GATE IT 2008 | Question: 29

Question

If $M$ is a square matrix with a zero determinant, which of the following assertion (s) is (are) correct?

S1: Each row of $M$ can be represented as a linear combination of the other rows
S2: Each column of $M$ can be represented as a linear combination of the other columns
S3: $MX = 0$ has a nontrivial solution
S4: $M$ has an inverse

• A. $S3$ and $S2$
• B. $S1$ and $S4$
• C. $S1$ and $S3$
• D. $S1, S2$ and $S3$

Comments

It should be false because of usage of "EACH"

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Both statements are true. If a matrix has determinant equal to $0$, then it means it has linearly dependent vectors.

A matrix $M$, has rows $< r_1, r_2, ...., r_n>$ then $M$ is said to be LI if and only if $c_1r_1 + c_2r_2 + .....c_nr_n = 0$ AND $c_1 = c_2 = ..... = c_n = 0$.

OR in other words, you can say that if at least one of $c_1,c_2,...c_n$ is not equal to zero in order to satisfy above equation, then M has Linearly dependent vectors and hence have determinant equal to zero.

Similarly, apply for columns also.

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3    5    7 

11  13 17

6    10  14

See here @Shubhanshu 1st row is not the linear combination of 2nd row but the det is zero.

Can you help me out where I'm wrong

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Here put $c_1=-2, c_2 = 0, c_3 = 1$. Since at least one of the "c" is non zero thus it has det = 0.

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For the same combination of contants can we write R2 as linear combination of others..no right since c2=0..
Question saying Each row can be written as linear combinatiin of others..is it saying at same time or using different combination of cobstants because for same combination there might be some rows which cant be represented as linear combination of other...where i m wrong ?

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@Shubhanshu.....

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Here you need to understand that R2 is also present in making the determinant equal to zero by taking C2 as 0.

Actually in the equation which I commented earlier, each row has to be multiplied by some constant and if any constant is equal to zero then it doesn't mean that, that row is not in linear combination. That row is present in linear combination but we took its constant as zero.

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Here put c1=−2,c2=0,c3=1. Since at least one of the "c" is non zero thus it has det = 0.

Here you said that this combination leads ti equation c1R1 + c2R2 + c3R3 = 0

But it is not asked if det is zero then rows are linearly dependent or not ??

It is asked Can each row be written as linear combination of other...?? they are not same things m i right .really confused :(

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being represented as a linear combination $\neq$ being linearly dependent.

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being represented as a linear combination ≠≠ being linearly dependent.

@JashanArora bro, how can both be different ?

can u pls give some examples for that ?

 

Thanks in advance

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Try linear combination keeping other two rows too in mind

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None of the answers is correct.

Reason: EACH row/column in statements S1 and S2.

if a set of vectors is linearly dependent then we can represent ATLEAST one vector as a linear combination of other vectors, not EACH vector.

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@Sachin Mittal 1 sir only statement  S3 is correct but it is don't in option

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only S3 statement is correct.

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@Sachin Mittal 1 sir thank youu soo much for confermation.

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@Sachin Mittal 1 sir plz elliminate happy mittal sir answer is it just copy paste    of madeeasy pyq book which is wrong…..

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Actually i think option C is correct only, AX=0 has a non-trivial solution if only if A is a  singular matrix.Remaining options are wrong, they might be correct if any one of the vectors can be represented as linear combinations of others.

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yes!!.

For matrices:{ [1 2 3], [3 6 9], [4 8 10] }  ----> Linearly dependent set

But, we cannot represent third vector as a linear combination of others

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Counter Example for each row/ col can expressed as linear combinations of other rows/columns

1. Determinant == 0 for any square matrix --> Rows/Columns are Linearly dependent.

2. Linearly Dependent vectors can be expressed as -

$c_0v_0 + c_1v_1 + c_2v_2 + ... + c_nv_n = 0$

Now, we can expressed $v_i$ as linear combination of other vectors only if it's corresponding $c_i \ne 0$

3. Consider the following vector set -

$\begin{bmatrix}2\\3\\4\\0\\\end{bmatrix},\begin{bmatrix}2\\3\\4\\0\\\end{bmatrix},\begin{bmatrix}0\\0\\0\\1\\\end{bmatrix}$

Here $v_1 - v_2 + 0v_3 = 0$ where $c_1 = 1, c_2 = -1, c_3 = 0$

We can find infinite values of $c_1,c_2$ that satisfy the equation but $c_3$ needs to be zero.

From point 2 and above statement, we can conclude that $v_3$ can never be expressed as linear combination of other vectors.

Let me know if there is anything wrong in the above argument.

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@Sachin Mittal 1

@Deepak Poonia

Sir I used necessary and sufficient condtiions to check if S1 and S2 are correct or not. And yes they are.

For a square matrix to have a zero determinant it is _______ to have all rows/columns a linear combination of other row/column.

We can say for square matrix determinant to be zero it is sufficient to say we can make all row/column as linear combination of other rows/columns as atleast one linearly dependent row or column can also make determinant 0. 

 

From this we can say,

If matrix determinant is 0 than  we can have all rows/columns as linear combination of each other.

So yes statement 1 and 2 are true.

 

And if can take an example can't we take zero matrix?

 

Please correct me if I am wrong.

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I think they haven't specified non trivial soln for statement 1 & 2, that's y it's correct @chetan-naik

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@Soham Kamat 

A square matrix with zero determinant (given) will have linearly dependent vectors.

The definition of linear dependent set of vectors is that - "Atleast one of the vectors from the set of vectors is a linear combination of other vectors".

However, S1 and S2 says that "Each" of the vector from the set is a linear combination of other. It could be the case that one of the vector has 0 as it's coefficient (as mentioned in my above comment) and we can't express it as LC of other vectors.

Hence, S1 and S2 are not necessarily true.

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got it, thanks dost, aur atb gate ke liye @chetan-naik

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Same to you brother! Will meet you soon! @Soham Kamat

Answer 1

None of the answers are correct for this question, and GATE should have given marks to all.
Only S3 is correct statement.

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Let's understand this using three simple examples:

 

			\[ \begin{array}{|c|c|c|c|} \hline  & \text{Example 1} & \text{Example 2} & \text{Example 3} \\ \hline \text{Suppose we have:} & v_1 + 2v_2 + v_3 = 0 & v_1 + v_2 + 0v_3 = 0 & 0v_1 + 2v_2 + 6v_3 = 0 \\ \hline \text{Can we say} & \text{Yes, because } v_3 & \text{Yes, because } v_2 & \text{Yes, because } v_2 = -3v_3, \\ v_1, v_2, v_3 & \text{is a linear combination} & \text{is a linear combination} & \text{showing } v_2 \\ \text{are linearly dependent?} & \text{of others:} & \text{of others:} & \text{is a linear combination} \\ \hline \text{Represent:} & v_3 = -(v_1 + 2v_2) & v_2 = -(v_1 + 0v_3) & v_2 = -3v_3 \\  & & \text{or} \quad v_2 = -v_1 & \\ \hline \end{array} \]

 

As we can see, if we can represent any of the vectors as a linear combination of others, then we say the vectors are linearly dependent.

Now, let's take Example 2 and 3 again:

In Example 2: Can we represent v₃ as a linear combination of the other vectors?

Answer: No

This is fine. As long as we have at least one vector that can be represented as a linear combination of the others, we are good to go.

In Example 3: Can we represent v₁ as a linear combination of the other vectors?

Answer: No

This is fine too.

However, in S1 and S2, the statement says "ALL," which makes both S1 and S2 false.

S3: MX=0 has a nontrivial solution

Answer: This is TRUE. Because if M has zero determinant, it means rows of M are linearly dependent, hence there are infinitely many solutions.

S4: M has an inverse

Answer: This is False because det(M) = 0

Answer 2

Since, $M$ has zero determinant, its rank is not full i.e., if $M$ is of size $3*3,$ then its rank is not $3.$ So there is a linear combination of rows which evaluates to $0$ i.e.

$k_{1}R_{1}+k_{2}R_{2}+\ldots +k_{n}R_{n}=0$

and there is a linear combination of columns which evaluates to $0$ i.e.,

$k_{1}C_{1}+k_{2}C_{2}+\ldots +k_{n}C_{n}=0$

Now any row $R_{i}$ can be written as linear combination of other rows as :

$R_{i}=-\dfrac{k_{1}}{k_{i}}R_{1}-\dfrac{k_{2}}{k_{i}}R_{2}-\ldots -\dfrac{k_{i-1}}{k_{i}}R_{i-1}-\dfrac{k_{i+1}}{k_{i}}R_{i+1}-\ldots-\dfrac{k_{n}}{k_{i}}R_{n} $

Similar is the case for columns.

Now $MX = 0$ always has one solution $: X = 0$ (which is called trivial solution). Now if $|M| = 0,$ then $MX = 0$ has non-trivial solutions also.

So, $(S1)$, $(S2)$, and $(S3)$ are true. So, option $D$ is correct.

Comments

You are dividing all the terms with ki to get a linear combination.
What if there's a row for which Ki = 0?

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Option D is correct only when in S1 and S2 they say ‘at least one’ instead of ‘each’.

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i think S1 and S2 is not true, as it is not necessary that each row /column is linearly independent of other row/column. as rank(M)<n we know that column and rows are not linearly independent so at least one row/coumn can be written as linear combination of other rows/column.
consider the case that first n-1 rows are linearly independent and last rows is zero, then these n rows are linearly dependent , but we can not write any one of those n-1 rows as linear combination of others.

Answer 3

If M is a square matrix with zero determinant, then the rows must be linearly dependent and the columns also must be linearly dependent.

We can take small examples to see how above fact works.

consider $\begin{bmatrix} 1 &2 \\ 2 & 4 \end{bmatrix}$

Here row2 is a linear combination of row1, and also you can see, columns are also dependent.

Now I take an example where columns are dependent,

$\begin{bmatrix} 3 &9 \\ 7 & 21 \end{bmatrix}$

But , row 2 is also a linear combination of Row1 as $7R_1-3R_2 \rightarrow R_2$ will result in $R_2$ being zero.

The matrix with zero determinant cannot have an inverse.

$MX=0$ has a non-trivial solution means for $X \not= 0$(the zero vector), $MX=0$ and yes this is always possible in a matrix with zero determinant because the columns will always be a linear combination of one another.

Why a matrix with linear rows and columns produces determinant 0?

if You decompose your matrix A(singular) into $LDU$ form where L is the lower triangular form, D is the diagonal form and U is the upper triangular form,your D matrix, which holds n pivots would look like this

$\begin{bmatrix} p_1& & & & & \\ & p_2 & & & & \\ & & &. & & \\ & & & &0 & \\ & & & & &.. \\ & & & & & p_n \end{bmatrix}$

one of the pivots would be zero because row elimination would cause so while you are generating U.

Now, the determinant of this D=$p_1.p_2.....p_k....0...p_n=0$

and hence, the determinant of the matrix A turns out to be 0.

Answer 4

Please someone verify my answer

For option A and B these could be correct if they have used word "Atleast One"  instead of "Each One" 

Take example of 3*3 matrix with rank 

1	2	3
0	0	4
0	0	0

R1 =  k * R2    only possible if k = 0   

R1 = k * R2     only possible if k = 0 

This means R1 is can not be represented as leaner  combination of  R2 and R3 

Hence only possible answer is C

Please correct if my answer is wrong 

Comments

this is absolutely right. it should be “atleast”….      bcz even if “all rows are linearly dependent ”

we can’t say that “each row can be represented as a linear combination of others” . good example

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I also think that there should be atleast, can anyone please clarify

Answer 5

Answer : None of the options are correct  ( Only statement 3 always correct )

Det ( M ) = 0 => Both columns & rows of square matrix M are Linearly Dependent.

Statement( 1 & 2 ) : A set of vectors are Linearly Dependent does’t mean that each one of them can be represented as linear combinations of other vectors. eg: 

0   0   0

0   1   2

1   0   0       // This matrix obviously have determinant 0, so columns and rows are linearly dependent, but 3^rd row ( 1   0   0  )  and first column (  [ 0   0   1 ]T ) cannot be represented as linear combinations of other rows/columns. So both S1 and S2 false.

Statement( 3 ) : This is correct, Ax = 0 will have a non trivial solution when the set of column vectors is Linearly dependent.

Statement( 4 ) : This is obviously false, A matrix have an inverse iff Det( M ) != 0

Answer 6

https://stats.stackexchange.com/questions/430671/singular-matrix-and-linear-dependency

Value of determinant = 0 ⇒ Singular matrix .

 

“Suppose M is a square matrix which is singular. The following are all equivalent:

1. The equation Ax=b has 0 or ∞many solutions depending on b.

2. det(A) = 0.

3. A does not have an inverse.

      4. The equation Ax=0 has solutions other than x=0.

      5. The columns of A are linearly dependent as vectors.

      6. The rows of A are linearly dependent “

         - (taken from page 44) http://www.ohiouniversityfaculty.com/youngt/IntNumMeth/lecture10.pdf

Comments

You put everything together. Thanks

7 – 0 will be one of it’s eigen value.