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If $$\Delta$$$$\left|\begin{array}{rrr} {2} & {-3} & {5} \\ {6} & {0} & {4} \\ {1} & {5} & {-7} \end{array}\right|$$ and $$\Delta_{1}=$$ $$\left|\begin{array}{rrr} {2} & {-3} & {5} \\ {1} & {5} & {-7}\\ {6} & {0} & {4} \end{array}\right|$$
then  $$\Delta=-\Delta_{1}$$
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## QuestionMathsClass 12

If $$\Delta$$$$\left|\begin{array}{rrr} {2} & {-3} & {5} \\ {6} & {0} & {4} \\ {1} & {5} & {-7} \end{array}\right|$$ and $$\Delta_{1}=$$ $$\left|\begin{array}{rrr} {2} & {-3} & {5} \\ {1} & {5} & {-7}\\ {6} & {0} & {4} \end{array}\right|$$
then  $$\Delta=-\Delta_{1}$$

$$\Delta=\left|\begin{array}{ccc} {2} & {-3} & {5} \\ {6} & {0} & {4} \\ {1} & {5} & {-7} \end{array}\right|$$ = – 28 (The value of the determinant remains unchanged if its rows and columns are interchanged.)
Interchanging rows R2 and R3 i.e., R2 $$\leftrightarrow$$ R3, we have
$$\Delta_{1}=\left|\begin{array}{ccc} {2} & {-3} & {5} \\ {1} & {5} & {-7} \\ {6} & {0} & {4} \end{array}\right|$$
Expanding the determinant $$\Delta$$1 along first row, we have
$$\Delta_{1}=2\left|\begin{array}{rr} {5} & {-7} \\ {0} & {4} \end{array}\right|-(-3)\left|\begin{array}{rr} {1} & {-7} \\ {6} & {4} \end{array}\right|+5\left|\begin{array}{ll} {1} & {5} \\ {6} & {0} \end{array}\right|$$
= 2 (20 – 0) + 3 (4 + 42) + 5 (0 – 30)
= 40 + 138 – 150 = 28
Clearly $$\Delta_{1}=-\Delta$$
Hence, If any two rows (or columns) of a determinant are interchanged, then sign of determinant changes.