Question

Use properties of determinants to find the value of each determinant if it is known that \(\left|\begin{array}{lll}x & y & z \\ u & v & w \\ 1 & 2 & 3\end{array}\right|=4\). \(\left|\begin{array}{ccc}x+3 & y+6 & z+9 \\ 3 u-1 & 3 v-2 & 3 w-3 \\ 1 & 2 & 3\end{array}\right|\)

Short answer

The value of the determinant is 12.

Step-by-step solution

- Understanding the given problem

Given is a determinant with known value: \[ \begin{vmatrix} x & y & z \ u & v & w \ 1 & 2 & 3 \end{vmatrix} = 4 \]We need to find the value of a new determinant: \[ \begin{vmatrix} x+3 & y+6 & z+9 \ 3u-1 & 3v-2 & 3w-3 \ 1 & 2 & 3 \end{vmatrix} \]

- Factor out constants from rows

Notice that the second row of the new determinant can be factored. Factoring out a 3 gives: \[ \begin{vmatrix} x+3 & y+6 & z+9 \ 3( u - \frac{1}{3} ) & 3( v - \frac{2}{3} ) & 3( w - 1 ) \ 1 & 2 & 3 \end{vmatrix} \]Because of the factor of 3 in the entire second row, we can factor out a 3 from the determinant: \[ 3 \begin{vmatrix} x+3 & y+6 & z+9 \ u - \frac{1}{3} & v - \frac{2}{3} & w - 1 \ 1 & 2 & 3 \end{vmatrix} \]

- Row operations

We can simplify the first row by subtracting 3 times the third row from it: This gives: \[ \begin{vmatrix} x+3 - 3 \cdot 1 & y+6 - 3 \cdot 2 & z+9 - 3 \cdot 3 \ \u - \frac{1}{3} & v - \frac{2}{3} & w - 1 \ 1 & 2 & 3 \end{vmatrix} = \begin{vmatrix} x & y & z \ u - \frac{1}{3} & v - \frac{2}{3} & w - 1 \ 1 & 2 & 3 \end{vmatrix} \]

- Linear combination of rows

Notice that the determinant now has its first row back to the original form: \[ 3 \begin{vmatrix} x & y & z \ u - \frac{1}{3} & v - \frac{2}{3} & w - 1 \ 1 & 2 & 3 \end{vmatrix} \] Consider the effect of adding a linear combination of rows. Since adding a multiple of the last row to the other rows does not change the determinant: This simplifies to: \[ 3 \begin{vmatrix} x & y & z \ u & v & w \ 1 & 2 & 3 \end{vmatrix} \]

- Substitute the known determinant value

Since we are given that \[ \begin{vmatrix} x & y & z \ u & v & w \ 1 & 2 & 3 \end{vmatrix} = 4 \]Therefore, \[ 3 \cdot 4 = 12 \]

Key concepts

properties of determinants

Determinants have several interesting properties that can simplify complex calculations.
One key property is that if we factor out a constant from any row or column, the determinant value is multiplied by that factor. For instance, if we factor a 3 out of a whole row, the determinant’s value gets multiplied by 3.
This principle is particularly useful when simplifying matrices by breaking them down into smaller, more manageable components. Knowing these properties can help us solve problems faster and with more accuracy.

factorization

Factorization involves breaking down an expression into a product of its factors. In the case of determinants, factorization helps streamline the matrix by dividing out common factors.
Imagine you have a row in your matrix where each element can be divided by 3. By factoring out the 3, you reduce the complexity of the determinant, making it easier to handle.
In our exercise, we factored out 3 from the entire second row, which allows us to simplify operations without changing the fundamental properties of the determinant.

row operations

Row operations are essential tools in linear algebra for manipulating matrices. Three main types of row operations include row swapping, row multiplication, and adding or subtracting rows.
In our exercise, we used a combination of these operations to transform the given matrix into a simpler form. Specifically, subtracting multiples of one row from another facilitated easier calculations.
Understanding and applying these operations correctly ensures we maintain the determinant’s properties while achieving a simpler or desired form of the matrix.

linear combinations of rows

A linear combination of rows involves adding or subtracting multiples of rows from each other. This technique can effectively simplify the structure of the determinant.
During our solution, subtracting three times the third row from the first row brought us back to the determinant's original form. This insight allows us to connect modified determinants with their known values.
Such operations do not affect the determinant’s value if correctly applied, preserving the equivalence and aiding in solving the problem efficiently.