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Chapter 1: Problem 3

Find the given determinant. \(\operatorname{det}\left[\begin{array}{rrr}1 & -2 & 3 \\ -4 & 5 & -6 \\ 7 & -8 & 9\end{array}\right]\)

Short Answer

Expert verified
The determinant is 0.

Step by step solution

01

Understand the Matrix and Determinant

You are given a 3x3 matrix: \(\begin{bmatrix}1 & -2 & 3 \-4 & 5 & -6 \7 & -8 & 9\end{bmatrix}\).The task is to find the determinant of this matrix, which is a scalar value that signifies certain properties of the matrix, like invertibility.
02

Recognize the Formula

For a 3x3 matrix \( A = \begin{bmatrix} a & b & c \ d & e & f \ g & h & i \end{bmatrix} \), the determinant is calculated using the formula:\[\operatorname{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg).\]
03

Substitute Values

Substitute the elements of the given matrix into the formula:\[\operatorname{det}\left(\begin{bmatrix} 1 & -2 & 3 \ -4 & 5 & -6 \ 7 & -8 & 9 \end{bmatrix}\right) = 1(5 \cdot 9 - (-6) \cdot (-8)) - (-2)(-4 \cdot 9 - (-6) \cdot 7) + 3(-4 \cdot (-8) - 5 \cdot 7).\]
04

Calculate the Determinant

Compute each part of the expression:- First: \( 5 \times 9 - (-6) \times (-8) = 45 - 48 = -3 \).- Second: \( -4 \times 9 - (-6) \times 7 = -36 + 42 = 6 \).- Third: \( -4 \times (-8) - 5 \times 7 = 32 - 35 = -3 \).Substituting back, the determinant becomes:\[1(-3) + 2(6) + 3(-3).\]
05

Simplify the Expression

Calculate the expression step by step:- \( 1(-3) = -3 \).- \( 2(6) = 12 \).- \( 3(-3) = -9 \).Adding these results together: \(-3 + 12 - 9 = 0\).
06

Conclusion

The determinant of the given matrix is calculated to be \( 0 \). This implies that the matrix is singular and does not have an inverse.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

3x3 Matrix
A 3x3 matrix is a mathematical concept that consists of numbers arranged in three rows and three columns. These numbers can be real or complex numbers. In our specific example, we have a 3x3 matrix:
  • The first row is [1, -2, 3],
  • The second row is [-4, 5, -6],
  • The third row is [7, -8, 9].
This arrangement helps in organizing data and performing operations like addition, subtraction, multiplication, and, importantly, calculating the determinant. Understanding the structure of the 3x3 matrix is key, as the position of each element influences computations such as the determinant.
Matrix Invertibility
Matrix invertibility refers to the ability to find a matrix's inverse. An inverse of a matrix, denoted as \(A^{-1}\) for a matrix \(A\), is a matrix that, when multiplied with the original matrix, results in the identity matrix. This is possible only for non-singular matrices, where the determinant is not equal to zero.
For the given matrix, the determinant was found to be zero. This means the matrix is singular, implying:
  • It does not have an inverse.
  • It is not invertible.
  • This property affects many applications, such as solving systems of equations, where invertibility ensures unique solutions.
Recognizing when a matrix has an inverse is crucial, especially in practical fields like engineering and computer science.
Scalar Value of Matrix
In the context of determinants, the scalar value of a matrix is succinctly the determinant itself. It is a single number that offers insight into the matrix's characteristics.
For our example with the 3x3 matrix, the determinant calculated was zero. This scalar value:
  • Determines matrix invertibility; zero means non-invertible.
  • Indicates linear dependence among rows or columns; there鈥檚 redundancy in data.
  • Impacts calculations like volumes in geometric transformations, where a zero determinant suggests a collapse in dimensions.
Thus, this scalar value is a powerful tool in linear algebra, providing a quick summary of a matrix's key properties.

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Most popular questions from this chapter

Let \(T: \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}\) be the linear transformation whose matrix with respect to the standard bases is \(A=\left[\begin{array}{cc}0 & 1 \\ 1 & 0\end{array}\right]\). Describe \(T\) geometrically.

For Exercises \(1.1-1.4,\) let \(\mathrm{x}\) and \(\mathbf{y}\) be the vectors \(\mathbf{x}=(1,2,3)\) and \(\mathbf{y}=(4,-5,6)\) in \(\mathbb{R}^{3}\). Also, \(\mathbf{0}\) denotes the zero vector, \(\mathbf{0}=(0,0,0)\). Find \(x+y, 2 x,\) and \(2 x-3 y\)

Show that the dot product satisfies the given property. The properties are true for vectors in \(\mathbb{R}^{n},\) though you may assume in your arguments that the vectors are in \(\mathbb{R}^{2}\), i.e., \(\mathbf{x}=\left(x_{1}, x_{2}\right), \mathbf{y}=\left(y_{1}, y_{2}\right),\) and so on. The proofs for \(\mathbb{R}^{n}\) in general are similar. \(w \cdot(x+y)=w \cdot x+w \cdot y\)

Find the indicated matrix products. \(A B\) and \(B A\), where \(A=\left[\begin{array}{lll}1 & 2 & 3 \\ 0 & 4 & 5 \\\ 0 & 0 & 6\end{array}\right]\) and \(B=\left[\begin{array}{lll}1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{array}\right]\)

Let \(\mathbf{x}\) and \(\mathbf{y}\) be points in \(\mathbb{R}^{n}\). (a) Show that the midpoint of line segment xy is given by \(\mathbf{m}=\frac{1}{2} \mathbf{x}+\frac{1}{2} \mathbf{y}\). (Hint: What do you add to \(x\) to get to the midpoint? \()\) (b) Find an analogous expression in terms of \(\mathbf{x}\) and \(\mathbf{y}\) for the point \(\mathbf{p}\) that is \(2 / 3\) of the way from \(x\) to \(y\) (c) Let \(\mathbf{x}=(1,1,0)\) and \(\mathbf{y}=(0,1,1) .\) Find the point \(\mathbf{z}\) in \(\mathbb{R}^{3}\) such that \(\mathbf{y}\) is the midpoint of line segment \(\mathrm{xz}\).
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