Question

Multiplication of a determinant by a scaler

Answer 1

No option provided as this is a descriptive question.

Answer 2

When a determinant is multiplied by a scalar $k$, the scalar multiplies only one row or only one column of the determinant. The value of the determinant is then $k$ times the original value.

Let $D$ be an $n \times n$ determinant:

a_{11} & a_{12} & \dots & a_{1n} \\ a_{21} & a_{22} & \dots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \dots & a_{nn} \end{vmatrix}$$ If we multiply the determinant $ D $ by a scalar $ k $, the result $ k \cdot D $ can be represented by multiplying any single row (say, the $ i $-th row) by $ k $: $$k \cdot D = \begin{vmatrix} a_{11} & a_{12} & \dots & a_{1n} \\ \vdots & \vdots & \ddots & \vdots \\ k a_{i1} & k a_{i2} & \dots & k a_{in} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \dots & a_{nn} \end{vmatrix}$$ Alternatively, it can be represented by multiplying any single column (say, the $ j $-th column) by $ k $: $$k \cdot D = \begin{vmatrix} a_{11} & \dots & k a_{1j} & \dots & a_{1n} \\ a_{21} & \dots & k a_{2j} & \dots & a_{2n} \\ \vdots & \ddots & \vdots & \ddots & \vdots \\ a_{n1} & \dots & k a_{nj} & \dots & a_{nn} \end{vmatrix}$$ This property distinguishes scalar multiplication of a determinant from scalar multiplication of a matrix, where every element of the matrix is multiplied by the scalar. **Example:** Let $ D = \begin{vmatrix} 1 & 2 \\ 3 & 4 \end{vmatrix} $. The value of $ D = (1)(4) - (2)(3) = 4 - 6 = -2 $. If we multiply $ D $ by a scalar $ k=5 $, then $ k \cdot D = 5 \times (-2) = -10 $. Applying the rule, we can multiply the first row by 5: $$\begin{vmatrix} 5 \times 1 & 5 \times 2 \\ 3 & 4 \end{vmatrix} = \begin{vmatrix} 5 & 10 \\ 3 & 4 \end{vmatrix} = (5)(4) - (10)(3) = 20 - 30 = -10$$ Or, we can multiply the second column by 5: $$\begin{vmatrix} 1 & 5 \times 2 \\ 3 & 5 \times 4 \end{vmatrix} = \begin{vmatrix} 1 & 10 \\ 3 & 20 \end{vmatrix} = (1)(20) - (10)(3) = 20 - 30 = -10$$ Both results match $ k \cdot D $. The final answer is $\boxed{No\ option\ provided}$ **Explanation of the solution:** When a determinant is multiplied by a scalar, the scalar multiplies only one row or only one column of the determinant.