GATE CSE 2005 | Question: 46

Question

Consider the set $H$ of all $3 * 3$ matrices of the type $$\left( \begin{array}{ccc} a & f & e \\ 0 & b & d \\ 0 & 0 & c \end{array} \right)$$ where $a,b,c,d,e$ and $f$ are real numbers and $abc ≠ 0.$ Under the matrix multiplication operation, the set $H$ is:

• A. a group
• B. a monoid but not a group
• C. a semi group but not a monoid
• D. neither a group nor a semi group

Answer 1

Given Information: Matrix is upper triangular. It's determinant is multiplication of principle diagonal elements. i.e., $abc.$
It is given that $abc \neq 0.$ So, Inverse for every such matrix exists.

Now this set is

1. Closed - You can see after multiplication matrix is in same format and $|AB| = |A||B| \neq 0$ as $|A|,|B| \neq 0$
2. Associative - Matrix multiplication is associative
3. Existence of Identity - Identity Matrix is present
4. Existence of Inverse - as determinant is non zero there exist inverse for every matrix

So, it is group.

Correct Answer: $A$

Comments

I think You should also check that " A inverse " is of the form of  " Upper triangular matrix ". Because it should belong to the set. Am I right ?

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Yes you should do that and fortunately in this case it does.

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inverse of the matrix will be as follows:

1/a   -f/ab   (fd-be)/abc

0     1/b        0

0       0         1/c

Since a,b,c,d,e and f are real numbers so, all the numbers in the above matrix is also real and since,      a*b*c != 0, means a!=0 and b!=0 and c!=0 So, 1/a=!0 , 1/b !=0, 1/c!=0  as well , So (1/a)*(1/b)*(1/c)!=0  . Hence , inverse matrix also belongs to set H

Answer 2

$abc\neq0$

The product of diagonal elements in a triangular matrix is the determinant.

=> determinant of such matrices $\neq0$

=> Matrices are non-singular

=> Matrices are invertible. -----------> #1

 

Closure holds.

Associativity holds. Matrix Chain Multiplication in Dynamic Programming is an example of this.

Identity holds. (The identity matrix)

Inverse holds. // From #1

Commutativity doesn't hold. As A.B $\neq$ B.A for matrices.

 

So, this is a group. Option A

Comments

The Rank of matrix is 3 , this become Invertible then it has inverse which means it is definitely group

Answer 3

It is given that $a,b,c,d,e,f$ are real numbers and $abc \neq 0$.
Consider the matrix
\[
A =
\begin{pmatrix}
a & f & e \\
0 & b & d \\
0 & 0 & c
\end{pmatrix}.
\]

Since $abc \neq 0$, we conclude that
\[
a \neq 0,\quad b \neq 0,\quad c \neq 0.
\]

Now, compute the determinant of the matrix:
\[
\det(A)
=
a
\begin{vmatrix}
b & d \\
0 & c
\end{vmatrix}
- f
\begin{vmatrix}
0 & d \\
0 & c
\end{vmatrix}
+ e
\begin{vmatrix}
0 & b \\
0 & 0
\end{vmatrix}.
\]

Simplifying,
\[
\det(A) = a(bc - 0) - f(0) + e(0) = abc.
\]

Since $abc \neq 0$, the determinant is non-zero. Hence, the matrix $A$ is non-singular and its inverse exists.

The identity matrix is
\[
I =
\begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{pmatrix},
\]
Which also belongs to this set of matrices.

Matrix multiplication is associative, but not commutative.  
The set is closed under matrix multiplication and every element has an inverse.Therefore, under matrix multiplication, the given set of matrices forms a Group. Hence, the correct option is Option A.

Answer 4

A group

Comments

let A b a null matrix..
A * Identity_Matrix = A = Identity_Matrix * A
I(n*n) satisfy above condition for every square matrix of order n..
for null or singular matrix , inverse not exist .. but it doesn't mean identity matrix not exist..

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But can we prove for this matrix identity matrix is  1  0  0

                                                                          0  1  0

                                                                          0   0  1

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In the matrix a,b,c,d,e,f all are present. so it is closed

matrix multiplication is associative

H * identity matrix= H ,so identity property satisfied

It is non singular matrix so inverse exist

So it is a group

But matrix multiplication is non commutative ,so it is not abelian group

Answer 5

if any matrix is non singular means it has non zero determinant then it is

-closure

-Associativity

-Identity

-Inverse

so it is group 

Answer 6

As we know that multipication ensures the associative and commutiative property  but one thing matric multipication over real number ensures only associative property not commutiative

check -->

1) closure property satisfy bcz there tell the real number and real number multipication we get the real numbers

2) as well we tell that that is associative

3) now come to identity element is the identity matrix 3*3

4) so it has determinant abc not equal to 0 so we can get the inverse of the matrix

but we can not get the commutiative property so it is a only a group .