Matrices & determinants

MATRIX:
Definition: A matrix is defined as an ordered
rectangular array of numbers. They can be used to
represent systems of linear equations, as will be
explained below.
Here are a couple of examples of different types of
matrices:

Upper Lower
Symme Diago Identit
Triang Triangul Zero
tric nal y
ular ar

And a fully expanded m×n matrix A, would look
like this:

... or in a more compact form:

The horizontal lines in a matrix are called rows and
the vertical lines are called columns. A matrix with
m rows and n columns is called an m-by-n matrix (or
m×n matrix) and m and n are called its dimensions.
The places in the matrix where the numbers are, are
called entries. The entry of a matrix A that lies in the
row number i and column number j is called the i,j
entry of A. This is written as A[i,j] or aij.
We write to define an m × n matrix A with
each entry in the matrix called aij for all 1 ≤ i ≤ m
and 1 ≤ j ≤ n.

Example
The matrix

is a 4×3 matrix. This matrix has m=4 rows, and n=3
columns.

The element A[2,3] or a23 is 7.
Operations
Addition
The sum of two matrices is the matrix, which (i,j)-th
entry is equal to the sum of the (i,j)-th entries of two
matrices:

The two matrices have the same dimensions. Here A +
B = B + A is true.
Subtraction

If A and B are matrices of the same type then the
subtraction is found by subtracting the corresponding
elements aij − bij.
Here is an example of subtracting matrices.

Multiplication of two matrices
The multiplication of two matrices is a bit more
complicated:

So with Numbers:

Two matrices can be multiplied with each other even
if they have different dimensions, as long as the
number of columns in the first matrix is equal to the
number of rows in the second matrix.
The result of the multiplication, called the product, is
another matrix with the same number of rows as the
first matrix and the same number of columns as the
second matrix.
the multiplication of matrices is not commutative, this
means, in general that
the multiplication of matrices is associative, this
means
Special matrices
There are some matrices that are special.
Square matrix
A square matrix has the same number of rows as columns,
so m=n.
An example of a square matrix is

This matrix has 3 rows and 3 columns: m=n=3.
Identity
Every square dimension set of a matrix has a special
counterpart called an "identity matrix". The identity

matrix has nothing but zeroes except on the main
diagonal, where there are all ones. For example:

is an identity matrix. There is exactly one identity matrix
for each square dimension set. An identity matrix is
special because when multiplying any matrix by the
identity matrix, the result is always the original matrix
with no change.
Inverse matrix
An inverse matrix is a matrix that, when multiplied by
another matrix, equals the identity matrix. For example:

is the inverse of .
One column matrix
A matrix, that has many rows, but only one column, is
called a column vector.

Transpose of Matrices
DEFINITION: The transpose of a matrix is found by
exchanging rows for columns i.e. Matrix A = (aij) and the
transpose of A is:

AT = (aji) where j is the column number and i is the
row number of matrix A.
For example, the transpose of a matrix would be:

In the case of a square matrix (m = n), the transpose
can be used to check if a matrix is symmetric. For a
symmetric matrix A = AT.

trace(A) The trace of a matrix is simply the
summation of its main diagonal.

AT The transpose of a matrix is switching the
rows and columns. For example:
a b c a d g
T
A =d e f A =b e h
g h i c f i

The adjoint of a square matrix [aij] is defined as
the transpose of the matrix [Aij] where Aij are the
cofactors of the elements aij.
Adjoint of A is denoted by adj A. Adjoint of a
square matrix

Minors and Co-factors

Minor : Minor of an element Aij of a
determinant is the determinant obtained by
deleting its ith row and jth column in which
element aij lies. Minor of an element aij is
denoted by Mij.

Co-factor : Cofactor of an element Aij ,
denoted by Aij is defined by Aij =(-1) i+j Mij

Properties of adjoint of a matrix , if A is an
invertible square matrix of order n.

1. A.(adj A) = (adj A). A = |A| In
( Note: |A(adjA)| = |A|n, if |A| ≠0, then |adjA| =
|A|n-1 )

2. adj (AB) = (adj B) . (adj A).

3. adj(adjA) = |A|n-2 .A, if a is an invertible square
matrix of order n.

4. If A is an invertible symmetric matrix, then A-1
is also sym.

5. Every skew- sym. Matrix of odd order is
singular

Example:

Find the adjoint of the matrix.

A11 = + = 4 – 5 = -1

A12 = - = -4

A13 = + =8

A21 = - = 19

A22 =+ = 14

A23 = - =3

A31 = + = -8

A32 = - = -1

A33 = + =2

Inverse of a matrix
If A is any square matrix of order n and there
exist another square matrix B of the same
order n, such that AB=BA =I , then B is called
an inverse matrix of A and is denoted by A-1 =

The Inverse of a Matrix

DEFINITION: Assuming we have a square matrix A,
which is non-singular (i.e. det(A) does not equal zero),
then there exists an n×n matrix A-1 which is called the
inverse of A, such that this property holds:
AA-1 = A-1A = I, where I is the identity matrix.
The inverse of a 2×2 matrix
Take for example an arbitrary 2×2 Matrix A whose
determinant (ad − bc) is not equal to zero.

where a,b,c,d are numbers, The inverse is:

Singular Matrix : Any matrix whose
determinant is zero, is singular matrix.

Non-singular Matrix : Any matrix whose
determinant is not zero, is non-singular matrix.

Note 1:
Only a square matrix can have its inverse.
Note 2:
From the definition, it is clear that if B is the
inverse of A, then A is the inverse of B.

Note 3:
Inverse of A is denoted by A-1, thus B = A-1 and
AA-1 = A-1A=I.
Inverse matrix
An inverse matrix is a matrix that, when multiplied
by another matrix, equals the identity matrix. For
example:

is the inverse of .

matrix Q is orthogonal if its transpose is equal to its
inverse:

which entails

where I is the identity matrix.
An orthogonal matrix Q is necessarily square and
invertible, with inverse Q−1 = QT.

Question: Give an examples of matrices A, B and C
such that (i) AB ≠ BA (ii) AB = O = BA, A≠ O, B ≠ O
(iii) AB = O but BA ≠O (iv) AB=AC but B≠ C,A≠ O.

Answer: (i) Let A = and B = then AB
= BA=

(ii) Let A= and B = (iii)

Let A= and B = then AB = O, BA =

(iv) Let A= ,B= and C =
then AB = = AC .

Give an example of matrices A, B such that AB=BA

A= B= .

Question: If A = then A5 .

Answer: A = 2I therefore A5 = (2I)5 = 32 I = 16(2I)
= 16A.

**Question: Given that A = and X

be the solution set of equation Ax = A, where xЄ N-{1}.

Evaluate .

Answer: A2 = A (by multiplying) and A3 =
A and so on......
Thus An = A for n = 2,3,4......

Now, =
= ( . ..... . )( . .... )

= 3/2.

Question: (i) ≠ , why? [

Not same order]

(ii) Is the equation -

= - Valid? [No]

Question: (i) Let = and =

, then - =0

[Hint = ]

(ii) If f(x) = , then

which is correct f(a)=0 , f(b)=0, f(0)=0 and f(1)=0
[ Hint f(0)=0 ∵ det.(skew-symm. matrix)=0].

(iii) Let f(t) = , then

is equal to 0,-1,2,3.

[Hint 0, = →

as t→ ].

(iv) There are two values of a which makes

determinant = = 86, then sum

of these numbers is 4,5,-4,9. [Hint a=-4,
operate R2 – 2R1]

Question: If A = , show that A2 – 6A +17I =
0. Hence find A-1 .

Solution: A2 = A.A = =

A2 – 6A +17I = -6 + 17

= =0

A2 – 6A +17I = 0 ⇨ 17I = - A2 +6A ⇨ IA-1 = -
1/17 A2 A-1 + 6/17 AA-1 = -1/17A+6/17I

A-1 = 1/17 +6/17 = 1/17
.

NOTE If A is symm. As well as skew-symm., then A
is a null matrix.( if A = AT then A is Symm. And if A = -
AT then A is skew- symm.)

A= is symmetric and B =

is skew-symmetric.

NOTE: (i) If A and B are symmetric matrices, then
BA-2AB is neither symm. nor skew-symm.
(ii) If A is symm. matrix then BTAB is
symm.
(iii) If A and B are symmetric matrices of
same order, then AB is symm. iff AB=BA.
(iv) Zero matrix is both symm. and skew-
symm.
(v) Sum of two skew-symm. matrices is
always skew-symm.
(vi) If A is a symm., then A3 is a symm. and
if A is skew-symm., then A2 is a symm.

Theorem:
The inverse of a square matrix if it exists, is
unique.

Let A be an invertible square matrix. If possible,
let B and C be two inverse of A.
Then AB = BA = I.
AC = CA = I (by def. of inverse)
Now,
B = BI = B(AC)
= (BA)C [ Matrix multiplication is associative]
= IC = C
i.e., B = C
Hence the inverse of A is unique.

Theorem
If A and B are two invertible matrices of the
same order, then (AB)-1 = B-1A-1.
Proof:
From the definition of inverse of a matrix, we
have
(AB)(AB)-1 = I
or A-1 (AB)(AB)-1 = A-1 I (Pre-multiplying both
sides by A-1)
or (A-1A) B (AB)-1 = A-1 (Since A-1 I = A-1)
or I B (AB)-1 = A-1
or B (AB)-1 = A-1
or (B-1B)(AB)-1 =B-1A-1
or I(AB)-1= B-1A-1
or (AB)-1 = B-1A-1

Properties of Inverse of Matrix

In other words, a square matrix A is invertible if
and only if A is a non-singular matrix.

(c) If A and B are invertible square matrices,
then
(AB)-1 = B-1 A-1
(d) If A and B are two non-singular square
matrices of the same order, then AB and BA are
also non-singular matrices of the same order.

The Determinant of a Matrix
DEFINITION: Determinants play an important role in
finding the inverse of a matrix and also in solving systems
of linear equations. In the following we assume we have a
square matrix (m = n). The determinant of a matrix A will
be denoted by det(A) or |A|. Firstly the determinant of a
2×2 and 3×3 matrix will be introduced, then the n×n case
will be shown.
Determinant of a 2×2 matrix

Assuming A is an arbitrary 2×2 matrix A, where the
elements are given by:

then the determinant of a this matrix is as follows:

The determinant of a 3×3 matrix is a little more tricky and
is found as follows (for this case assume A is an arbitrary
3×3 matrix A, where the elements are given below).


Consistent and Inconsistent
Solutions
Consistent system : A system of
equation is said to be consistent if
its solution ( one or more ) exists.
Inconsistent system : A system of

equation is said to be inconsistent if
its solution does not exist.

Working rule to check consistency: Case I When
A 0
System is consistence and has unique
solution.
Case II When A =0 .
Find Adj(A) and then find Adj(A) .B
If Adj(A) .B 0 then system is
inconsistence .
Case III If Adj(A).B=0 Then it may have
infinite solutions then it is consistence or
have no solution then it is inconsistence.

Solving Systems of Equations using Matrices
DEFINITION: A system of linear equations is a set of
equations with n equations and n unknowns, is of the form
of

The unknowns are denoted by x1, x2, ..., xn and the
coefficients (a and b above) are assumed to be given.
In matrix form the system of equations above can be
written as:

A simplified way of writing above is like this: Ax = b
Inverse Matrix Method
DEFINITION: The inverse matrix method uses the
inverse of a matrix to help solve a system of
equations, such like the above Ax = b. By pre-
multiplying both sides of this equation by A-1 gives:

or alternatively

So by calculating the inverse of the matrix and
multiplying this by the vector b we can find the
solution to the system of equations directly. And from
earlier we found that the inverse is given by

From the above it is clear that the existence of a
solution depends on the value of the determinant of A.
There are three cases:
1. If the det(A) does not equal zero then
solutions exist using
2. If the det(A) is zero and b=0 then the
solution will be not be unique or does not
exist.

3. If the det(A) is zero and b=0 then the
solution can be x = 0 but as with 2. is not
unique or does not exist.
Looking at two equations we might have that

Written in matrix form would look like

and by rearranging we would get that the solution
would look like

Three simultaneous equations in x, y and z

ax + by + cz = p
dx + ey + fz = q
gx + hy + iz = r
To solve use the following By cramer’s rule

Solve the system using matrices.
−x + 5y = 4
2x + 5y = −2
Always check your solutions!

Answer

, and
To solve the system, we need the inverse of A,
which we write as A-1.
Swap leading diagonal:

Change signs of the other 2 elements:

Now we find the determinant of A:
|A| = -5 - 10 = -15
So

So the solution to the system is given by:

This answer means that we have found the
solution x = -2 and y = 2/5.
Is the solution correct?
We check it in the original set of equations:

Substituting x = -2 and y = 2/5, we get:

−(−2) + 5×(2/5) = 2 + 2 = 4

2×(−2) + 5×(2/5) = −4 + 2 = −2
So the solution to the original system of
equations is
x = -2, y = 2/5.
Solve the system using matrix methods.

Did I mention? It's a good idea to always check your
solutions.

Answer

Use adjoint of A , we find the inverse of A to be:

So the solution to the system of equations is:

Check:

22 + 2(-16) - (-16) = 6

3(22) + 5(-16) - (-16) = 2

-2(22) - (16) - 2(-16) = 4
So the solution is x = 22, y = -16 and z = -16.

The Determinant of a Matrix
DEFINITION: Determinants play an important role
in finding the inverse of a matrix and also in solving
systems of linear equations. In the following we
assume we have a square matrix (m = n). The
determinant of a matrix A will be denoted by det(A)
or |A|. Firstly the determinant of a 2×2 and 3×3
matrix will be introduced, then the n×n case will be
shown.

Assuming A is an arbitrary 2×2 matrix A, where the
elements are given by:


The determinant of a 3×3 matrix is a little more
tricky and is found as follows (for this case assume A
is an arbitrary 3×3 matrix A, where the elements are
given below).


Properties of Determinants:

Property 1. If each element of a row (
column)
of a determinant is zero , then
value of determinant is zero.

Property 2. Value of a determinant is not
changed by changing the rows
into columns and columns into rows.

Property 3. If two adjacent rows (columns)of
a determinant are interchanged , then the
sign of the determinant is changed but its
numerical value is unchanged.

Property 4. If two rows (columns) are
identical, then the value of the determinant is
zero.

Property 5. If every element of a row
(column) is multiplied by some constant k,
the value of the determinant is multiplied by k.

Property 6 .If each element in any row
(column) consist of two terms , then the
determinant can be expressed as the sum of
the determinants of same order.

Property 7 . The value of a determinant
remain unchanged if to each element of a
row (column) be add ( or subtracted)

equimultiplies of the corresponding elements
of one or more rows (columns) of the
determinant.

Property 8. The value of the determinant of a
diagonal matrix is equal to the product of the
diagonal elements.

Property 9. The value of the determinant of a
skew-symmetric matrix of odd order is
always zero.

Property 10. The determinant of a symmetric
matrix of even order is always a perfect
square.

Notations
Let be the given determinant. Then

(i)R1, R2, R3 stand for first, second and third
rows of .

(ii) C1, C2, C3 stand for first, second and third
columns of .

(iii) By R2 R2 - R3 we mean that third row is to
be subtracted from 2nd row.

(iv) By C1 C1 + 2C2 - 3C3, we mean that we
are to add in first column, the two times of C2
and subtract three times C3.

The given determinant is

y+z x y
z+x z x =
x+y y z

By ( R1 + R2 + R3 ) and, then, taking (x+y+z) out
from R1,

2 1 1
z+x z x · (x+y+z) =
x+y y z

By [ C1 - ( C2 + C3 ) ] and [ C2 - C3 ],

0 0 1
0 z-x x · (x+y+z) =
x-z y-z z

Now, expanding along R1,

= { 0 - 0 + 1 [ (0) - (z-x)(x-z) ] } • (x+y+z)

= (x-z)(x-z)·(x+y+z)

Applications of Determinants

1. Area of a Triangle The area of a triangle
with vertices (x1,y1), (x2,y2), (x3,y3) is given
by

=1/2 [ x1(y2-y3)+x2(y3-y1)+x3(y1-y2)]

In determinant form

= 1/2

Because area is always positive so we take
absolute value of determinant.

Question.1 Prove that the points P (a, b+c), Q(b,
c+a), R(c, a+b) are collinear.

Answer : If P,Q and R are collinear then

=0

By applying C2 → C2+C1

= (a+b+c) =0 (

∵ C2, C3 are identical)

Question.2 Find the value of k if the area of the
triangle with vertices (-2,0),(0,4) and (0,k) is 4 square
units.

Answer: Area of =½ =4

⇨ the absolute value of ½ (-2)(4 – k) = 4 ⇨
the absolute value of (k – 4) = 4

⇨ |k – 4| = 4 ⇨ k – 4 = 4, -4 ⇨ k = 8,
0.

Question.3 Without expanding, show that

(i) =0

Operating C1 → C1+C2+C3, we get

= 0.

(ii) =0

Taking out (-1) from C1,C2 and C3, we get =

(-1)(-1)(-1) = -1 =-

(by interchanging rows and columns)

2 =0 ⇨ =0

(iii) =0 ⇨

=

= abc

( Operating C3 → C3+C1)

abc(ab+bc+ac) x 0 = 0 ( two cols. Are identical)

(iv) =0 ⇨ -

= -

- =0

(Pass C3 over the first two columns.)

(v) =

R.H.S. =

( applying C1 ↔ C2)

= -

(apply C3 → C3 – (ab+bc+ca))C1)

=

= (apply C2 ↔ C3

and C1 ↔ C2)

If a,b,c are +ve and are the pth,qth and rth
terms resp. Of a G.P.,show without expanding that

(vi) =0 (put a=xyp-1

,b=xyq-1 ,c=xyr-1 , apply C1 →C1-logx.C3
,C1→C1+C3)

(vii) = (same

method as given below)

(viii) = (

Multiply by abc as R1 with a,R2 by b and R3 by c
then divide with abc )

Find the values of:

(ix) (Operate C2

→ C2. and value is 0

(x)

(Operating C3 → C3 – cos .C2+ sin .C1 and value
is 0)

Prove that :

(a)

= (1+a2+b2)3 ( Apply C1 → (C1 - bC3) and C2 →
(C2+aC3)

(b) =1

(Apply R3 → sin R3 + cos R1)

(C) =

xyz(x+y+z)3 (Apply R1 → x R1, R2 → y R2, R3
→ z R3 and take

x,y,z common from C1,C2,C3 resp.)

(d) = (a-b)(b-c)(c-

a)(a+b+c)(a2+b2+c2) ( Apply C1 → (C1+C2 – 2C3)

(e) =

2(a+b)(b+c)(c+a) (Apply C1 →(C1+C3 )and C2
→(C2+C3))

(f) =

(a+b+c)(a2+b2+c2)

(g) = 2

(apply C1→C1-C2-C3, C2→C2-C1,C3→C3-C1,
C2↔C3)

(h) = abc ( + +

+1) = (ab+bc+ca+abc).

(Hint taking a,b,c common from each row ,
apply R1→R1+R2+R3 then expand along first row).

(i)

= )3

Apply C1→C1-b C3, C2→C2+a C3, we get

=(

expand along C1, We get ( .

(h) Evaluate where

=C(x,1) ( binomial coefficient)

Solution: =

( taking

x,y,z common from

R1,R2,R3 resp. and ½!,1/3!

From C2,C3 resp.)
( by formula of C(n,r) = )

Apply C3→C3 + C2 and put a= x-1, b=y-1,
c=z-1

= (a-b)(b-c)(c-a)

= (x-y)(y-z)(z-x).

Question: If x,y,z are all different and if

= 0 , prove that xyz = -1.

Solution: =

+ = + xyz =

0

(1+xyz) = 0 ⇨ (x-y)(y-

z)(z-x)(1+xyz) = 0 ⇨ xyz=-1 ∵ x ≠y≠ z.

Elementary Transformation
Elementary transformations are of the following
three types:
Interchange of any two rows (or columns)
The multiplication of the elements of a row (or
column) by a non-zero number.

The addition to the elements of any row (or
column) the corresponding elements of any
other row (or column) multiplied by any
number.
Any elementary operation is called a row
transformation or a column transformation
according as it applies to rows or columns.
Definition
Let Ri denotes the ith row of the matrix A = [aij]
then the elementary row operations on the
matrix A are defined as:

3. Ri → Ri + kRj means multiply each element
of jth row by k and add it to the corresponding
elements of ith row.
The corresponding column transformations are

4. In applying one or more row operations
while finding A-1 by elementary row operations,
we obtain all zeros in one or more, then A-1
does not exist.

Q.15 Of Ex.3.4
Step 2: Transform the matrix to the reduced row echelon form

-
3 1
- 3
2 3 1 0 0 1 0 0
3
Row 2 2
multiply the 1st row by 2
Operation 2 2 3 0 1 0
1/2
1: 2 2 3 0 1 0
-
3 2 0 0 1
2 -
3 2 0 0 1
2

-
- 3 1
3 1 3
3 1 0 0
1 0 0
2 2
Row 2 2 2
2 add -2 times the 1st row
Operation
to the 2nd row -
2: 2 2 3 0 1 0 0 5 0 1 0
1
-
3 2 0 0 1 -
2 3 2 0 0 1
2

-
3 1
- 3
3 1 1 0 0
3
1 0 0 2 2
2
2 2
Row 2 add -3 times the 1st -
Operation 0 5 0 1 0
- row to the 3rd row 1
3: 0 5 0 1 0
1 - -
5
- 5 3
3 2 0 0 1 0 0 1
2
2
2 2

Row - 3 1 - 3 1
multiply the 2nd row
Operation 1 3 0 0 1 3 0 0
by 1/5
4: 2 2 2 2

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

- -
3 1 3 1
3 3
1 0 0 1 0 0
2 2 2 2
2 2
- -
Row 1 1
1 add -5/2 times the 2nd 1
Operation 0 1 0 0 0 1 0 0
row to the 3rd row
5: 5 5
5 5
- - - -
5
5 3 5 - 1
0 0 1 0 0 1
1
2
2 2 2 2

- -
3 1 3 1
3 3
1 0 0 1 0 0
2 2 2 2
2 2
- -
Row 1 1
1 multiply the 3rd row 1
Operation 0 1 0 0 0 1 0 0
by -2/5
6: 5 5
5 5
- - -
2 1
5 - 1 2
0 0 1 0 0 1
1
5 5
2 2 5

- -
3 1 -1 -3 3
3 3
1 0 0 1 0
Row add -3/2 times the 10 10 5
2 2 2
Operation 2 3rd row to the 1st
7: - 1 row -1 1
0 1 0 1 0 0 1 0 0
5 5 5

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

-
- 3
-1 -3 3 2
3 1 0 0 0
1 0
5
10 10 5 5
2
-
Row -1 1 1
add 3/2 times the 2nd 1
Operation 0 1 0 0 0 1 0 0
row to the 1st row
8: 5 5 5
5
-
2 1 -
2 2 1
0 0 1 2
0 0 1
5 5
5 5 5
5

Q. 16 Of Ex.3.4
Step 1: Adjoin the identity matrix to the given matrix

1 3 -2 1 0 0
Adjoining I3 to the given matrix, we obtain the 3x6 matrix: -3 0 -5 0 1 0
2 5 0 0 0 1


-
1 3 1 0 0 1 3 -2 1 0 0
2
Row
add 3 times the 1st row -
Operation - - 0 9 3 1 0
0 0 1 0 to the 2nd row 11
1: 3 5
2 5 0 0 0 1
2 5 0 0 0 1

1 3 -2 1 0 0
1 3 -2 1 0 0
Row -
- add -2 times the 1st 0 9 3 1 0
11 row to the 3rd row
2: - -
2 5 0 0 0 1 0 4 0 1
1 2

1 3 -2 1 0 0
1 3 -2 1 0 0 -
1 1
Row - 11
0 9 3 1 0 multiply the 2nd 0 1 0
Operation 11
row by 1/9 3 9
3: 9
- -
0 4 0 1
1 2 - -
0 4 0 1
1 2

1 3 -2 1 0 0
1 3 -2 1 0 0 -
1 1
- 11
1 1 0 1 0
Row 11
0 1 0 add 1 times the 2nd 3 9
Operation 9
3 9 row to the 3rd row
4: 9 -
25 1
- - 5
0 4 0 1 0 0 1
1 2
9 9
3

1 3 -2 1 0 0 1 3 -2 1 0 0
- -
1 1 1 1
11 11
0 1 0 0 1 0
Row
3 9 multiply the 3rd 3 9
Operation 9 9
row by 9/25
5: - -
25 1 1 9
5 3
0 0 1 0 0 1
9 9 25 25
3 5

1 3 -2 1 0 0 -
1 3 1 0 0
- 2
Row 1 1 add 11/9 times the
Operation 11 3rd row to the 2nd -
0 1 0 4 11
6: 3 9 row 2
9 0 1 0
25 25
0 0 1 - 1 9 5

3 -
1 9
25 25 3
0 0 1
5
25 25
5

-
2 18
- 1
1 3 1 0 0 1 3 0
2
25 25
5
-
4 11
2 -
Row 0 1 0 4 11
add 2 times the 3rd 2
Operation 25 25 0 1 0
5 row to the 1st row
7: 25 25
5
-
1 9
3 -
0 0 1 1 9
3
25 25 0 0 1
5
25 25
5

-
2 18 -2 -3
1
1 3 0 1 0 0 1
25 25 5 5
5
-
- 4 11
Row 4 11 add -3 times the 2
2 0 1 0
Operation 0 1 0 2nd row to the 1st
25 25
8: 25 25 row 5
5
-
- 1 9
1 9 3
3 0 0 1
0 0 1
25 25
25 25 5
5

Step 1: Adjoin the identity matrix to the given matrix

2 6 -2 1 0 0
Adjoining I3 to the given matrix, we obtain the 3x6 matrix: 1 4 0 0 1 0
2 -5 1 0 0 1


1
- -
2 6 1 0 0 1 3 0 0
2 1
Row 2
multiply the 1st row
by 1/2 1 4 0 0 1 0
1: -
2 1 0 0 1 -
5 2 1 0 0 1
5

1
-
1 3 0 0
1 1
- 2
1 3 0 0
1
Row 2 -
add -1 times the 1st
Operation 1
1 4 0 0 1 0 row to the 2nd row 0 1 1 1 0
2:
- 2
2 1 0 0 1
5
-
2 1 0 0 1
5

1 1
- -
1 3 0 0 1 3 0 0
1 1
2 2
Row - -
add -2 times the 1st
Operation 1 1
0 1 1 1 0 row to the 3rd row 0 1 1 1 0
3:
2 2
- - -
2 1 0 0 1 0 3 0 1
5 11 1

1
1
- 1 3 -1 0 0
1 3 0 0
1 2
2
-1
Row - add 11 times the
0 1 1 1 0
Operation 1 2nd row to the 3rd
0 1 1 1 0 2
4: row
2 -
13
- - 0 0 14 11 1
0 3 0 1
11 1
2

Row 1 1
multiply the 3rd -
Operation 1 3 -1 0 0 1 3 0 0
row by 1/14 1
5: 2 2

-1 -1
0 1 1 1 0 0 1 1 1 0
2 2
- -
11 1
13 13
0 0 14 11 1 0 0 1
14 14
2 28

1 1
- -
1 3 0 0 1 3 0 0
1 1
2 2
-1 -1 3 -1
Row add -1 times the
0 1 1 1 0 0 1 0
Operation 3rd row to the
2 28 14 14
6: 2nd row
- -
11 1 11 1
13 13
0 0 1 0 0 1
14 14 14 14
28 28

1 1 11 1
-
1 3 0 0 1 3 0
1
2 28 14 14
-1 3 -1 -1 3 -1
Row add 1 times the
0 1 0 0 1 0
Operation 3rd row to the
28 14 14 28 14 14
7: 1st row
- -
11 1 11 1
13 13
0 0 1 0 0 1
14 14 14 14
28 28

1 11 1 1 1 2
1 3 0 1 0 0
28 14 14 7 7 7
-1 3 -1 -1 3 -1
Row add -3 times the
0 1 0 0 1 0
Operation 2nd row to the
28 14 14 28 14 14
8: 1st row
- -
11 1 11 1
13 13
0 0 1 0 0 1
14 14 14 14
28 28

Q. 14 OF EX. 4.6

- -
1 2 7 1 2 7
1 1
Row
- add -3 times the 1st row to the - -
Operation 3 4 -5 0 7
5 2nd row 11 26
1:
- -
2 3 12 2 3 12
1 1

-
1 2 7 -
1 1 2 7
1
Row
- - add -2 times the 1st row to the
Operation 0 7 - -
11 26 3rd row 0 7
2: 11 26
-
2 3 12 0 1 -1 -2
1

1 -1 2 7
1 -1 2 7
Row -11 -26
Operation 0 7 -11 -26 multiply the 2nd row by 1/7 0 1
3: 7 7
0 1 -1 -2
0 1 -1 -2

-
1 2 7
- 1
1 2 7
1 - -
Row - - 11 26
add -1 times the 2nd row to the 0 1
Operation 11 26
0 1 3rd row
4: 7 7
7 7 4 12
0 1 -1 -2 0 0
7 7

1 -1 2 7
1 -1 2 7
-11 -26
Row 0 1 -11 -26
Operation 7 7 multiply the 3rd row by 7/4 0 1
5: 7 7
4 12
0 0 0 0 1 3
7 7

Row
- add 11/7 times the 3rd row to the -
1 2nd row 1
6:

- - 0 1 0 1
11 26
0 1 0 0 1 3
7 7
0 0 1 3

1 -1 2 7 1 -1 0 1
Row
Operation 0 1 0 1 add -2 times the 3rd row to the 1st row 0 1 0 1
7: 0 0 1 3 0 0 1 3

-
Row 1 0 1 1 0 0 2
1
Operation add 1 times the 2nd row to
0 1 0 1
8: 0 1 0 1 the 1st row
0 0 1 3
0 0 1 3

ANSWER IS X=2,
Y=1, Z=3.

Mathematics reference
Rules for matrices
Basic properties of matrices.
A, B, a and n are scalars.
Basic.
-A == (-1) A
A - B == A + (-B)

1A=A
0A=O
A+O=O+A=A
IA=AI=A
A-A=O
Addition and scalar product.
A+B=B+A
(A + B) + C = A + (B + C)
r (A + B) = r A + r B
(r + s) A = r A + s A
(r s) A = r (s A)
Matrix product.
A0 == I
A2 == A A
An = A An - 1
(A B) C = A (B C)
A (B + C) = A B + A C
(A + B) C = A C + B C
Transpose and inverse.
IT = I

(AT)T = A
(A + B)T = AT + BT
(r A)T = r AT
(A B)T = BT AT
I-1 = I
A A-1 = A-1 A = I
(A B)-1 = B-1 A-1
(A-1)T = (AT)-1
Trace.
tr (A + B) = tr A + tr B
tr (r A) = r tr A
tr (A B) = tr (B A)
Determinant and adjoint.
det O = 0
det I = 1
det A = det AT
det (A B) = (det A) (det B)

Definitions

Match the following terms with their definitions.

____address>_______________1. Diagonal from the
upper left corner entry to the bottom right corner entry
____ determinant of 2x2 matrix>____________2. A
rectangular array of numbers enclosed in brackets
____ dimensions>____________3. Variation in size of a
matrix
____ main diagonal>__________4. Any matrix that has
the same number of rows as it does columns
____ matrix>________________5. Matrix in which all
of the entries are zero
____ scalar>________________6. The difference of the
products of the diagonals
____ square matrix>__________7. Number located
outside of a single matrix which is multiplied by each
entry of the matrix
____ zero matrix>____________8. Describes where
each value, or entry, of a matrix lives

Matrices & determinants

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