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60 votes

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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

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15 votes

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

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

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

$R1\rightarrow R1 + R2$

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

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

$|A|= 0$

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

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

$R1\rightarrow R1 + R2$

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

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

$|A|= 0$

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.