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**Unformatted text preview: **det(A B ) = det(A) × det(B ) = 1 × 6 = 6.
(b) Computing the matrix A B yields 111
100
653
AB = 0 1 1 2 2 0 = 5 5 3 .
001
333
333
To compute the determinant det(A B ), we use that, from Lemma 10.35 (iii), we
may subtract the second row of the matrix from the ﬁrst row without changing the
new
value of the determinant (that is, r1 = r1 − r2 ). Thus
det(A B ) = 653
553
333 = 6−5
5
3 5−5
5
3 3−3
3
3 = 100
553.
333 Now we expand about the ﬁrst row and get det(A B ) = 100
553
333 =1× 53
33 We get indeed the same result. 10.4 = 1 × (5 × 3 − 3 × 3) = 15 − 9 = 6. 2 Invertible Matrices In this section, we introduce the inverse matrix of an (invertible) square n × n
matrix A. The inverse matrix A−1 of A satisﬁes A A−1 = A−1 A = In . If we compare 10. Matrices 283 invertible matrices A to real numbers x = 0, then the inverse matrix A−1 corresponds
to the number 1/x. We learn that a matrix A has an inverse matrix if and only
if det(A) = 0. We introduce the adjoint matrix, and we learn a formula for the
inverse matrix which involves the adjoint matrix and the determinant.
Deﬁnition 10.40 (invertible and inverse matrix)
Let A be an n × n matrix. If there exists an n × n matrix B such that
A B = In and B A = In , (10.10) then A is invertible. The matrix B satisfying (10.10) is called the inverse
(matrix) of A, and usually the inverse matrix of A is denoted by A−1 .
In Deﬁnition 10.40 we would prefer that, if there exists a matrix B satisfying (10.10),
then this matrix is the only matrix satisfying (10.10). In other words, we would
like the inverse matrix to be unique. To show that this is indeed true, assume that
there exits another n × n matrix C satisfying
A C = In and C A = In . (10.11) Multiplying the ﬁrst formula in (10.11) from the left by the matrix B yields
B (A C ) = B In ⇒ (B A) C = B ⇒ In C = B ⇒ C = B, where we have used the associative law of matrix multiplication (see Theorem 10.20)
and (10.1). Since C = B , we see that the inverse matrix is indeed uniquely determined.
Example 10.41 (matrix and its inverse)
Consider the matrices A and B , given by
A= 23
34 and B= −4
3
3 −2 . Show that the matrix B is the inverse matrix to the matrix A.
Solution: We compute A B and B A, and ﬁnd
AB =
= 23
34 −4
3
3 −2 2 × (−4) + 3 × 3
3 × (−4) + 4 × 3 2 × 3 + 3 × (−2)
3 × 3 + 4 × (−2) 284 10. Matrices =
BA = 10
01 , −4
3
3 −2 23
34 = (−4) × 2 + 3 × 3
3 × 2 + (−2) × 3 = 10
01 (−4) × 3 + 3 × 4
3 × 3 + (−2) × 4 . Since A B = B A = I2 , we know that B is the inverse matrix to A. We also see that
A is the inverse matrix of B .
2
Remark 10.42 (not every matrix is invertible)
It is easy to see that the zero matrix
O= 00
00 does not have an inverse and is thus not invertible. Indeed, if B is any other 2 × 2
matrix, then
OB = 00
00 B= 00
00 and BO =B 00
00 = 00
00 . We see that not every square matrix is invertible.
If an n × n matrix A has an inverse A−1 , then taking the determinant on both sides
of the equation
In = A−1 A
and using (10.9) yields
det(In ) = det(A−1 A) = det(A−1 ) det(A).
From (10.7), we know that det(In ) = 1, and thus
1 = det(A−1 ) det(A). (10.12) From (10.12), we know that neither det(A) nor det(A−1 ) can be zero, s...

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