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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 ______.
asked in Linear Algebra by Veteran (99.8k points)
retagged by | 1.4k views
0
how to solve this simple question ?

5 Answers

+29 votes
Best answer
Hi,

For this kind of matrices Determinant is zero.

A will be a $3$x$3$ 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 $|A| = 0$.

References: https://www.youtube.com/watch?v=aKX5_DucNq8&list=PL221E2BBF13BECF6C&index=19
answered by Active (4.7k points)
edited by
+14
we know that
$\begin{align}
    \left. \begin{array}{l}
       \operatorname{rank}(AB) \leq \operatorname{rank}(A) \\
        \operatorname{rank}(AB) \leq \operatorname{rank}(B)
    \end{array}\right\} \implies \operatorname{rank}(AB) \leq \operatorname{min}(\operatorname{rank}(A), \operatorname{rank}(B)).
\end{align}$

Therefore deteminant of given matrix has to be zero to make its $ \operatorname{rank} \leq 1$
0
perfect use of properties :P
0
Very intuitive question

if you notice

our matrix A will have the first row as $2*[1\,\,9\,\,5]$

Second row as $-4 *[1\,\,9\,\,5]$

Third row as $7* [1\,\,9\,\,5]$

So, all three rows will be the linear combination of $[1\,\,9\,\,5]$ and hence det(A)=0
+7 votes
$A=\begin{bmatrix} 2\\ -4\\ 7 \end{bmatrix}\begin{bmatrix} 1 &9 &5 \end{bmatrix}$

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

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

$R1\rightarrow R1 + R2$

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

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

$|A|= 0$
answered by Boss (40.3k points)
+4

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 

+2 votes
mutiply matrices after that find determinant..

its zero..
answered by Veteran (55.1k points)
+1 vote
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.
answered by Loyal (8.3k points)
edited by
0 votes
$A= \begin{bmatrix} 2\\ −4\\7\end{bmatrix}\begin{bmatrix}1& 9& 5\end{bmatrix}$

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

So, only 1 row in the matrix

determinant must be 0
answered by Veteran (91.8k points)


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