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**Unformatted text preview: **ince zero
multiplied with any other numbers is always zero. Thus we know that any matrix 10. Matrices 285 which has an inverse matrix satisﬁes det(A) = 0. If A has in inverse matrix
then we have from (10.12) that
det(A−1 ) = 1
.
det(A) (10.13) In fact, for any square matrix A, the condition det(A) = 0 guarantees that A has
an inverse matrix.
Theorem 10.43 (criterion for having an inverse matrix)
A square matrix A is invertible, and thus has an inverse matrix, if and only
if its determinant det(A) is non-zero, that is,
det(A) = 0.
Now that we have a criterion for checking whether a matrix is invertible, that is,
has an inverse, it remains to learn how to compute the inverse matrix. This
leads to the deﬁnition and the theorem below.
Deﬁnition 10.44 (adjoint matrix)
The adjoint of an n × n square matrix A is that n × n square matrix adj(A)
whose entry in the ith row and j th column is the cofactor (−1)j +i Cj,i of
aj,i (note the changing order of the subscripts!). In formulas,
[adj(A)]i,j = (−1)j +i Cj,i for 1 ≤ i, j ≤ n. With the help of the adjoint matrix, we can now easily give a formula for the inverse
matrix of a square matrix.
Theorem 10.45 (formula for inverse matrix)
Suppose that A is an n × n square matrix with det(A) = 0. Then A is invertible
and the inverse matrix A−1 of A is given by
A−1 = 1
adj(A).
det(A) Example 10.46 (special case: inverse of a 2 × 2 matrix)
Suppose that we have a 2 × 2 matrix
A= ab
cd . 286 10. Matrices Then det(A) = a d − b c. The matrix of cofactors of A is
d −c
−b
a . So the adjoint of A is
adj(A) = T d −c
−b
a = d −b
−c
a . So if a d = b c, then the inverse matrix A−1 of A is
ab
cd −1 = A−1 = 1
1
adj(A) =
det(A)
ad − bc d −b
c
a . (10.14) The formula (10.14) for the inverse of a 2 × 2 matrix with det(A) = a d − b c = 0 is
easy to remember and worth knowing from memory.
2
Example 10.47 (inverse of a 2 × 2 matrix)
We use the formula (10.14) to determine the inverse of the matrix
A= 23
34 from Example 10.41. The determinant is det(A) = 2 × 4 − 3 × 3 = 8 − 9 = −1.
Thus from (10.14)
A−1 = 1
(−1) 4 −3
−3
2 = −4
3
3 −2 , and we have the matrix B from Example 10.41. 2 Example 10.48 (inverse of a 2 × 2 matrix)
Determine whether the 2 × 2
32
A=
53
is invertible, and if yes ﬁnd its inverse matrix A−1 .
Solution: Since det(A) = 3 × 3 − 2 × 5 = 9 − 10 = −1 = 0, we know that the matrix
A is invertible. From (10.14), its inverse is given by
A−1 = 1
(−1) 3 −2
−5
3 = −3
2
5 −3 . 2 10. Matrices 287 Example 10.49 Determine whether the matrix 2
3
A=
1 1 −3
1
0 .
2
3 is invertible, and if yes, ﬁnd its inverse matrix.
Solution: From the computations in Example 10.36, we can deduce that this matrix
has the determinant
det(A) = −18.
Thus A is invertible. Furthermore, the matrix (Ci,j ) of cofactors of A is 1×3−0×2
(Ci,j ) = −(1 × 3 − (−3) × 2)
1 × 0 − (−3) × 1 3 −9
5
= −9
9 −3 .
3 −9 −1
Hence −(3 × 3 − 0 × 1)
2 × 3 − (−3) × 1
−(2 × 0 − (−3) × 3) 3×2−1×1
−(2 × 2 − 1 × 1) 2×1−1×3 T 3 −9
3
3 −9
5
adj(A) = −9
9 −9 .
9 −3 = −9
5 −3 −1
3 −9 −1 It fo...

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