Properties of Determinants

Properties of Determinants

(1) The value of the determinant remains unchanged if its rows and columns are interchanged.
$\begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix} = \begin{vmatrix} a & d & g \\ b & e & h \\ c & f & i \end{vmatrix}$
Verification
LHS=a(ei - fh) -b(di - fg) + c(dh - eg)
RHS (expanding along column C1)=a(ei - fh) -b(di - fg) + c(dh - eg)

(2)If we interchange any two rows (or columns), then sign of determinant changes.
$\begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix} = -\begin{vmatrix} d & e & f \\ a & b & c \\ g & h & i \end{vmatrix}$
Verification
We can easily proof this by expanding
LHS=a(ei - fh) -b(di - fg) + c(dh - eg)=aei-afh=bdi+bfg+cdh-ceg
RHS= d(bi - ch) -e(ai - cg) + f(ah - bg)=dbi-dch-aei+ecg+fah-fbg=-(aei-afh=bdi+bfg+cdh-ceg)

(2)If any two rows (or columns) of a determinant are identical (all corresponding elements are same), then value of determinant is zero.
Proof If we interchange the identical rows (or columns) of the determinant $\Delta$, then $\Delta$ does not change. However, by above property, it follows that $\Delta$ has changed its sign
$\Delta= - \Delta$
hence $\Delta=0$

(3)If we multiply each element of a row or a column of a determinant by constant k, then value of determinant is multiplied by k.
$A= \begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix}$ $B= \begin{vmatrix} ka & kb & kc \\ d & e & f \\ g & h & i \end{vmatrix} = k \begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix} =kA$
Verification
A=a(ei - fh) -b(di - fg) + c(dh - eg)=aei-afh=bdi+bfg+cdh-ceg
B=ka(ei - fh) -kb(di - fg) + kc(dh - eg)=k(aei-afh=bdi+bfg+cdh-ceg)
Similarly
$C=|kA|= \begin{vmatrix} ka & kb & kc \\ kd & ke & kf \\ kg & kh & ki \end{vmatrix} = k^3 \begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix} =k^3A$

(4) If some or all elements of a row or column of a determinant are expressed as sum of two (or more) terms, then the determinant can be expressed as sum of two (or more) determinants
$\begin{vmatrix} a+k & b+k & c+k \\ d & e & f \\ g & h & i \end{vmatrix} = \begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix} + \begin{vmatrix} k & k & k \\ d & e & f \\ g & h & i \end{vmatrix}$
Verification
We can easily proof this by expanding
LHS=(a+k)(ei - fh) -(b+k)(di - fg) + (c+h)(dh - eg)=a(ei - fh) -b(di - fg) + c(dh - eg) + k(ei - fh) -k(di - fg) + k(dh - eg)
RHS=a(ei - fh) -b(di - fg) + c(dh - eg) + k(ei - fh) -k(di - fg) + k(dh - eg)

(5)If to each element of a row or a column of a determinant the equimultiples of corresponding elements of other rows or columns are added, then value of determinant remains same
$A= \begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix} = \begin{vmatrix} a+kd & b+ke & c+kf \\ d & e & f \\ g & h & i \end{vmatrix}$
Verification
RHS= $\begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix} + \begin{vmatrix} kd & ke & kf \\ d & e & f \\ g & h & i \end{vmatrix}$
$=\begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix} + k\begin{vmatrix} d & e & f \\ d & e & f \\ g & h & i \end{vmatrix}$
As second determinant is having same values, so it is zero
Hence
$=\begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix}$

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