Let |A| = |■(a_1&b_1&c_1@a_2&b_2&c_2@a_3&b_3&c_3 )|  And, let |B| be the determinant obtained by multiplying the elements of 2nd row of A by k and adding these products to the elements of the 1st row. Then, |B| = |■(a_1+ka_2&b_1+kb_2&c_1+kc_2@a_2&b_2&c_2@a_3&b_3&c_3 )|           (by prop. 5.)        = |■(a_1&b_1&c_1@a_2&b_2&c_2@a_3&b_3&c_3 )| + |■(ka_2&kb_2&kc_2@a_2&b_2&c_2@a_3&b_3&c_3 )|            (by prop. 4.)                = |A| + k |■(a_2&b_2&c_2@a_2&b_2&c_2@a_3&b_3&c_3 )|               = |A| + k × 0 (∵ R1 = R2, by prop. 3.)       = |A| Hence verified.
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