2.1k views
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 ______.
retagged | 2.1k views
+1
how to solve this simple question ?
+1
easiest question that you expect

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 $|A| = 0$.

answered by Active (4.6k points)
edited by
+24
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
+4
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
+1

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

0
@Jatin-you can directly tell that determinant of such matrix is 0. How?

$A=\begin{bmatrix} 2.1 & 2.9 &2.5 \\ -4.1 & -4.9 &-4.5 \\ 7.1 & 7.9 & 7.5 \end{bmatrix}$

Please note that $.$ operator shown above is actually multiplication operator.

You can directly see all combinations of matrix are a linear combination of $\begin{bmatrix} 2\\ -4\\ 7 \end{bmatrix}$

Hence $rank(A)=1$ and $det(A)=0$
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

0
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.
0

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

I think $rank(A)=1?$

0
@Lakshman-Yes buddy thanks for correction. I am bit tired by now so made mistake.
$A=\begin{bmatrix} 2\\ -4\\ 7 \end{bmatrix}\begin{bmatrix} 1 &9 &5 \end{bmatrix}$

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

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

$R1\rightarrow R1 + R2$

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

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

$|A|= 0$
answered by Boss (40.9k points)
+10

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.

answered by Boss (24.9k points)
0
$det(row\;matrix*column\;matrix)=0$

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

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

its zero..
answered by Veteran (59.9k 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 (9.3k points)
edited
0
$A= \begin{bmatrix} 2\\ −4\\7\end{bmatrix}\begin{bmatrix}1& 9& 5\end{bmatrix}$

$A=-56\begin{bmatrix}1& 9& 5\end{bmatrix}$ [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 (112k points)
0

$A=-56[1\; 9\; 5]$   How??

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

1
2