GATE CSE 2014 Set 2 | Question: 4

Question

If the matrix $A$ is such that $$A= \begin{bmatrix} 2\\ −4\\7\end{bmatrix}\begin{bmatrix}1& 9& 5\end{bmatrix}$$ then the determinant of $A$ is equal to ______.

Comments

Can't we use following logic??

|A.B|=|A|*|B|

and since determinant is defined only for square matrices...leading to |A|=0 and |B|=0.

Thus, |A.B|=0 

Correct me if I am wrong..

Thanks.

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There are some point below which need to be understood in matrices:

1. Most of the students have answered below by calculating the 3*3 matrices, which is correct way but not optimal here.  If you’re confused how 3*3 matrix is even possible, please note A is 3*1 and B is 1*3. The rule to check if multiplication is even possible or not, there should always be n=p while multiplying m*n and p*q matrices.
    
2. The optimal way here is a CONCEPT. The concept says that given matrices A and B, if we have n Linearly Independent Vectors in A then we can generate at most n Linearly Independent Vectors in multiplication of both matrices i.e A*B.

Therefore, in first matrix we have given column [2 -4 7] which is a LI column, and in 3*3 matrix of A*B we have at most one LI Column possible. Rest 2 columns will be Linearly Dependent i.e redundant columns. So Determinant will always be zero.

 

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here we have 3*1 and 1*3 matrices ,and we are getting 3*3 matices first column of matrix is [2 -4 7] second column is [18  -36  63] and third column is [10 -20 35] if we have one pair is LD in the marix then the whole matrix is LD and we know that the determinant of LD matrix is 0 . second this is 5[2 -4 7]  = [10 -20 35] so [2 -4 7]

is not LI.If  i am wrong then please correct me

Answer 1

A is given as the product of 2 matrices which are of order 3 x 1 and 1 x 3 respectively.

So after multiplication of these matrices, matrix A would be a square matrix of order 3 x 3. So, matrix A is : 2 18 10 -4 -36 -20 7 63 35

Now, we can observe by looking at the matrix that row 2 can be made completely zero by using row 1, this is to be done by using the row operation of matrix which here is : R2 <- R2 + 2R1

After applying above row operation in the matrix, the resultant matrix would be: 2 18 10 0 0 0 7 63 35 i.e. Row 2 has become zero now.

And if a square matrix has a row or column with all its elements as 0, then its determinant is 0. ( A property of a square matrix ) Hence answer is 0.

Note: Determinant is defined only for square matrices, and it is a number which encodes certain properties of a matrix, for ex: a square matrix with determinant 0 does not has its inverse matrix.

Comments

Refer : https://www.geeksforgeeks.org/gate-gate-cs-2014-set-2-question-14/

Answer 2

$A= \begin{bmatrix} 2\\ −4\\7\end{bmatrix}\begin{bmatrix}1& 9& 5\end{bmatrix}$

  $A=-56\begin{bmatrix}1& 9& 5\end{bmatrix}$ [1st matrix is just common factor, so multiply it's terms]

So, only 1 row in the matrix

determinant must be 0

Comments

@srestha 

$A=-56[1\;  9\;  5]$   How??

A results in a 3*3 matrix so it can't be a row matrix.

Answer 3

a little deep dive to the easy problem…

Answer 4

Answer is 0