If each element of a row (or a column) of a determinant is multiplied by a constant k, then its value gets...A... by k. Here, A refers to

Question

If each element of a row (or a column) of a determinant is multiplied by a constant k, then its value gets...A... by k. Here, A refers to (A) added (B) subtracted (C) divided (D) multiplied

Tardigrade Admin · Accepted Answer

If each element of a row (or a column) of determinant is multiplied by a constant

k

, then its value gets multiplied by

k

.
Verification Let

Δ = ∣ ∣ a_{1} a_{2} a_{3} b_{1} b_{2} b_{3} c_{1} c_{2} c_{3} ∣ ∣

and

Δ_{1}

be the determinant obtained by multiplying the elements of the first row by

k

.
Then,

Δ_{1} = ∣ ∣ k a_{1} a_{2} a_{3} k b_{1} b_{2} b_{3} k c_{1} c_{2} c_{3} ∣ ∣

By expanding along tirst row, we get

Δ_{1} = k a_{1} (b_{2} c_{3} - b_{3} c_{2}) - k b_{1} (a_{2} c_{3} - c_{2} a_{3}) + k c_{1} (a_{2} b_{3} - b_{2} a_{3})

= k [a_{1} (b_{2} c_{3} - b_{3} c_{2}) - b_{1} (a_{2} c_{3} - c_{2} a_{3}) + c_{1} (a_{2} b_{3} - b_{2} a_{3})]

Hence,

∣ ∣ k a_{1} a_{2} a_{3} k b_{1} b_{2} b_{3} k c_{1} c_{2} c_{3} ∣ ∣ = k ∣ ∣ a_{1} a_{2} a_{3} b_{1} b_{2} b_{3} c_{1} c_{2} c_{3} ∣ ∣

By this property, we can take out any common factor from anyone row or anyone column of a given determinant.

Q. If each element of a row (or a column) of a determinant is multiplied by a constant k, then its value gets...A... by k. Here, A refers to

added

subtracted

divided

multiplied

Q. If each element of a row (or a column) of a determinant is multiplied by a constant $k$ , then its value gets...A... by $k$ . Here, A refers to