol := means true by definition.) If A is a 2 2 matrix with non-zero determinant, we can solve Ax = b by x = 1 [ ] d b b. det A c a The solution is uniquely determined by the above equation. If det A = 0, then either Ax = b has no solution or infinitely many solutions. To emphasize the dependency on the rows of the matrix, we let u and v be the first and second rows of a 2 2 matrix, respectively, and write the determinant function as det(u, v). With this notation, we can readily check from (1) that the determinant function satisfies the following properties: 1. det(u, v) = det(v, u). 2. det(u 1 + u 2, v) = det(u 1, v) + det(u 2, v). 3. det(u, v 1 + v 2 ) = det(u, v 1 ) + det(u, v 2 ). 4. det(cu, v) = c det(u, v), for any constant c. 5. det(u, cv) = c det(u, v), for any constant c. 1

6. det([1 0], [0 1]) = 1. Here, u, u 1, u 2, v, v 1 and v 2 are row vectors of length two, and c is a real constant which may equal zero. We want to generalize the notion of determinant to general n n matrices, so that we can efficiently determine whether an n n linear system has unique solution. 2 How to define n n determinant? There are several approaches in defining determinant beyond the 2 2 case. We shall give two of them below. The first one is recursive, and the second one is not. In the following, the matrix A = [a ij ] is an n n matrix. The (i, j)-entry of A is denoted by a ij. [ ] a b Expansion on the first row. For 2 2 matrices, we define the determinant as ad bc. c d Recursively, suppose that we know how to calculate (n 1) (n 1) determinant for some integer n 3. In order to simplify notations, for positive integers between 1 and n, we let A ij be the determinant of the submatrix obtained by removing the i-th row and the j-th column. The value A ij is called the minor corresponding to the (i, j)-entry of A. In terms of the minors, the determinant of the n n matrix A can be defined by det(a) := a 11 A 11 a 12 A 12 + a 13 A 13 + ( 1) n+1 a 1n A 1n. (2) This definition is usually called the Laplace expansion on the first row. Example: det 1 2 3 8 9 4 = 1 9 4 6 5 2 8 4 7 5 + 3 8 9 7 6 = (21) 2(12) + 3( 15) = 48. 7 6 5 Definition in terms of permutations There are n! ways to write 1 to n in some order. Each of them is called a permutation of 1, 2,..., n. For example, (2, 5, 1, 4, 3) and (1, 4, 5, 2, 3) are two permutations of 1 to 5. For a sequence of distinct numbers, (a 1, a 2,..., a n ), we define the corresponding inversion number as the number of pairs which are not in the natural order; we count the number of pairs a i and a j such that a i > a j and i < j. Mathematically, we let inv(a 1, a 2,..., a n ) := #{(i, j) : a i > a j and i < j}. The symbol #S stands for the number of elements in a set S. In other words, the inversion number of permutation (p 1, p 2,..., p n ) counts the number of integers r and s such that r is to the left of s and r is larger than s. A permutation is in ascending order if and only if the inversion number is equal to zero. Example: The inversion number of (2, 5, 1, 4, 3) is 5, because we can find five pairs (2, 1), (5, 1), (5, 3), (4, 3) and (5, 4). The inversion number of (1, 4, 5, 2, 3) is 4, because of (4, 2), (5, 2), (4, 3) and (5, 3). Inversion number is defined for any sequence of distinct numbers, not necessarily a permutation of 1 to n. For example, the inversion number of (1, 16, 4, 9) is 2. For an n n matrix A, we can define the determinant of A by det A := ( 1) inv(p 1,p 2,...,p n ) a 1,p1 a 2,p2 a 3,p3 a 4,p4 a n,pn, (3) with the summation over all permutations of 1 to n. There are n! terms, and each term consists of a sign ( 1) inv(p1,p2,...,pn) and the product of n entries in A. This formula is sometime called the Leibniz formula. We note that the determinant only needs the information whether the inversion number is even or odd; if the inversion number is even, then the corresponding term has positive sign, otherwise the sign is negative. 2

Example: For n = 4, we check that inv(1, 2, 3, 4) = 0, inv(2, 1, 3, 4) = 1, inv(3, 1, 2, 4) = 2, inv(4, 1, 2, 3) = 3, inv(1, 2, 4, 3) = 1, inv(2, 1, 4, 3) = 2, inv(3, 1, 4, 2) = 3, inv(4, 1, 3, 2) = 4, inv(1, 3, 2, 4) = 1, inv(2, 3, 1, 4) = 2, inv(3, 2, 1, 4) = 3, inv(4, 2, 1, 3) = 4, inv(1, 3, 4, 2) = 2, inv(2, 3, 4, 1) = 3, inv(3, 2, 4, 1) = 4, inv(4, 2, 3, 1) = 5, inv(1, 4, 2, 3) = 2, inv(2, 4, 1, 3) = 3, inv(3, 4, 1, 2) = 4, inv(4, 3, 1, 2) = 5, inv(1, 4, 3, 2) = 3, inv(2, 4, 3, 1) = 4, inv(3, 4, 2, 1) = 5, inv(4, 3, 2, 1) = 6. Using the inversion number of the 24 permutations of 1,2,3,4, we can write the determinant of a 4 4 matrix a 11 a 12 a 13 a 14 A = a 21 a 22 a 23 a 24 a 31 a 32 a 33 a 34 a 41 a 42 a 43 a 44 as det A := a 11 a 22 a 33 a 44 a 12 a 21 a 33 a 44 + a 13 a 21 a 32 a 44 a 14 a 21 a 32 a 43 a 11 a 22 a 34 a 43 + a 12 a 21 a 34 a 43 a 13 a 21 a 34 a 42 + a 14 a 21 a 33 a 42 a 11 a 23 a 32 a 44 + a 12 a 23 a 31 a 44 a 13 a 22 a 31 a 44 + a 14 a 22 a 31 a 43 + a 11 a 23 a 34 a 42 a 12 a 23 a 34 a 41 + a 13 a 22 a 34 a 41 a 14 a 22 a 33 a 41 + a 11 a 24 a 32 a 43 a 12 a 24 a 31 a 43 + a 13 a 24 a 31 a 42 a 14 a 23 a 31 a 42 a 11 a 24 a 33 a 42 + a 12 a 24 a 33 a 41 a 13 a 24 a 32 a 41 + a 14 a 23 a 32 a 41. We note that in each term, the first subscripts are in ascending order, while the second subscripts form a permutation of 1, 2, 3 and 4. There are 24 terms. Half of them have positive sign and half of them have negative sign. Proposition 1. The two definitions given above are equivalent. Proof We want to show that, for any n n matrix, the calculations in the two definitions of determinant give the same result. We proceed by mathematical induction. [ ] a11 a According to the first definition, the determinant 2 2 matrix 12 is a a 21 a 11 a 22 a 12 a 21. It is in 22 accordance with the second definition, because the inversion number of the permutation (1, 2) is 0 and the inversion number of (2, 1) is 1, [ ] a11 a det 12 = ( 1) inv(1,2) a a 21 a 11 a 22 + ( 1) inv(2,1) a 12 a 21 = a 11 a 22 a 12 a 21. 22 Let n be an integer larger than or equal to 3. Suppose that the two definitions of determinant agree with each other for matrices of size (n 1) (n 1). We will use the fact that if (p 1, p 2,..., p n ) is a permutation of 1 to n, then there are exactly p 1 1 pairs of number (p 1, p k ) with p 1 > p k. Therefore, inv(p 1, p 2,..., p n ) = (p 1 1) + inv(p 2, p 3,..., p n ) (4) if (p 1, p 2,..., p n ) is a permutation of 1 to n. In (4), inv(p 2, p 3,..., p n ) is the inversion number of (p 2, p 3,..., p n ), which is a permutation of {1, 2,..., p 1 1, p 1 + 1,..., n}. From the definition in (3), we can group the terms according to p 1, and get det A = ( 1) inv(p 1,p 2,...,p n ) a 1,p1 a 2,p2 a 3,p3 a 4,p4 a n,pn, = n ( 1) p1 1 a 1,p1 p 1=1 (p 2,p 3,...,p n ) ( 1) inv(p2,p3,...,pn) a 2,p2 a 3,p3 a 4,p4 a n,pn. 3

By the induction hypothesis, the summation ( 1) inv(p2,p3,...,pn) a 2,p2 a 3,p3 a 4,p4 a n,pn (p 2,p 3,...,p n ) is precisely the minor corresponding to the (1, p 1 )-entry of A. Therefore, det A = which coincides with the first definition in (2). n ( 1) p1 1 A 1,p1, p 1=1 At this point, a computer scientist would say: the computational complexity involved in each of the two definitions given above are so high that they are not suitable for calculating determinant of large size. Indeed, the Leibniz formula has n! terms, and n! increases exponentially with n. If we apply the recursive formula in the straightforward way, the computational complexity is also in the order of n!. For example the calculation of a 10 10 matrix using the first recursive definition directly requires more than three million multiplications. A more efficient method of calculating the determinant will be given in the next section. 3 Axioms of determinant It turns out that the properties of determinant are more important then how to calculate the determinant. We shall identify several basic and desirable properties of determinant, called axioms, and verify that the two equivalent ways of calculating determinant in the last section satisfy these axioms. We write the desired function on matrices as a function which whose inputs are n row vectors of length n, and the output is a real number, det(v 1, v 2,..., v n ). As in the 2 2 case, we want to construct a function which (at least) satisfies the following properties. (Axiom 1) If any two row vectors in det(v 1, v 2,..., v n ) are exchanged, then the value of the determinant is multiplied by 1. (Axiom 2) det(u + u, v 2,..., v n ) = det(u, v 2,..., v n ) + det(u, v 2,..., v n ) for any two row vectors u and u of length n. (Axiom 3) det(cv 1, v 2,..., v n ) = c det(v 1, v 2,..., v n ) for any constant c. (Axiom 4) For i = 1, 2,..., n, let e i be the standard basis of R n, i.e., Then det(e 1, e 2,..., e n ) = 1. e i := (0, 0,..., 0, 1, 0, 0,..., 0). } {{ } i These four properties are called the axioms of determinant. We first derive some immediate properties from the four axioms. Proposition 2. If det is a function which satisfies Axioms 1 to 4, then 1. If any two row vectors are identical, then det(v 1, v 2,..., v n ) is equal to 0. 4

2. Suppose that the i-th row vector can be decomposed as the sum u + u. Then 3. For any constant c, we have det(v 1,..., v i 1, u + u,..., v n ) = det(v 1,..., v i 1, u,..., v n ) + det(v 1,..., v i 1, u,..., v n ). det(v 1,..., v i 1, cv i,..., v n ) = c det(v 1,..., v i 1, v i,..., v n ). 4. For any two distinct row indices i j, adding any constant multiple of the r j to r i does not change the value of determinant. 5. If one of the input row vector is an all-zero vector, then the determinant is equal to zero. Proof (1) Suppose that the i-th row and the j-th row are identical. Let the determinant function value be x. Since the i-th and the j-th row are the same, exchanging them gives the same determinant value x. However, by Axiom 1, if the i-th and the j-th row are exchanged, the value of the resulting determinant is x. Hence, x = x. This is possible only if x = 0. (2) We substitute v i by u + u, det(v 1,..., v i 1, u + u,..., v n ) = det(u + u,..., v i 1, v 1,..., v n ) = det(u,..., v i 1, v 1,..., v n ) det(u,..., v i 1, v 1,..., v n ) = det(v 1,..., v i 1, u,..., v n ) + det(v 1,..., v i 1, u,..., v n ). The first and third equalities follow from Axiom 1, while the second equality follows from Axiom 2. (3) The proof of part 3 is similar to part 2, and is omitted. (4) For any constant c, det(v 1,..., v i + cv j,..., v j..., v n ) = det(v 1,..., v i,..., v j..., v n ) + det(v 1,..., cv j,..., v j..., v n ) = det(v 1,..., v i,..., v j..., v n ) + c det(v 1,..., v j,..., v j..., v n ) = det(v 1,..., v i,..., v j..., v n ). The first and second equalities follows from Axiom 2 and 3, respectively. The last equality follows from the first part of the proposition. (5) Suppose that the components of v i are all zero. Let c be any real number which is not equal to 1. We have cv i = v i = [0 0... 0], and by part (3) of the proposition, we get Therefore det(v 1,..., v i 1, v i,..., v n ) = det(v 1,..., v i 1, cv i,..., v n ) This implies that det(v 1,..., v i 1, v i,..., v n ) = 0. = c det(v 1,..., v i 1, v i,..., v n ). (1 c) det(v 1,..., v i 1, v i,..., v n ) = 0. We now verify that the second definition of determinant satisfies the four axioms. Proposition 3. The function defined in (3) satisfies Axioms 1 to 4. 5

Proof In order to avoid potential confusion, we let v i = [a i1 a i2... a in ] be row vector of length n, for i = 1, 2,..., n, and let f(v 1, v 2,..., v n ) be a function defined as f(v 1, v 2,..., v n ) := ( 1) inv(p 1,p 2,...,p n ) a 1,p1 a 2,p2 a n,pn, (5) where the summation is over all possible permutations (p 1,..., p n ) of 1 to n. (Verification of Axiom 4) Let v i be the i-th basis vector e i. Then { 1 if i = j, a ij = 0 otherwise. Hence, all terms in (5) are zero except the term corresponding to (p 1, p 2,..., p n ) = (1, 2,..., n). This gives f(e 1, e 2,..., e n ) = inv(1, 2,..., n)a 11 a 22 a nn = 1 1 1 1 = 1. (Verification of Axiom 3) Suppose that the i-th row v i is multiplied by a constant c. We have f(v 1, v 2,..., cv i,..., v n ) = ( 1) inv(p 1,p 2,...,p n ) a 1,p1 a 2,p2 (ca i,pi ) a n,pn, = c ( 1) inv(p1,p2,...,pn) a 1,p1 a 2,p2 a i,pi a n,pn, = cf(v 1, v 2,..., cv i,..., v n ). (Verification of Axiom 2) The verification of Axiom 2 is similar to the verification of Axiom 3, and is omitted. (Verification of Axiom 1) For Axiom 1, we need the property that if we swap two numbers in (p 1, p 2,..., p n ), then the parity of the inversion number changes, i.e., if the inversion number is even before the swapping, then the inversion after the swapping is odd, and vice versa. Suppose that we interchange row i and row j, with i j. To express the idea algebraically, we let v i = [a j1 a j2... a jn ] v j = [a i1 a i2... a in ] p i = p j, and p j = p i. The determinant of the matrix after exchanging rows i and j is f(v 1, v 2,..., v i,..., v j,..., v n ) = ( 1) inv(p 1,p 2,...,p n ) a 1,p1 a 2,p2 a j,pi a i,pj a n,pn = = (p 1,..., p i,..., p j,...,p n) ( 1) inv(p1,p2,...,pn) a 1,p1 a 2,p2 a i,pj a j,pi a n,pn ( 1) ( 1) inv(p 1,..., p i,..., p j,...,p n ) = f(v 1, v 2,..., v i,..., v j,..., v n ). a 1,p1 a 2,p2 a i, pi a j, pj a n,pn 6

This completes the proof of the proposition. 4 Calculation of determinant by row reductions We have verified that we can construct a function which satisfies the four axioms of determinant. The next question we want to ask is Can we construct two different functions, both of which satisfy the four axioms of determinant? The answer is no, because the function value is uniquely determined by the four axioms. Given an n n matrix A, we can calculate the determinant of A by row-reducing it to an upper triangular or a lower triangular matrix, and use the property that the determinant of an upper triangular or a lower triangular matrix is equal to the product of the diagonal entries. The effect of elementary row operations on the determinant is summarized in the following table. Elementary row operation Effect on determinant Exchange two rows Multiply by 1 Multiply a row by a non-zero constant c Multiply by c Add a constant multiple of a row to another row No change Example: Find the determinant of 2 0 1 2 A = 0 1 2 2 1 2 2 0. 2 2 0 1 We first exchange row 1 and row 3, in order to make the top-left corner equal to 1, 1 2 2 0 det(a) = det 0 1 2 2 2 0 1 2. 2 2 0 1 The value is multiplied by 1, because of Axiom 1. After subtracting two times row 1 from row 3, and subtracting two times row 1 from row 4, we get 1 2 2 0 det(a) = det 0 1 2 2 0 4 3 2. 0 2 4 1 The value does not change after these two row operations. Now, we add four times row 2 to row 3, and 2 times row 2 to row 4, 1 2 2 0 det(a) = det 0 1 2 2 0 0 5 10. 0 0 0 5 The matrix on the right-hand side is upper triangular. The entries above the diagonal becomes zero after a few more row reductions of the third type, 1 0 0 0 det(a) = det 0 1 0 0 0 0 5 0 0 0 0 5 7

We can pull out the two factor of 5 from the determinant by Axiom 3, and obtain 1 0 0 0 det(a) = 5 5 det 0 1 0 0 0 0 1 0 = 25. 0 0 0 1 The last equality follows from Axiom 4. Example: Find the determinant of B = 2 3 0 0 1 1 2 4 1 by row reduction. After subtracting row 1 and row 2 from row 3, we obtain 2 3 0 0 1 1. 0 0 0 As the last row is all-zero, the determinant is zero by part (5) of Proposition 2, det B = det 2 3 0 0 1 1 = 0. 0 0 0 We note that the calculations in these two examples only depend on the properties of determinant. In general, we have the following Proposition 4. There is one and only one function which satisfies the four axioms of determinant. From now on, any function which satisfies the four axioms is called the determinant function. 5 Using determinant to test linear independence Given a set of n vectors of length n, we can test whether they are linearly independent by putting these n vectors together to form an n n matrix, and calculating the determinant. If the determinant is non-zero, then the n vectors are linearly independent, otherwise they are linearly dependent. Proposition 5. Let u 1, u 2,..., u n be column vectors of length n, and let M be the n n matrix whose columns are precisely u 1, u 2,..., u n. We have det M = 0 if and only if u 1, u 2,..., u n are linearly dependent. Proof Let c 1, c 2,..., c n be real numbers such that c 1 u 1 + c 2 u 2 +... + c n u n = 0. The real numbers c 1, c 2,..., c n are solution to c 1 0 c 2.. M = 0.. (6) 0 c n 8

We want to test whether we can find a non-trivial solution to (6). We row-reduce the matrix M to reduced row echelon form (RREF). Recall that whether the determinant of M is zero or not is invariant under elementary row operations. Therefore, the determinant of M is non-zero if and only if the determinant of the RREF is nonzero. We count the number of pivots in the RREF and consider two cases: (i) there are exactly n pivots in the RREF, (ii) there are strictly less than n pivots in the RREF. We we call that in RREF, each row contains at most one pivot, and each column contains at most one pivot. Case 1, there are exactly n pivots: Since each pivot occupies one row and one column, there is exactly one pivot in each row and each column. The RREF of M is the n n identity matrix, which has determinant equal to 1 by Axiom 4. Hence det(m) 0 in this case. Since M can be row-reduced to the identity matrix, the only choice for c 1 to c n is c 1 = c 2 = = c n = 0. Therefore u 1, u 2,..., u n are linearly independent. Case 2, there are strictly less than n pivots: The last row of the RREF of M is an all-zero row. The determinant of the RREF is zero by part (5) of Proposition 2. By part (5) of Proposition 2, det(m) = 0 in this case. Since the last row is zero, the number of variables, n, is strictly larger than the number of essentially different equations. We can find (infinitely many) non-trivial solution to the system of linear equations in (6). This implies that u 1, u 2,..., u n are linearly dependent. Example: We want to test whether the following four vectors are linearly independent, 2 0 1 2 0 1, 1 2, 2 2, 2 0. 2 2 0 1 Concatenate the four vectors and form the matrix 2 0 1 2 0 1 2 2 1 2 2 0. 2 2 0 1 We have already calculated that the determinant of this matrix is 25, which is non-zero. Therefore, these four vectors are linearly independent. 6 Computing determinant by column expansion We used several times the property that if a matrix contains an all-zero row, then the determinant is zero. Similar thing are true for columns. Proposition 6. If a matrix contains an all-zero column, then the determinant of this matrix is zero. Proof Let the columns of the matrix be denoted by u 1, u 2,..., u n. Suppose that the j-th column, u j, is the zero vector. Then, columns vectors u 1, u 2,..., u n are linearly dependent, because the linear combination c 1 u 1 + c 2 u 2 +... + c j u j +... + c n u n is equal to the zero vector by setting c j = 1 and the rest of the coefficients to be zero. From Proposition 5, we conclude that the determinant of the matrix is zero. 9

We can now show that the determinant can be calculated by Laplace expansion on the first column. Consider the 4 4 case as an example: a e The first row can be decomposed as a sum of two vectors [a ] = [a 0 0 0] + []. By the second axiom of determinant, we get a a 0 0 0 e = e + e By Laplace expansion on the first row, the first determinant on the right hand side can be simplified to a 3 3 determinant, a e = a v w x + e Repeat the same procedure to the second row of e, we obtain a e = a v w x + e 0 0 0 + 0 e 0 0 0 = a v w x + 0 = a v w x e v w x + 0. We have used the property that the determinant is multiplied by 1 after exchanging two rows. Apply the 10

same steps to rows 3 and 4, we get a v w x e v w x + q v w x + 0 0 = a v w x e v w x + q v w x u + 0 0 0 v w x = a v w x e v w x + q v w x u. In the last step, we have used the property that if there is a zero column vector in a matrix, then the determinant is zero. In general, we have the following Proposition 7. Let A be an n n matrix and A ij be the determinant of the submatrix obtained by deleting row i and column j. The determinant of A can be computed by expansion on the first column, Example: det A = a 11 A 11 a 21 A 21 + a 31 A 31... + ( 1) n+1 a n1 A n1. det 2 3 1 0 1 4 = 2 1 4 2 0 0 3 1 2 0 + 1 3 1 1 4 = 2(8) + (13) = 29. 1 2 0 7 Duality between rows and columns Proposition 8. Let T denote the transpose operator. We have for square matrix A. Proof det(a T ) = det(a) We can proceed by mathematical induction. The statement is true for 2 2 matrix, [ ] [ ] a b a c det = ad bc = det. c d b d Suppose that det(b T ) = det(b) for all (n 1) (n 1) matrices B. Let A be an n n matrix, and let A ij be the minor corresponding to the (i, j)-entry. We let a ij = a ji be the (i, j)-entry of matrix A T, and A ij be the minor corresponding to the (i, j)-entry of AT. By the induction hypothesis, we have A ij = A ji. We calculate the determinant of A by expanding on the first column, det A T = a 11A 11 a 21A 21 + a 31A 31... + ( 1) n+1 a n1a n1 = a 11 A 11 a 12 A 12 + a 13 A 13... + ( 1) n+1 a 1n A 1n. The last line is precisely the Laplace expansion of det(a) on the first row. Therefore det A T = det A. Using Proposition 8, every row property of determinant can be translated to a column property. For example, if we exchange two columns of a matrix, the determinant is multiplied by 1. If we add a scalar multiple of a column to another column, the determinant does not change. Combining Propositions 5 and 8, we obtain the following important theorem 11

Theorem 9. For an n n matrix A, the followings are equivalent: 1. the columns of A are linearly independent. 2. det(a) 0. 3. det(a T ) 0. 4. the rows of A are linearly independent. ( Equivalence means that if one statement is true, then all the other statements are also true, but if one statement is false, then all the other statements are also false.) Proof The equivalence between (1) and (2) is precisely the content of Proposition 5. The equivalence between (2) and (3) follows directly from Proposition 8. The equivalence between (3) and (4) follows from the dual of Proposition 5, by replacing all columns by rows, and by A by A T. The following corollary is an immediate consequence of Theorem 9. Corollary 10. Let A be an n n matrix. The system of linear equation has a non-trivial solution if and only if det(a) = 0. Example Consider the system of linear equations Ax = 0 x + y + z = 0 sx 2y z = 0 x y + 2z = 0 where s is a parameter. We want to find the value of s so that we can find a non-trivial solution to the above linear system. By Corollary 10, we solve det 1 1 1 s 2 1 = 0 1 1 2 for the parameter s. The above determinant is equal to det 1 1 1 s 2 1 = det 1 1 1 s 2 1 1 1 2 0 2 1 [ 2 1 = 1 det 2 1 = 4 3s. ] s det [ 1 ] 1 2 1 The first equality follows by subtracting row 1 from row 3, and the second by expansion on the first column. By setting the determinant to zero, we see that the system of linear equations has non-trivial solution if and only if s = 4/3. Indeed, x = 2427, y = 809, z = 1618 is a non-trivial solution when s = 4/3. When s 4/3, the only solution is x = y = z = 0. 12