Question

If \(\overline{\mathrm{a}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}, \overline{\mathrm{b}}=\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}, \overline{\mathrm{c}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}}\), then value of \(\left|\begin{array}{ccc}\bar{a} \cdot \bar{a} & \bar{a} \cdot \bar{b} & \bar{a} \cdot \bar{c} \\ \bar{b} \cdot \bar{a} & \bar{b} \cdot \bar{b} & \bar{b} \cdot \bar{c} \\ \bar{c} \cdot \bar{a} & \bar{c} \cdot \bar{b} & \bar{c} \cdot \bar{c}\end{array}\right|=\) (1) 16 (2) 24 (3) 256 (4) 144

Short answer

The value of the determinant is 16.

Step-by-step solution

- Compute the dot products

Calculate the dot products for all pairs of vectors. \(\overline{\mathrm{a}} \cdot \overline{\mathrm{a}} = (1 \hat{\mathrm{i}} + 1 \hat{\mathrm{j}} + 1 \hat{\mathrm{k}}) \cdot (1 \hat{\mathrm{i}} + 1 \hat{\mathrm{j}} + 1 \hat{\mathrm{k}}) = 1^2 + 1^2 + 1^2 = 3\)\(\overline{\mathrm{a}} \cdot \overline{\mathrm{b}} = (1 \hat{\mathrm{i}} + 1 \hat{\mathrm{j}} + 1 \hat{\mathrm{k}}) \cdot (1 \hat{\mathrm{i}} - 1 \hat{\mathrm{j}} + 1 \hat{\mathrm{k}}) = 1 - 1 + 1 = 1\)\(\overline{\mathrm{a}} \cdot \overline{\mathrm{c}} = (1 \hat{\mathrm{i}} + 1 \hat{\mathrm{j}} + 1 \hat{\mathrm{k}}) \cdot (1 \hat{\mathrm{i}} + 2 \hat{\mathrm{j}} - 1 \hat{\mathrm{k}}) = 1 + 2 - 1 = 2\)\(\overline{\mathrm{b}} \cdot \overline{\mathrm{a}} = \overline{\mathrm{a}} \cdot \overline{\mathrm{b}} = 1\)\(\overline{\mathrm{b}} \cdot \overline{\mathrm{b}} = (1 \hat{\mathrm{i}} - 1 \hat{\mathrm{j}} + 1 \hat{\mathrm{k}}) \cdot (1 \hat{\mathrm{i}} - 1 \hat{\mathrm{j}} + 1 \hat{\mathrm{k}}) = 1 + 1 + 1 = 3\)\(\overline{\mathrm{b}} \cdot \overline{\mathrm{c}} = (1 \hat{\mathrm{i}} - 1 \hat{\mathrm{j}} + 1 \hat{\mathrm{k}}) \cdot (1 \hat{\mathrm{i}} + 2 \hat{\mathrm{j}} - 1 \hat{\mathrm{k}}) = 1 - 2 - 1 = -2\)\(\overline{\mathrm{c}} \cdot \overline{\mathrm{a}} = \overline{\mathrm{a}} \cdot \overline{\mathrm{c}} = 2\)\(\overline{\mathrm{c}} \cdot \overline{\mathrm{b}} = \overline{\mathrm{b}} \cdot \overline{\mathrm{c}} = -2\)\(\overline{\mathrm{c}} \cdot \overline{\mathrm{c}} = (1 \hat{\mathrm{i}} + 2 \hat{\mathrm{j}} - 1 \hat{\mathrm{k}}) \cdot (1 \hat{\mathrm{i}} + 2 \hat{\mathrm{j}} - 1 \hat{\mathrm{k}}) = 1 + 4 + 1 = 6\)

- Construct the matrix

Form the matrix using the computed dot products: \(\begin{pmatrix} 3 & 1 & 2 \ 1 & 3 & -2 \ 2 & -2 & 6 \end{pmatrix}\)

- Compute the determinant

Calculate the determinant of the matrix: \( \left| \begin{matrix} 3 & 1 & 2 \ 1 & 3 & -2 \ 2 & -2 & 6 \end{matrix} \right| \)Using cofactor expansion along the first row:\[\begin{align*} 3 \cdot \left( \begin{matrix} 3 & -2 \ -2 & 6 \end{matrix} \right) - 1 \cdot \left( \begin{matrix} 1 & -2 \ 2 & 6 \end{matrix} \right) + 2 \cdot \left( \begin{matrix} 1 & 3 \ 2 & -2 \end{matrix} \right) \end{align*}\]Compute each minor determinant:\[= 3(3 \cdot 6 - (-2) \cdot (-2)) - 1(1 \cdot 6 - (-2) \cdot 2) + 2(1 \cdot (-2) - 3 \cdot 2)\]\[= 3(18 - 4) - 1(6 - (-4)) + 2(-2 - 6)\]\[= 3 \cdot 14 - 1 \cdot 10 + 2 \cdot (-8)\]\[= 42 - 10 - 16\]\[= 16\]

Key concepts

Vector Dot Product

In mathematics, the dot product is an operation that takes two equal-length sequences of numbers and returns a single number. In the context of vectors, the dot product combines two vectors to produce a scalar. This scalar provides information about the angle between the vectors and their magnitudes. To calculate the dot product, multiply each corresponding pair of vector components and sum these products. For example, given vectors \( \bar{a} \) and \( \bar{b} \):

\[ \bar{a} \cdot \, \bar{b} = a_1b_1 + a_2b_2 + a_3b_3 \]

This is calculated step-by-step for each pair of components. For instance, using vectors from the exercise: \[ \bar{a} = \hat{\mathrm{i}} + \hat{\mathrm{j}} + \hat{\mathrm{k}}, \bar{b} = \hat{\mathrm{i}} - \hat{\mathrm{j}} + \hat{\mathrm{k}} \]

The dot product is: \[ (1 \cdot 1) + (1 \cdot (-1)) + (1 \cdot 1) = 1 - 1 + 1 = 1 \]
Here, each pair of components from \( \bar{a} \) and \( \bar{b} \) are multiplied and summed, resulting in a scalar value. This operation is key in forming matrices involving vector spaces.

Matrix Formation

Matrices are a rectangular array of numbers, symbols, or expressions arranged in rows and columns. In this exercise, we form a matrix using dot products of vectors.

Matrices help in organizing and manipulating linear transformations and systems of linear equations. The matrix for our exercise is formed as follows:

\[ \begin{pmatrix} \bar{a} \cdots \bar{a} & \bar{a} \cdocs \bar{b} & \bar{a} \cdots \bar{c} \bar{b} \cdocs \bar{a} & \bar{b} \cdocs \bar{b} & \bar{b} \cdocs \bar{c} \bar{c} \cdocs \bar{a} & \bar{c} \cdocs \bar{b} & \bar{c} \cdocs \bar{c} \end{pmatrix} \]

Using the computed dot products:

\[ \begin{pmatrix} 3 & 1 & 2 \ 1 & 3 & -2 \ 2 & -2 & 6 \end{pmatrix} \]
This matrix effectively combines the relationships between the vectors. Calculating dot products ensures each element accurately represents the interaction between vectors. This formation extends to many applications such as computer graphics, physics simulations, and solutions to simultaneous equations.

Cofactor Expansion

Cofactor expansion is a method used to compute the determinant of a square matrix. This helps in solving systems of linear equations, finding inverses of matrices, and understanding properties like linear independence.

To perform cofactor expansion, select a row or column to expand along. Compute small determinants (minors) of all elements in this chosen row or column, considering their respective sign (positive or negative). Here’s the step-by-step approach for our matrix:

1. Choose the first row for expansion:
\[ 3 \cdot \begin{pmatrix} 3 & -2 \ -2 & 6 \end{pmatrix} \ - 1 \cdot \begin{pmatrix} 1 & -2 \ 2 & 6 \end{pmatrix} \ + 2 \cdot \begin{pmatrix} 1 & 3 \ 2 & -2 \end{pmatrix} \ \]
2. Compute the minors:
\[ 3 (3 \cdot 6 \ - (-2) \cdot (-2)) \ - 1 (1 \cdot 6 \ - (-2) \cdot 2) + 2 (1 \cdot (-2) \ - 3 \cdot 2) \ \]
3. Simplify to find the determinant:
\[ 3 (18 - 4) - 1 (6 + 4) + 2 (-2 - 6) = 3 \cdot 14 - 1 \cdot 10 + 2 \cdot (-8) = 42 - 10 - 16 = 16 \]
Mentally breaking this process into smaller steps makes it more approachable. Remember that the determinant is vital for understanding the unique solutions and properties of a matrix.