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Determine if \(\mathbf{w}=\left[\begin{array}{r}{5} \\ {-3} \\\ {2}\end{array}\right]\) is in Nul \(A,\) where $$ A=\left[\begin{array}{rrr}{5} & {21} & {19} \\ {13} & {23} & {2} \\ {8} & {14} & {1}\end{array}\right] $$

Short Answer

Expert verified
Yes, \( \mathbf{w} \) is in Nul \(A\).

Step by step solution

01

Understand Nul A

The null space of a matrix \( A \) is the set of all vectors \( \mathbf{x} \) such that \( A\mathbf{x} = \mathbf{0} \). We need to check if \( \mathbf{w} = \begin{bmatrix} 5 \ -3 \ 2 \end{bmatrix} \) satisfies this condition.
02

Perform the Matrix-Vector Multiplication

Calculate \( A \mathbf{w} \) by multiplying matrix \( A \) by vector \( \mathbf{w} \). We need to multiply each row of \( A \) with \( \mathbf{w} \):\[ A\mathbf{w} = \begin{bmatrix} 5 & 21 & 19 \ 13 & 23 & 2 \ 8 & 14 & 1 \end{bmatrix} \begin{bmatrix} 5 \ -3 \ 2 \end{bmatrix} \]
03

Compute Each Row's Multiplication Result

Calculate the individual dot products for each row with the vector \( \mathbf{w} \):First row: \[ 5 \cdot 5 + 21 \cdot (-3) + 19 \cdot 2 = 25 - 63 + 38 = 0 \]Second row:\[ 13 \cdot 5 + 23 \cdot (-3) + 2 \cdot 2 = 65 - 69 + 4 = 0 \]Third row:\[ 8 \cdot 5 + 14 \cdot (-3) + 1 \cdot 2 = 40 - 42 + 2 = 0 \]
04

Verify the Result

Since each row multiplication resulted in 0, the product \( A\mathbf{w} \) equals \( \begin{bmatrix} 0 \ 0 \ 0 \end{bmatrix} \). This confirms that \( \mathbf{w} \) is indeed in the null space of \( A \).

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Key Concepts

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

Matrix-Vector Multiplication
Matrix-vector multiplication is a basic operation in linear algebra where a matrix is multiplied by a vector. This operation involves taking each row of the matrix and performing a dot product with the vector. The result is a new vector.

To perform a matrix-vector multiplication, consider a matrix \( A \) with dimensions \( m \times n \) and a vector \( \mathbf{w} \) of size \( n \). The number of columns in the matrix must match the number of entries in the vector.
  • Take each row of the matrix.
  • Multiply corresponding elements in the row and the vector.
  • Sum these products to get an element of the resulting vector.
This process is repeated for each row of the matrix, producing a vector of dimension \( m \). Matrix-vector multiplication is essential not only for calculating the null space but also in various applications like transformations and solving systems of equations.
Dot Product
The dot product is a way to multiply two vectors to get a scalar. It is executed by multiplying corresponding elements from the vectors and then summing up these products. It's a central concept in linear algebra when dealing with matrices and vectors.

To compute the dot product of two vectors \( \mathbf{a} \) and \( \mathbf{b} \), follow these steps:
  • Multiply each pair of corresponding elements of \( \mathbf{a} \) and \( \mathbf{b} \).
  • Sum all the multiplied values to get a single number (scalar).
For example, the dot product of \( [x_1, y_1, z_1] \) and \( [x_2, y_2, z_2] \) is calculated as \( x_1x_2 + y_1y_2 + z_1z_2 \). The dot product is instrumental in matrix-vector multiplication, projecting vectors, and measuring angles between vectors.
Linear Algebra
Linear algebra is the branch of mathematics concerning vector spaces and linear equations. It uses concepts like matrices, vectors, determinants, and eigenvalues to solve problems in various fields such as engineering, physics, and computer science.

Linear algebra focuses on:
  • Solving systems of linear equations using matrix representations.
  • Transforming geometrical data with matrices and vectors.
  • Studying vector spaces, dependencies, and bases.
  • Exploring properties of matrices such as rank, determinant, and eigenvalues.
Understanding linear algebra necessitates familiarity with critical operations like matrix-vector multiplication and concepts such as the null space, which helps solve systems of equations and transform data.
Null Space of a Matrix
The null space of a matrix, often referred to as the kernel, is the set of all vectors that result in the zero vector when the matrix is multiplied by them. It is a fundamental concept in linear algebra that provides insight into the solutions of homogeneous systems of linear equations.

To find the null space of a matrix \( A \):
  • Identify all vectors \( \mathbf{x} \) such that \( A\mathbf{x} = \mathbf{0} \).
  • Perform a matrix-vector multiplication and use row operations to achieve a reduced form.
  • Determine the vectors that satisfy the condition resulting in the null space.
In practical terms, the null space tells us about the failure of the matrix \( A \) to be invertible and provides solutions to \( A\mathbf{x} = \mathbf{0} \). It's indispensable in understanding the relationships between coefficients and solutions in linear equations.

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Most popular questions from this chapter

Let \(y_{k}=k^{2}\) and \(z_{k}=2 k|k|\) . Are the signals \(\left\\{y_{k}\right\\}\) and \(\left\\{z_{k}\right\\}\) linearly independent? Evaluate the associated Casorati ma- \(\operatorname{trix} C(k)\) for \(k=0, k=-1,\) and \(k=-2,\) and discuss your results.

Let \(\mathrm{S}_{0}\) be the vector space of all sequences of the form \(\left(y_{0}, y_{1}, y_{2}, \ldots\right),\) and define linear transformations \(T\) and \(D\) from \(\mathrm{S}_{0}\) into \(\mathrm{S}_{0}\) by $$ \begin{array}{l}{T\left(y_{0}, y_{1}, y_{2}, \ldots\right)=\left(y_{1}, y_{2}, y_{3}, \ldots\right)} \\ {D\left(y_{0}, y_{1}, y_{2}, \ldots\right)=\left(0, y_{0}, y_{1}, y_{2}, \ldots\right)}\end{array} $$ Show that \(T D=I\) (the identity transformation on \(\mathrm{S}_{0} )\) and yet \(D T \neq I .\)

Suppose a nonhomogeneous system of nine linear equations in ten unknowns has a solution for all possible constants on the right sides of the equations. Is it possible to find two nonzero solutions of the associated homogeneous system that are not multiples of each other? Discuss.

In Exercises \(15-18,\) find a basis for the space spanned by the given vectors, \(\mathbf{v}_{1}, \ldots, \mathbf{v}_{5}\) $$ \left[\begin{array}{r}{1} \\ {0} \\ {-3} \\\ {2}\end{array}\right],\left[\begin{array}{r}{0} \\ {1} \\ {2} \\\ {-3}\end{array}\right],\left[\begin{array}{r}{-3} \\ {-4} \\ {1} \\\ {6}\end{array}\right],\left[\begin{array}{r}{1} \\ {-3} \\ {-8} \\\ {7}\end{array}\right],\left[\begin{array}{r}{2} \\ {1} \\ {-6} \\\ {9}\end{array}\right] $$

(RLC circuit) The circuit in the figure consists of a resistor \((R \text { ohms ), an inductor }(L \text { henrys), a capacitor }(C \text { farads) }\) and an inital voltage source. Let \(b=R /(2 L),\) and suppose \(R, L,\) and \(C\) have been selected so that \(b\) also equals 1\(/ \sqrt{L C} .\) This is done, for instance, when the circuit is used in a voltmeter.) Let \(v(t)\) be the voltage (in volts) at time \(t\) measured across the capacitor. It can be shown that \(v\) is in the null space \(H\) of the linear transformation that maps \(v(t)\) into \(L v^{\prime \prime}(t)+R v^{\prime}(t)+(1 / C) v(t),\) and \(H\) consists of all functions of the form \(v(t)=e^{-b t}\left(c_{1}+c_{2} t\right) .\) Find a basis for \(H .\) figure can't copy

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