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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 ______.
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0
how to solve this simple question ?

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 $|A| = 0$.

edited
+17
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
$A=\begin{bmatrix} 2\\ -4\\ 7 \end{bmatrix}\begin{bmatrix} 1 &9 &5 \end{bmatrix}$

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

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

$R1\rightarrow R1 + R2$

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

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

$|A|= 0$
+6

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

mutiply matrices after that find determinant..

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

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

So, only 1 row in the matrix

determinant must be 0