in Linear Algebra edited by
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19 votes

Consider the following statements:

  • S1: The sum of two singular $n \times n$ matrices may be non-singular
  • S2: The sum of two $n \times n$ non-singular matrices may be singular


Which one of the following statements is correct?

  1. $S1$ and $S2$ both are true
  2. $S1$ is true, $S2$ is false
  3. $S1$ is false, $S2$ is true
  4. $S1$ and $S2$ both are false
in Linear Algebra edited by
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2 Comments

I think that without taking any example we can get the answer.   Following is the explaination for the two statements.

statement 1 : let us consider any two matrices of n-by-n which are singular, where n >= 2 (beacuse for n = 1 we can only have 1 matrix which is singular). Now, the addition of these two matrices may be singular or nonsingular. Assume that all such possible pairs of matrices yield the sum as non-singular, then certainly statement 1 is true. Now, assume only some possible pairs of n-by-n matrices yield the sum as non-singular while others as singular, then also statement 1 is true, due to the keyword "may be" in the statement. Now, assume the last possibility that all pairs of such matrices yield as singular matrix , then also statement 1 is true again due to "may be" keyword. Hence, statement 1 is true for all cases.

statement 2 : we can argue similarly that statement 2 is also true in all cases.

Hence option A is right answer. Please correct me if I am wrong.

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6 Answers

21 votes
A singular matrix is a square matrix whose determinant is 0. Whenever you solve question related to determinant, think of triangular matrix, like upper triangular matrix, because its determinant is just product of diagonal entries, and so easy to visualize. So definitely, in a singular matrix, one of the entries in diagonal must be zero. Similarly, in non-singular, none of the entries should be zero.

Now take any matrices and just check .

So we see that both S1 and S2 are true. So option (A) is correct.
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This answer has been copied from here.

When you copy, paste information from somewhere add source in your answer.

1
9 votes
ans should be A.

for S1 :  singular matrices matrix[0,0,0,1] + matrix[1,0,0,0] = matrix[1,0,0,1], which is non singular

for S2 : non-singular matrices matrix[1,0,0,1] + matrix[0,1,1,0] = matrix[1,1,1,1], which is singular

3 Comments

but matrices should be square ..?
0
yes, that is true
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edited by
They are square matrices--

$\begin{matrix} 0&0 \\ 0&1\end{matrix}\quad +$ $\quad\begin{matrix} 1&0 \\ 0&0\end{matrix}$

$=\begin{matrix} 1&0\\0&1\end{matrix}$

$\Rightarrow (singular+singular)\;$ may be $non\; singular$
4
4 votes

det (A + B)  is not equal to det (A) + det (B), it means that

if A and B both are singular then, the determinant of sum of A and B may not to be equal to 0.

S1 is correct

Similarly, if A and B are non singular matrix, then det(A+B) may not be Non Zero, it can be Zero means Singular.

S2 is also correct

3 votes
I think there is a contradiction  Let me explain with example
 If A = [ 2 10
            1  5]
B = [ 3 6
        2  4]
 Sum [ 5 16
           3  9]
| sum| = 9×5_16×3=- 3 which is not equal to zero . So it is non singular
edited by

1 comment

I too, took a specific example and found that “ The sum of two singular n×n matrices was resulting into singular matrix ”.

 

I think, this is not contradiction to given statement “ The sum of two singular n×n matrices may be non-singular ”.

If the statement would have been given as “ The sum of two singular n×n matrices must be non-singular ”, then contradiction would have been present and hence this statement would be marked as false.

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2 votes

$\boldsymbol{\textbf{A. Is correct option }}$

2 Comments

@

Yes, you are right.

Using counterexample is a good way to solve such type of problems.

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Question is not asking for "always" the statements has to be true,so no  need to find counter example ..it is mentioned as "may be" even if one example statisfies ,statement will be true !!
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0 votes

So $S_1$ and $S_2$ both are true,

 

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