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Magazine Chapter 3: Problem 11
Show that if \(\operatorname{det}(A) \neq 0\) for \(A\) an \(n \times n\) matrix, it follows that if \(A X=0,\) then \(X=0\).
Short Answer
Expert verified
If \(A X = 0\), then \(X = 0\).
Step by step solution
01
- Understand the problem
We are given that \(\text{det}(A) e 0\) for an \(n \times n\) matrix \(A\) and need to show that if \(A X = 0\), then \(X = 0\).
02
- Recall properties of determinants
When \(\text{det}(A) e 0\), matrix \(A\) is invertible. This means there exists an inverse matrix \(A^{-1}\) such that \(A A^{-1} = I\), where \(I\) is the identity matrix.
03
- Write the given equation
We start with the given equation \(A X = 0\).
04
- Apply the inverse matrix
Multiply both sides of the equation \(A X = 0\) by \(A^{-1}\). This gives us: \ A^{-1} A X = A^{-1} 0 \.
05
- Simplify the equation
Since \(A^{-1} A = I\) and \(A^{-1} 0 = 0\), we have: \ I X = 0 \.
06
- Conclusion
The identity matrix \(I\) times any vector \(X\) is just \(X\). Therefore, \(X = 0\). This shows that if \(A X = 0\), then \(X = 0\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Determinants
A key concept in linear algebra is the determinant of a matrix. The determinant is a scalar value that can help determine whether a matrix is invertible.
In simpler terms, it's a special number that provides useful properties about the matrix.
For a matrix \(A\), the determinant, denoted as \(\operatorname{det}(A)\), is calculated using specific rules depending on the size of the matrix.
One critical property of determinants is that if \(\operatorname{det}(A) eq 0\), the matrix \(A\) is invertible (i.e., it has an inverse). Otherwise, if \(\operatorname{det}(A) = 0\), the matrix is not invertible. This property is fundamental in solving linear equations and performing other matrix operations.
In simpler terms, it's a special number that provides useful properties about the matrix.
For a matrix \(A\), the determinant, denoted as \(\operatorname{det}(A)\), is calculated using specific rules depending on the size of the matrix.
One critical property of determinants is that if \(\operatorname{det}(A) eq 0\), the matrix \(A\) is invertible (i.e., it has an inverse). Otherwise, if \(\operatorname{det}(A) = 0\), the matrix is not invertible. This property is fundamental in solving linear equations and performing other matrix operations.
Matrix Inverse
Another fundamental concept is the inverse of a matrix. If a matrix \(A\) is invertible, there exists another matrix, denoted as \(A^{-1}\), such that \(A A^{-1} = A^{-1} A = I\), where \(I\) is the identity matrix.
This means you can 'undo' the multiplication by \(A\) by multiplying by its inverse; this is similar to how multiplying by a number and then dividing by the same number returns to the original value.
In the problem, since \(\operatorname{det}(A) eq 0\), matrix \(A\) is invertible. Therefore, we can use the inverse to solve the equation \(A X = 0\). By multiplying both sides by \(A^{-1}\), we isolate \(X\) to show that \(X = 0\).
This is a step-by-step process of simplifying the equation using properties of matrix multiplication and inverses.
This means you can 'undo' the multiplication by \(A\) by multiplying by its inverse; this is similar to how multiplying by a number and then dividing by the same number returns to the original value.
In the problem, since \(\operatorname{det}(A) eq 0\), matrix \(A\) is invertible. Therefore, we can use the inverse to solve the equation \(A X = 0\). By multiplying both sides by \(A^{-1}\), we isolate \(X\) to show that \(X = 0\).
This is a step-by-step process of simplifying the equation using properties of matrix multiplication and inverses.
Identity Matrix
The identity matrix, denoted as \(I\), plays a crucial role in linear algebra. It is a square matrix with ones on the diagonal and zeros elsewhere. For example, for a 3x3 matrix, it looks like this:
\[ I = \begin{pmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{pmatrix} \]
When any matrix \(A\) is multiplied by the identity matrix \(I\), the result is just \(A\). In other words, \(A I = I A = A\). This property makes the identity matrix the multiplicative identity in the matrix world, much like the number 1 is for regular multiplication.
In the context of the original exercise, when it's shown that multiplying the inverse matrix \(A^{-1}\) by \(A\) results in the identity matrix \(I\), it simplifies the equation and ensures that the only solution to \(A X = 0\) is \(X = 0\).
\[ I = \begin{pmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{pmatrix} \]
When any matrix \(A\) is multiplied by the identity matrix \(I\), the result is just \(A\). In other words, \(A I = I A = A\). This property makes the identity matrix the multiplicative identity in the matrix world, much like the number 1 is for regular multiplication.
In the context of the original exercise, when it's shown that multiplying the inverse matrix \(A^{-1}\) by \(A\) results in the identity matrix \(I\), it simplifies the equation and ensures that the only solution to \(A X = 0\) is \(X = 0\).
Linear Algebra Proof
Proving statements in linear algebra often involves understanding and using core concepts like determinants, matrix inverses, and identity matrices.
To prove that if \(A X = 0\) then necessarily \(X = 0\) when \(\operatorname{det}(A) eq 0\), we start by leveraging the fact that \(\operatorname{det}(A) eq 0\) implies the existence of an inverse matrix \(A^{-1}\).
Multiplying both sides of the equation \(A X = 0\) by \(A^{-1}\) eliminates \(A\) on one side, leaving us with:
\[ A^{-1} A X = A^{-1} 0 \]
Since \(A^{-1} A\) equals the identity matrix \(I\), and \(A^{-1} 0 = 0\), it simplifies to:
\[ I X = 0 \]
This results in \(X = 0\), which completes the proof. This approach shows the step-by-step utilization of these properties to arrive at the final conclusion systematically.
To prove that if \(A X = 0\) then necessarily \(X = 0\) when \(\operatorname{det}(A) eq 0\), we start by leveraging the fact that \(\operatorname{det}(A) eq 0\) implies the existence of an inverse matrix \(A^{-1}\).
Multiplying both sides of the equation \(A X = 0\) by \(A^{-1}\) eliminates \(A\) on one side, leaving us with:
\[ A^{-1} A X = A^{-1} 0 \]
Since \(A^{-1} A\) equals the identity matrix \(I\), and \(A^{-1} 0 = 0\), it simplifies to:
\[ I X = 0 \]
This results in \(X = 0\), which completes the proof. This approach shows the step-by-step utilization of these properties to arrive at the final conclusion systematically.
