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The property states that,

"If to each element of any row (or column) of a matrix, product of a scalar and a corresponding element of any other row (or column) is added, the determinant of the new matrix is same as that of the original matrix"

First of all, I calculated the determinant of the original third-order non-zero matrix, say $A$.

Input: I first tested this property by adding the "product of a scalar $k_1$ and the corresponding elements of any column, say $C_2$" to only one column, say $C_1$, of the original matrix and then calculated the determinant of the new matrix $B$.

Result: $\det{B}=\det{A}$ (Property verified)

Input: Then, I tested this property by adding the "product of a scalar $k_1$ and the corresponding elements of column $C_2$" to $C_1$ and by adding the "product of a different scalar $k_2$ and the corresponding elements of another column $C_3$ to the column $C_2$ as well. A new matrix appeared, say $C$.

Result: $\det {C} = \det {A}$ (Property verified)

Input: Then, ultimately, I tested this property by taking matrix $C$ and adding the "product of a new scalar $k_3$ and the corresponding elements of the column $C_1$" to column $C_3$. A new matrix $D$ came up.

Result: $\det {D}= (\det {A})(1+k_1k_2k_3)$

$ \implies \det {D} \neq \det {A}$, unless $k_1=0$ or $k_2=0$ or $k_3=0$ or $k_1=k_2=k_3=0$ (Property invalid)

Note: In first two inputs, changes are made with matrix A directly however in third input changes are made to C in order to save time.

I think, my understanding to this property is flawed. Ensure me if it is the case.

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  • 3
    $\begingroup$ To which definition of $\det$ do you refer? $\endgroup$ – Michael Hoppe Oct 20 '15 at 8:38
  • $\begingroup$ One associated with Minor and co-factor method. $\endgroup$ – Sufyan Naeem Oct 23 '15 at 14:10
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The problem is in your incorrect applying the property for Input 3. You take the matrix with columns $C_k$ $$ A=\left[\matrix{C_1 & C_2 & C_3}\right]. $$

  1. $B=\left[\matrix{C_1+k_1\color{blue}{C_2} & \color{blue}{C_2} & C_3}\right]$. Correct.
  2. $C=\left[\matrix{C_1+k_1C_2 & C_2+k_2\color{green}{C_3} & \color{green}{C_3}}\right]$. Correct.
  3. $D=\left[\matrix{\color{magenta}{C_1+k_1C_2} & C_2+k_2C_3 & C_3+k_3\color{red}{C_1}}\right]$. Incorrect (!!!)

The correct way to apply the property in Case 3 is $$ D_{\text{ok}}=\left[\matrix{\color{magenta}{C_1+k_1C_2} & C_2+k_2C_3 & C_3+k_3\color{magenta}{(C_1+k_1C_2)}}\right] $$

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  • 2
    $\begingroup$ @SufyanNaeem As stated, you may only add multiples of columns from the same matrix. Not preceding matrices, the same (!), it is important. The problem in 3 is that $C_1$ is not a column of $D$, but a column of other matrix $A$. So the property is not valid. $\endgroup$ – A.Γ. Oct 29 '15 at 16:02
  • $\begingroup$ @SufyanNaeem that's what I was trying to say that the statement has double meanings. A.G. is definitely right. "You may only add multiples of columns from the same matrix". +1 given. $\endgroup$ – Man_Of_Wisdom Oct 29 '15 at 19:14

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