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.
For these kind of matrices Determinant is zero. $A$ will be a $3\times3$ matrix where the first row will be $2 [1\;9\;5],$ second row will be $-4 [1\;9\;5]$ and third will be $7 [1\;9\;5].$ That is, all the rows of $A$ are linearly dependent which means $A$ is singular. When matrix is singular $|A| = 0$. Reference: https://www.youtube.com/watch?v=aKX5_DucNq8&list=PL221E2BBF13BECF6C&index=19
@Ayush Upadhyaya @Sachin Mittal 1 sir,
Can you please check ..whether i understood it right ??
here A = $\begin{bmatrix} 2 \\ -4\\ 7\\ \end{bmatrix} \begin{bmatrix} 1 & 9 & 5 \end{bmatrix}$
Rank of Column vector/matrix is always 1 because single column is independent
Rank of Row Vector/matrix is always 1 because single row is independent
Not sure about above 2 statements ?
And we know $rank(PQ) = min(rank(P),rank(Q))$
Hence here $rank(A) = min(1,1)$ ==> $rank(A) <= 1$
Size of Matrix A will be 3 X 3 and $rank <=1 $ ==> All Submatrices of A of size >1 has Determinant = 0
Hence $\left | A \right |$ should be 0
Yes..tht is easy..m just trying to understand concept mentioned in sachin sir's comment.
is this true ??
Rank of Column vector/matrix is always 1 because single column is independent Rank of Row Vector/matrix is always 1 because single row is independent
@Ayush Upadhyayay
Your above example how $rank(A)=0?$
I think $rank(A)=1?$
You have no need to solve it further
From here u can see two rows are identical when two rows are identical we have property
if two rows or column identical then determinant will be zero
It's a very simple question and answer can be directly given 0 without lifting your pen.
$\begin{bmatrix} 2\\ -4\\ 7 \end{bmatrix}.$$\begin{bmatrix} 1 & 9 & 5 \end{bmatrix}$
Column1 of Matrix A=$1 \times\begin{bmatrix} 2\\ -4\\ 7 \end{bmatrix}$
Column2 of Matrix A=$9 \times \begin{bmatrix} 2\\ -4\\ 7 \end{bmatrix}$
Column3 of Matrix A=$5 \times \begin{bmatrix} 2\\ -4\\ 7 \end{bmatrix}$
All columns of A are linear combinations of $\begin{bmatrix} 2\\ -4\\ 7 \end{bmatrix}$ and hence independent column is only column 1-->$\begin{bmatrix} 2\\ -4\\ 7 \end{bmatrix}$
When an nxn matrix does not have full set of n independent columns
(1)Surely it's determinant is 0.
(2)One of the Eigen-values of such matrix should be 0.
(3)Such matrix won't be diagonalizable.
(4)On Reducing such matrix to Echelon form(U), while reducing, you will find a 0 entry at somewhere along the diagonal and hence this matrix will never be LU Decomposible.
@srestha
$A=-56[1\; 9\; 5]$ How??
A results in a 3*3 matrix so it can't be a row matrix.