Maths8

Országok listájaHungaryBudapesti Corvinus EgyetemGazdálkodástudományi KarNemzetközi gazdálkodás (angol nyelven)Mathematics1JegyzetekMaths8

Matrices
8.1 Concept of matrices

In our everyday life we often meet tables of numbers containing statistical data, coecients of system of linear equations and so on. In linear algebra this tables of numbers are referred to as matrices.

matrix is

Denition 8.1

A rectangular shaped array of numbers is called matrix, that is, a
11 21 A= m1

...

12 22 m2

...

... ... ...

...

where (i = 1, . . . , m; j = 1, . . . , n) are numbers. If the matrix has m rows and n columns, then it is said to be of type m × n. If m = n then we call it square matrix of order m. Here is the entry in the intersection of the ith row and j th column.
ij ij

1n 2n , mn

...

If the entries of the matrix are real numbers, then we say that A is a real matrix, and the set of all real matrices of type m × n will be denoted by Rm×n . For short we often reer to a matrix A by its general entry and we denote it as [ij ]. If a matrix has only one column, then we call it column matrix or more often column vector and similarly if the matrix consists of only one row, then we call it row matrix or row vector. In some cases we partition the matrix into submatrices A11 A12 . . . A1k A21 A22 . . . A2k A= . . . .. . . . . . . .

A

1

A

2

...

A

k

In particular, we often use the partition, where the submatrices are the columns a1 a2 or rows of the matrix A = [a1 , a2 , . . . , an ] = . , where ai denotes the ith . . column and a j the j th row of A.
8.1.1 Operations on

am

Rm×n
A + B = [ij + ij ].
1

The sum of two matrices of the same type A = [ij ] and B = [ij ] is the matrix

8.1.

CONCEPT OF MATRICES

If R and A = [ij ] Rm×n then we dene the multiplication of a matrix by scalar A as follows: A = [ij ]. Combining the addition and multiplication by scalars we can form the linear combination of matrices of the same type: 1 , . . . , k R and A1 , . . . , Ak Rm×n , then their linear combination is

1 A1 + · · · + k Ak Rm×n .

Example 8.1

combination 2A - 3B is
2A - 3B

If A =

1 6

2 5

3 4 2 12

and
4 10
2A

B =

-1 2 -3 6

0 1 -2 3 0 3 -6 9

then the linear
=

=

6 8

-

3B

= =

2 - (-3) 12 - 6 5 6 4 16 3 -1

4-0 10 - (-6) .

6-3 8-9

=

Similarly to the operations of vectors, the addition of matrices is also led back to the addition of numbers, and the multiplication of matrices by scalars is led back to the multiplication of numbers, therefore the following properties are obviously hold true. If A, B, C are arbitrary matrices in Rm×n and , R, then 1. A + B = B + A, 2. (A + B) + C = A + (B + C), 3. O : A : O + A = A, 4. A : (-A) : A + (-A) = O. The addition of matrices is commutative, associative, there exists zero matrix (each entry is 0), and each matrix has an opposite (additive inverse). 1. (A + B) = A + B, 2. ( + )A = A + A, 3. ()A = (A), 4. 1A = A. The set Rm×n of matrices with the addition and scalar multiplication is a real vector space. While the addition of matrices was dened for matrices having the same type, the multiplication of two matrices is dened only if the number of columns in the left side matrix equals the number of rows in the right side one.

Multiplication of matrices

Denition 8.2

Let A = [ ] be an m × n matrix and B = [ ] be an n × p matrix, then their product AB = C = [ ] is the m × p matrix for which
ij jk ij

ij = i1 1j + i2 2j + · · · + in nj .
2

8.1.

CONCEPT OF MATRICES

Example 8.2
AB

If A =

is

1 6

2 5

3 4

and

-1 B = 2 0



0 -2 1

then their product
3 4 1 -6 .

1 · (-1) + 2 · 2 + 3 · 0 1 · 0 + 2 · (-2) + 3 · 1 6 · (-1) + 5 · 2 + 4 · 0 6 · 0 + 5 · (-2) + 4 · 1

=

Multiplying the matrix of 11 12 21 22 A= . . . . . .

m1
the product

m2

type m × n . . . 1n . . . 2n . .. . . . . . . mn


by the vector

x=

x1 x2 . . . xn



Ax = = x1

x1 11 + x2 12 + · · · + xn 1n x1 21 + x2 22 + · · · + xn 2n . . .

= ,

x1 m1 + x2 m2 + · · · + xn mn 1n 12 11 2n 22 21 . + x2 . + · · · + xn . . . . . . . mn m2 m1

is the linear combination of the columns of A, where the scalar coecients are the components of x. Similarly, the product of a row vector and a matrix is a row vector, which is a linear combination the rows of the matrix, but its proof remains a homework. of a1 a2 If A = . , where a i denotes the ith row of A and B = [b1 , b2 , . . . , bp ], . .

am where bj is the j th column of B, a 1 b1 a 1 b2 a 2 b1 a 2 b2 AB = . . . . . . a m b1 a m b2

then their product is . . . a 1 bp . . . a 2 bp = [Ab1 , Ab2 , . . . , Abp ] = . .. . . . . . . a m bp

=

a 1B a 2B . . . a mB



The ij th entry a i bj of the product is the product of the row matrix a i and column matrix bj , which is a scalar like the inner product of two vectors. Thus each column of the matrix product AB is the linear combination of the columns of A and each row of AB is the linear combination of the rows of B.

3

8.1.

CONCEPT OF MATRICES

Proposition 8.1

Properties of matrix multiplication: 1. noncommutative, 2. associative, 3. distributive with respect to the addition. Proof: 1) If A = 1 1 and B = 1 1 0 1 0 0

, then the product AB =

1 1 1 2 , while the product BA = , verifying the noncommutative prop0 0 0 0 erty of the matrix multiplication. 2) If A = [ij ] B = [jk ] and C = [k ] are matrices of type m × n, n × p and p × q respectively, then their product
p p n

[(AB)C]st =
u=1

[AB]su ut =
u=1 p n v=1

sv vu

ut =

=
u=1 v=1

(sv vu )ut ,
n p

and

n

[A(BC)]st =
v=1

sv [BC]vt =
v=1 p n

sv
u=1

vu ut

=

=
u=1 v=1

sv (vu ut ).

The proof of the distributive property remains a homework.

Using the multiplication of matrices we can dene the powers of matrices (number of rows and columns are equal) with non-negative integer exponents as follows: if A Rn×n then

square

where En =

1 0 . . .

A0 := En and k N : Ak+1 := Ak · A. 0 ... 0 1 ... 0 is the n × n identity matrix. Notice that for an . . ··· . . . .

0 0 ... 1 arbitrary matrix A Rm×n : AEn = Em A = A, that is, the identity matrix plays the same role among the matrices as the 1 among the numbers.

system. The following theorem says that the row vector system of a matrix could also be used to dene the rank of the matrix.

Denition 8.3 Let A = [a , a , . . . , a ] be a matrix. The rank (A) of the matrix is the rank of the vector system {a , a , . . . , a }. Remark: We dened the rank of a matrix to be the rank of its column vector
1 2 n 1 2 n

The system of column vectors and the system of row vectors of a matrix have the same rank.
Theorem 8.2

4

8.2.

TRANSPOSE OF MATRICES

8.2

Transpose of matrices
ij m×n

The transpose of a matrix A = [ ] R is the matrix A of type n × m is obtained by exchanging A s columns and rows. Properties:
Denition 8.4
1. (A + B)

=A +B ,

2. (A) = A , 3. (AB) = B A .
8.3 System of linear equations

Denition 8.5

The general form of a system of linear equations is
11 x1 + 12 x2 + · · · + 1n xn 21 x1 + 22 x2 + · · · + 2n xn = 1 = 2

...

m1 x1 + m2 x2 + · · · + mn xn

= m

ij i = 1, 2, . . . m, j = 1, 2, . . . , n xj j = 1, 2, . . . , n

where

are the unknowns.

and

i

i = 1, 2 . . . , m

are given scalars and

The system of linear equations is said to be if i = 0 for all i = 1, . . . , m, otherwise it is called Collecting the coecients in the matrix A = [ij ] the unknowns and the rightside scalars in the column vectors 1 x1 2 x2 x = . and b = . . . . .

homogeneous nonhomogeneous.

xn

m

respectively, the matrix equation Ax = b can be obtained. Since the left side of the above matrix equation is a linear combination of the columns of A the following statement can be obtained:

Proposition 8.3 The (8.5) linear system of equations is solvable if and only if b is the linear combination of columns of the coecient matrix A, that is, b is in the subspace spanned by the columns of A, or equivalently (A) = ([A, b]).
A solution of the system of linear equations Ax = b is an n-tuple (a vector) s = (s1 , s2 , . . . , sn ), such that substituting si for xi i = 1, 2, . . . , n each equation of the system (8.5) becomes true numerical statement. The solution set of (8.5) is the set of all solution vectors. To provide an eective method for solution of systems of linear equation rst we show a process for factoring matrices.

5

8.3.

SYSTEM OF LINEAR EQUATIONS

Basis factoring matrices

Let A = [a1 , a2 , . . . , an ] be a matrix having rank r. Then there is an r element linearly independent set of vectors ai1 , . . . , air such that any vector of the subspace spanned by the columns of A which is called the of A can be expressed as a linear combination of the vectors ai1 , . . . , air , that is, ai1 , . . . , air is a basis of the column space of A. Therefore

column space

a1 a2

= d11 ai1 + d21 ai2 + · · · + dr1 air = d12 ai1 + d22 ai2 + · · · + dr2 air . . . = d1n ai1 + d2n ai2 + · · · + drn air .

an

Form the matrices A1 = [ai1 , . . . , air ] and d11 d12 d21 d22 A2 = . . .

... ... ...

dr1

dr2

d1n d2n , drn

then the above equation can be written in the form:

A = A1 · A2 .
Notice that the j th column of the matrix A2 is just the coordinate vector of aj with respect to the basis {ai1 , . . . , air }.

Example 8.3

Factor the matrix

1 A= 2 -1



2 1 1

0 3 3 3 . -3 0

Using elementary basis transformation we nd the rank of A and a basis of its column space, starting with the standard basis {e1 , e2 e3 } of R3 .

Solution:
e1 e2 e3

a1 1 2 -1

a 2 a 3 a4 a1 2 0 3 - e2 1 3 3 1 -3 0 e3

a2 2 -3 3

a 3 a4 0 3 a - 1 a2 3 -3 e3 -3 3

a3 2 -1 0

a4 1 1 0

It can be red o from the last table, that the rst and second column of A form a basis of its column space and a3 = 2a1 - a2 and a4 = a1 + a2 . Therefore 1 2 1 0 2 1 , A1 = [a1 , a2 ] = 2 1 and A2 = 0 1 -1 1 -1 1 where we used the obvious fact that

1 0 and are the coordinate vectors of 0 1 a1 respectively a2 with respect to the basis {a1 , a2 }.

Solution of homogeneous linear system of equations

A homogeneous linear system of equation Ax = 0 always has a solution, namely the vector 0. By the concept of the linearly independent set of vectors it is obvious, that 0 is the only solution if the columns of the coecient matrix A are linearly 6

8.3.

SYSTEM OF LINEAR EQUATIONS

independent. In the general case there are non-zero solutions as well. For if the basis factored form of the coecient matrix is A = A1 A2 , then the equation takes the form A1 A2 x = 0 and taking into account, that the columns of A1 are linearly independent, a vector satises the original system of equations if and only if it satises the equation A2 x = 0. Just for the sake of simplicity let us assume, that the rank of A is r and its rst r column is linearly independent and these columns form the matrix A1 . Then as it is seen in the previous example the A2 factor contains the identity matrix of order r, that is, it has the form A2 = [E, D]. Thus x1 partitioning the vector x of unknowns correspondingly x = , the equation x2 can be rewritten:

A2 x = [E, D] ·

x1 x2

= x1 + Dx2 = 0.

It follows that the components of x2 can be chosen arbitrarily, while the components of x1 must be given by x1 = -Dx2 and then the vector

x=

x1 x2

=

-Dx2 x2

=

-D E

x2

is a solution of the homogeneous linear system of equations. The number of components in x2 is called the of the linear system of equations.

degree of freedom Example 8.4 Solve the homogeneous linear system of equations:
x1 + 2x2 + 3x4 2x1 + x2 + 3x3 + 3x4 -x1 + x2 - 3x3 = = = 0 0 0

The process is similar to the one performed at basis factoring matrices, the only change is that the columns of the matrix are labelled by the corresponding variables and the starting basis is usually not denoted in the left heading.

Solution:

x1 x2 1 2 2 1 -1 1

x3 x4 x1 0 3 - 3 3 -3 0

x2 x3 x4 2 0 3 x - 1 x2 -3 3 -3 3 -3 3

x3 2 -1 0

x4 1 1 0

The last table shows, that the rank of the coecient matrix is 2 and its rst two columns form a basis in the column space, the corresponding variables x1 and x2 are the components of the vector x1 and the free variables are x3 and x4 form the vector x2 . The coordinate vectors of the third and fourth column of the matrix with respect to the basis of the column space are the columns of the matrix D. The connection between the free and non-free variables is

x1 x2

=-

2 -1

1 1

·

x3 x4

,

therefore the solutions are x1 -2 x2 1 x= x3 = 1 x4 0

-1 -1 · 0 1

t1 t2

,

t1 , t2 R .

7

8.3.

SYSTEM OF LINEAR EQUATIONS

Exercise 8.1

equations! a) c) e) g) i) k) m) o) q) s)

Find all solutions of the subsequent homogeneous linear systems of
- 2x2 - x2 - 4x2 + - 4x2 x2 - 4x3 + 16x3 + 4x3 - 2x3 + 8x3 - x3 - - + + 2x3 4x3 4x3 2x3 = 0 = 0 = 0 = = = 0 0 0

2x1 -5x1

b) d) f) h) j) l)

-3x1 -3x1 3x1 -5x1 -x1 -4x1 -5x1 -5x1 2x1 x1 3x1 3x1 -3x1 -5x1

- 4x2 + x2 + 5x2 + 4x2 + 15x2 + 10x2 + 5x2 - 4x2 - x2 + 3x2 - 6x2 + 4x2 + 8x2 - x2

+ 9x3 + 13x3 - 10x3 + 10x3 - x3 - 3x3 + 5x3 + 10x3 - 2x3 - 2x3 - 12x3 - 3x3 + x3 + 3x3 - x3 + 2x3 + x3 - 3x3 + 2x3 - x3 - 3x3 + 5x3

= 0 = 0 = 0 = 0 + x4 + 3x4 - 5x4 = 0 = 0 = 0 = 0 - 15x4 + 9x4 + 19x4 - 3x4 + 4x4 - 16x4 + 10x4 + 18x4 = = = = = = = = 0 0 0 0 0 0 0 0 = 0 = 0 = 0

-x1 -4x1 x1

-14x1 2x1 -6x1 -x1 -2x1 -3x1 x1 -x1 -3x1 -2x1 -4x1 -x1 2x1

- x2 + 5x2 + x2 + 4x2 + 2x2 - 3x2 - 3x2 + x2 + 5x2 + 4x2 - 5x2 - 3x2 + 2x2 - 2x2 - 5x2 + 2x2 - 2x2

= 0 = 0 = 0 = 0 = = = = = = = = = = = = = = = = = = 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

- 8x3 + 5x3 + 3x3 - x3 - 12x3 - 14x3 - 4x3 + 2x3 - 4x3 + 4x3 - 2x3 + + 30x3 9x3

-2x1 5x1 -2x1 -4x1 -x1 -4x1 -5x1 x1 4x1

- 3x2 + 3x2 + 5x2

+ 4x2 + 15x2 + 10x2 + 3x2

+ x4 + 3x4 - 5x4 + 3x4 - 10x4 - 4x4 + 4x4 - 12x4 - 14x4

= 0 = 0 = 0 = 0 = 0 = 0 = = = 0 0 0

5x1 2x1 x1 -3x1 3x1 5x1 5x1

n) p) r) t)

+ 2x3 - 5x3

- 2x2 4x2 - 4x2 + 4x2 - 4x3 + 4x3

- 3x2 - x2 + 2x2 - x2 - 3x2 + 4x2 + 2x2 + 4x2 + 2x2 - x2 - 4x2

+ 3x3 - 6x3 - 6x3 - 3x3 - 2x3 - 3x3 - x3 + 5x3 + x3 - 4x3 + 4x3 - x3

4x1 5x1

3x1 -9x1 x1 -x1 4x1 -4x1 x1

= 0 = 0 = 0 = 0 = 0 = 0 = 0 = 0

-5x1 5x1 -5x1 -x1 -3x1 4x1 x1

- 10x2 + 9x2 - 15x2 + 3x2 + 4x2 - 4x2 - 5x2

- x3 - 5x3 - - + + 5x3 5x3 4x3 9x3

= 0 = 0 = 0 + 4x4 + 7x4 - 8x4 - 6x4 = 0 = 0 = 0 = 0

8

8.3.

SYSTEM OF LINEAR EQUATIONS

Solution of inhomogeneous linear system of equations

An inhomogeneous linear system of equations Ax = b has a solution if and only if b is in the column space of A. In this case any solution of Ax = b can be obtained as the sum of a particular solution of the inhomogeneous system and a solution of the corresponding homogeneous system Ax = 0. If Ax = b has a solution then all its solution is obtained as follows: Let A = A1 A2 be the basis factored form of the coecient matrix. Since b is in the column space of A it can be uniquely expressed as a linear combination of the columns of A1 , that is, there exists a vector d, such that A1 d = b. Therefore

Ax = b A1 A2 x = A1 d A2 x = d,
thus again assuming that A1 contains the rst r columns of A, the connection between the free and non-free variables is

x1 = d - Dx2 .
Choosing the free variables equal to 0 a solution

x=

x1 x2

=

d 0

of the inhomogeneous system is obtained and then the solution set is

d 0

+

-D E

t where t Rs ,

denoting by s the degree of freedom of the system.

Example 8.5

Solve the inhomogeneous linear system of equations:
x1 + 2x2 + 3x4 2x1 + x2 + 3x3 + 3x4 -x1 + x2 - 3x3 = = 4 5 = -1

The only change as compared to the solution of homogeneous linear system of equations is that the table is extended by the vector b and its coordinate vector is also computed when the basis is changed.

Solution:
x1 1 2 -1 x2 2 1 1

x3 x4 b x1 4 0 3 - 3 3 5 -3 0 -1 2 1

x2 2 -3 3

x3 x4 b 0 3 4 x - 1 x2 3 -3 -3 -3 3 3

x3 2 -1 0

x4 1 1 0

b 2 1 0

The last table shows the vector d = and the solution set x1 x2 x= x3 x4 is

, thus a particular solution is s = [2, 1, 0, 0] -1 -1 · 0 1

2 -2 1 1 = + 0 1 0 0



t1 t2

,

t1 , t2 R .

9

8.3.

SYSTEM OF LINEAR EQUATIONS

Exercise 8.2

of equations: a) b) c) d) e) f) g) h) i) j) k) l)

Determine the solutions of the folloving inhomodeneous linear systems
- x2 - 5x2 x2 - 2x2 + 3x2 - 4x2 + 4x2 5x2 - 2x2 - x2 + 5x2 - 2x3 - 10x3 + 2x3 - 4x3 - + - - 12x3 10x3 18x3 10x3 = -2 = -20 = 3 = -3 - + - - 18x4 15x4 27x4 15x4 = 12 = -10 = 18 = 10 = 12 = 9 = 2 = 3 = 5 = -7 = 8 = -8 = -2 = -10 = 2 = 0 = 4 = -5

x1 -5x1 3x1

-3x1 x1 -5x1 x1 2x1 x1 14x1 x1 4x1 x1 5x1 2x1 2x1 5x1 x1 5x1 -x1 -x1 -2x1 -4x1 4x1 3x1 -3x1 -5x1 x1 2x1 -4x1 6x1 -x1 -3x1 2x1 -x1 -4x1 2x1

+ 10x3 + 8x3 + 2x3 + 8x3 + 5x3 - 7x3 + 8x3 - 8x3 - 2x3 + x3 + 4x3 - 5x3 - 4x3 - 3x3

+ 5x4 + 4x4 + x4 + 4x4 - 2x4 + x4 - 3x4 + x4 + x4 - 11x4 - 5x4 - 15x4 - 7x4 - 10x4 = -4 = -1 = 2 = 4 = 8 = -8 = 15 = -3 = -15 - 23x4 - 16x4 + 17x4 - 3x4 + 8x4 - 6x4 - x4 = -13 = 23 = 11

+ 4x2 + 7x2 + 5x2 + 9x2 x2 + x2 - 2x2 - 15x2 + x2 - 20x2 4x2 - 3x2 + 4x2 - 2x2 - 4x2 + 3x2 + 3x2 - 4x2 - 2x2 + x2 + 3x2 + 3x2 - x2 - x2 + 3x2 - 13x2 - x2

- 12x3 + 6x3 + 12x3 - 3x3 + 9x3 + 10x3 - - + - 5x3 4x3 5x3 3x3

= 12 = 6 = -3 = -9 = -8 = 6 = 1

- 8x3 + 6x3 + x3 + 5x3 - 5x3 - 5x3

10

8.4.

MATRIX EQUATIONS

m) n) o) p) q) r) s) t)
8.4

-x1 2x1 5x1 2x1 5x1 -x1 x1 x1 5x1 -3x1 -3x1 -x1 -x1 -3x1 x1 -4x1 x1 -4x1 -3x1 4x1 -3x1 x1 -x1

- 2x2 - 3x2 + 4x2 - x2 - 2x2 - 2x2 - 2x2 - 7x2 - 2x2 + 5x2 - 4x2 - 3x2 - 5x2 + 2x2 + 5x2 + 4x2 2x2 - 10x2 3x2

- x3 - 12x3 - 7x3 - x3 - 2x3 - 2x3 + + - 5x3 4x3 x3

= 6 = 9 = -12 = 1 = 6 = 12 = -2 = 2 = 2 - 20x4 + 10x4 + 14x4 = 14 = -7 = -18 - x4 - 4x4 + 3x4 - 2x4 = 11 = -20 = 1 = 4 = -9 = 3 = -3 = 3 = 8 = 1 = 8 = -30 = 16 = 20

+ 5x3 + x3 - 5x3 + + + 2x3 2x3 6x3

- x3 - 2x3 - 2x3 - 3x3 + 9x3

+ 4x2 - 4x2 + 5x2 - 4x2 - 3x2 - 2x2 - 4x2

+ 24x3 - 24x3 - 6x3 - 7x3 - 11x3

Matrix equations

Denition 8.6

Let A and B be given matrices. The equations of the form AX = B or YA = B are called matrix equations.
Since YA = B A Y = B it is enough to deal with the rst type equations. Remembering the partitions of a matrix product AX = [Ax1 , Ax2 , . . . , Axp ], where xi i = 1, 2, . . . , p are the column vectors of the unknown matrix X it can be recognized that a matrix equation is equivalent to linear systems of equations having the same coecient matrix:

Ax1 = b1 , Ax2 = b2 , . . . , Axp = bp .

Proposition 8.4 The matrix equation AX = B is solvable if and only if each column of B is a linear combination of columns of A, that is, (A) = ([A, B]).
Inverse of matrices

Denition 8.7
A-1 .

The inverse of a square matrix A is the solution of the matrix equation AX = E, where E is the identity matrix. The inverse of A is denoted by
11

8.4.

MATRIX EQUATIONS

Proposition 8.5

order.

A square matrix has an inverse if and only if its rank equals to its regular

The invertible matrices are called , while the matrices having no inverse are said to be . The inverse A-1 of a regular matrix A is unique and

Remark:

singular

A-1 A = AA-1 = E.

Proposition 8.6

The product of regular matrices is regular and
(AB)-1 = B-1 A-1 .

The transpose of a regular matrix is regular and
(A )-1 = (A-1 ) .
To nd the inverse of a given square matrix A the solution method of inhomogeneous linear system of equations can be used, because the columns of the inverse matrix are the solutions of the equations Ax = e1 , . . . , Axen . Since the coecient matrix of these equations is the same the solutions can be performed together as the following example illustrates it. 1 0 1 Example 8.6 A = 1 1 1 . 0 1 1

Determining matrix inverse

Find the inverse of the matrix

We solve three linear system of equations at the same time as they have common coecient matrix:

Solution:

a1 1 1 0

a2 0 1 1

a3 1 1 1

e1 1 0 0 a3 1 0 1

e2 0 1 0 e1 1 -1 1

e3 a1 0 0 1

a2 0 1 1

a3 1 0 1 e1 0 -1 1

e1 1 -1 0

e2 e3 0 0 1 0 0 1

a1 a2

e2 e3 0 0 a 1 1 0 a2 a3 -1 1

e2 e3 1 -1 1 0 -1 1

from which the inverse matrix is

A-1

0 = -1 1

1 -1 1 0 -1 1

If the columns of A do not form a basis, then it indicates that A is singular and therefore has no inverse.

Exercise 8.3

a) d)

-1 A= 1 0 0 A = -1 -2



Find the inverseof the given matrices!
-1 -3 2 2 5 -4 1 -4 -1 1 -1 -1

b)

e)

1 3 3 A = -1 -4 -4 -1 3 2 1 1 1 A = -3 -4 -1 2 4 -3

c) f)

-1 -3 -4 4 -3 A= 1 1 4 -4 1 -2 -2 A = 3 -5 -5 1 0 1



12

8.4.

MATRIX EQUATIONS

g) j)

1 A= 0 3 1 A= 2 0



-3 -1 -4 1 3 0

5 3 -1 4 -2 1

h) k)

-1 A = -3 2 1 -2 0 0 A= 0 1 2 -1



-2 -5 -5 -3 3 -1 -5 -5 0 -1 3 3 -2 3

i) l)

0 -1 -1 A = -1 -3 -3 4 -1 -2 1 -2 0 1 -3 1 A= 2 -1 -4 0 3 0



1 1 2 1

13



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