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

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 $7$ [19 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 (4.9k points) edited +22 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 0 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. +9 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 (41k points) +9 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 +5 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.

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

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..
+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
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$A=-56[1\; 9\; 5]$   How??

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

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