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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

  1. $S3$ and $S2$
  2. $S1$ and $S4$
  3. $S1$ and $S3$
  4. $S1, S2$ and $S3$
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7 Answers

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40 votes
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.
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7 votes
7 votes
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.
3 votes
3 votes

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 

1 votes
1 votes

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.
  1. det(A) = 0.
  1. 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

Answer:

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