- •1 Solving Linear Equations
- •2 Matrices
- •3 Inverses of Matrices
- •4 Matrices and Row Operations
- •5 Finding the Inverse of a Matrix
- •6 The Determinant
- •7 The Determinant via Elimination
- •8 Span
- •9 Bases
- •10 Kernel and Image
- •11 Eigenvalues and Eigenvectors
- •12 Bases of Eigenvectors
- •13 Inner Product
- •14 The Spectral Theorem
- •15 Complex Vectors
- •Hints
- •Bibliography
- •List of Notation
- •Index
Chapter 7
The Determinant via Elimination
The fast way to compute the determinant of a large matrix is via elimination.
The fast formula for the determinant
Theorem 7.1. Via forward elimination,
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det A = |
± (product of the pivots) if there is a pivot in each column, |
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where |
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In particular, A is invertible just when det A 6= 0.
Forward elimination takes
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A = 0 |
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to U = |
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with one row swap so det A = −(2)(7) = −14.
The fast formula isn’t actually any faster for small matrices, so for a 2 × 2 or 3 × 3 you wouldn’t use it. But we need the fast formula anyway; each of the two formulas gives di erent insight.
Proof. We can see how the determinant changes during elimination: adding multiples of rows to other rows does nothing, swapping rows changes sign. 

7.1 Use the fast formula to find the determinant of
2 5 5
A = 2 5 |
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2 6 11
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The Determinant via Elimination |
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7.2 Just by looking, find |
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det 2343 |
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7.3 Prove that a square matrix is invertible just when its determinant is not |
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Review problems |
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7.4 Find the determinant of |
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7.5 Find the determinant of |
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7.6 Find the determinant of |
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7.7 Find the determinant of |
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7.8 Find the determinant of |
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7.9 Find the determinant of |
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Determinants Multiply |
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7.10Prove that a square matrix with a zero row has determinant 0.
7.11Prove that det P A = (−1)N det A if P is the permutation matrix of a product of N transpositions.
7.12Use the fast formula to find the determinant of
0 2 1
A = 3 |
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7.13 Prove that the determinant of any lower triangular square matrix
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L21 |
L22 |
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L11 |
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(with zeroes above the diagonal) is the product of the diagonal terms: det L =
L11L22 . . . Lnn.
Determinants Multiply
Theorem 7.2. det (AB) = det(A) det(B), for any square matrices A and B of the same size.
Proof. Suppose that det A = 0. By the fast formula, A is not invertible. Problem 5.10 on page 43 tells us that therefore AB is not invertible, and both det(AB) and det(A) det(B) are 0. So we can safely suppose that det A 6= 0. Via Gauss-Jordan elimination, any invertible matrix is a product of matrices each of which adds a multiple of one row to another, or scales a row, or swaps two rows. Write A as a product of such matrices, and peel o one factor at a time, applying lemma 6.4 on page 48 and proposition 6.7 on page 51. 
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then it is hard to compute out AB, and then compute out det AB. But det AB = det A det B = (1)(2)(3)(1)(2)(4) = 48.
56 The Determinant via Elimination
Transpose
The transpose of a matrix A is the matrix At whose entries are Atij = Aji (switching rows with columns). Flip over the diagonal:
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A = |
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Find the transpose of |
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Prove that |
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(The transpose of the product is the product of the transposes, in the reverse order.)
7.16 Prove that the transpose of any permutation matrix is a permutation matrix. How is the permutation of the transpose related to the original permutation?
Corollary 7.3.
det A = det At
Proof. Forward elimination gives U = V A, U upper triangular and V a product of permutation and strictly lower triangular matrices. Tranpose: Ut = AtV t. But V t is a product of permutation and strictly upper triangular matrices, with
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swaps as V , so det V t = det V = 1. The matrix Ut |
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problem 7.13 on the preceding page), which are the diagonal entries of U, so det Ut = det U. 
Expanding Down Any Column or Across Any Row
Consider the “checkerboard pattern”
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Theorem 7.4. We can compute the determinant of any square matrix A by picking any column (or any row) of A, writing down plus and minus signs from the same column (or row) of the checkboard pattern matrix, writing down the
Summary |
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entries of A from that column (or row), multiplying each of these entries by the determinant obtained from deleting the row and column of that entry, and adding all of these up.
For
A = |
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if we expand along the second row, we get
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det A = − (1) det |
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Proof. By swapping columns (or rows), we change signs of the determinant. Swap columns (or rows) to get the required column (or row) to slide over to become the first column (or row). Take the sign changes into account with the checkboard pattern: changing all plus and minus signs for each swap. 
7.17 Use this to calculate the determinant of |
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Summary
Determinants
(a)scale when you scale across a row (or down a column),
(b)add when you add across a row (or down a column),
(c)switch sign when you swap two rows, (or when you swap two columns),
(d)don’t change when you add a multiple of one row to another row (or a multiple of one column to another column),
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The Determinant via Elimination |
(e)don’t change when you transpose,
(f)multiply when you multiply matrices. The determinant of
(a)an upper (or lower) triangular matrix is the product of the diagonal entries.
(b)a permutation matrix is (−1)# of transpositions.
(c)a matrix is not zero just when the matrix is invertible.
(d)any matrix is det A = (−1)N det U, if A is taken by forward elimination with N row swaps to a matrix U.
7.18If A is a square matrix, prove that
det Ak = (det A)k
for k = 1, 2, 3, . . . .
7.19 Use this last exercise to find
det A2222444466668888
where
A = |
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7.20 If A is invertible, prove that
det A−1 = det1 A.
Review problems
7.21What are all of the di erent ways you know to calculate determinants?
7.22How many solutions are there to the following equations?
x1 + 1010x2 + 130923x3 = 2839040283 2x2 + 23932x3 = 2390843248 3x3 = 98234092384
7.23 Prove that no matter which entry of an n × n matrix you pick (n > 1), you can find some invertible n × n matrix for which that entry is zero.
Bases and Subspaces
59
