Exercises 2.4: Partitioned Matrix
In Exercises 5 and 7, find formulas for X, Y , and Z in terms of A, B, and C, and justify your
calculations. In some cases, you may need to make assumptions about the size of a matrix in order
to produce a formula.
5.
!
A B
C
0
"!
I
0
X Y
"
=
!
0
I
Z 0
"
S OLUTION : Compute the left side of the equation:
!
" !
" !
"
A B
I 0
AI + BX A0 + BY
=
=
C 0
X Y
CI + 0X C0 + 0Y
Set this equal to the right side of the equation:
!
"
A + BX BY
C
0
=
!
0
I
Z 0
"
so that
A + BX = 0 BY = I
C=Z
0=0
Clearly, Z = C. Assume B and Y are square. Then the equation BY = I implies that B is
invertible, and Y = B →1 . Since BX = →A, we have
B →1 BX = B →1 A,
7.
!
and X = →B →1 A.
" A Z
!
"
X 0 0
I 0
0 0 =
Y 0 I
0 I
B I
S OLUTION : Compute the left side of the equation:
!
" A Z
!
"
X 0 0
XA + 0 + 0B XZ + 0 + 0I
0 0 =
Y 0 I
Y A + 0 + IB Y Z + 0 + II
B I
1
Set this eqault to the right side of the equation:
!
"
XA
XZ
Y A + IB Y Z + I
=
!
I
0
0 I
"
so that
XA = I
^
XZ = 0
YA+B =0 YZ +I =I
Assume that A and X are square. The equation XA = I implies that A is invertible and
X = A→1 . Also X is invertible. Since XZ = 0, X →1 XZ = IZ = 0, so Z = 0. Finally,
∅
Y A = →B, and Y = →BA→1 .
9. Suppose A11 is an invertible matrix. Find matrices X and Y such that the product below has
the form indicated. Also, compute B22 .
I 0 0
A11 A12
B11 B12
X I 0 = A21 A22 = 0 B22
Y 0 I
A31 A32
0 B32
S OLUTION : Compute the left side of the equation:
I 0 0
A11 A12
IA11 + 0A12 + 0A31 IA12 + 0A22 + 0A32
X I 0 A21 A22 = XA11 + IA21 + 0A31 XA12 + IA22 + 0A32
Y
0 I
A31 A32
Y A11 + 0A21 + IA31
Y A12 + 0A22 + IA32
Set this equal to the right side of the equation:
A11
A12
B11 B12
XA11 + A21 XA12 + A22 = 0 B22
Y A11 + A31 Y A12 + A32
0 B32
so that
A11 = B11
A12 = B12
XA11 + A21 = 0 XA12 + A22 = B22
Y A11 + A31 = 0 Y A12 + A32 = B32
Since A11 is invertible, we have
X = →A21 A→1
11
and Y = →A31 A→1
11
Then
B22 = →A21 A→1
11 A12 + A22
2
0
