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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 ______.
in Linear Algebra retagged by
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3 Comments

how to solve this simple question ?
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easiest question that you expect
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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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6 Answers

59 votes
59 votes
Best answer

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

edited by

4 Comments

Yes you also feel it logially. You have only 1 row or 1 vector. It's independent because you don't have any other row or column available which can generate it right? So true.

You also check my answer on this one.
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@Ayush Upadhyayay

Your above example how $rank(A)=0?$

I think $rank(A)=1?$

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@Lakshman-Yes buddy thanks for correction. I am bit tired by now so made mistake.
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15 votes
15 votes
$A=\begin{bmatrix} 2\\ -4\\ 7 \end{bmatrix}\begin{bmatrix} 1 &9 &5 \end{bmatrix}$

$A=\begin{bmatrix} 2&18 &10 \\ -4 & -36 &-20 \\ 7&63 &35 \end{bmatrix}$

$|A|=2 \times 4 \times 7\begin{vmatrix} 1&9 &5 \\ -1 & -9 &-5 \\ 1&9 &5 \end{vmatrix}$

$R1\rightarrow R1 + R2$

$|A|=2 \times 4 \times 7\begin{vmatrix} 0&0 &0 \\ -1 & -9 &-5 \\ 1&9 &5 \end{vmatrix}$

$|A|=2 \times 4 \times 7 \times 0$

$|A|= 0$

1 comment

edited by

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 

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10 votes
10 votes

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.

1 comment

$det(row\;matrix*column\;matrix)=0$   

Where row and column matrices are of order n*1 and 1*n (n>1) respectively.

is it true always??
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2 votes
2 votes
mutiply matrices after that find determinant..

its zero..
Answer:

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