Problem 55
Question
If \(\operatorname{det}\left(A^{3}\right)=0,\) is it possible for \(A\) to be invertible? Justify your answer.
Step-by-Step Solution
Verified Answer
No, it is not possible for matrix \(A\) to be invertible. Based on the property \(\operatorname{det}(A^n) = \operatorname{det}(A)^n\), we have \(\operatorname{det}(A^3) = \operatorname{det}(A)^3 = 0\), which implies \(\operatorname{det}(A) = 0\). A matrix is invertible if and only if its determinant is non-zero, so in this case, matrix \(A\) is not invertible.
1Step 1: Invertibility and determinants
A matrix is invertible if and only if its determinant is non-zero.
2Step 2: Property of determinants with powers of a matrix
For a square matrix \(A\), we have the property:
\(\operatorname{det}(A^n) = \operatorname{det}(A)^n\), where \(n\) is a positive integer.
3Step 3: Apply the property to our given situation
Since we have:
\(\operatorname{det}(A^3) = \operatorname{det}(A)^3 = 0\), we can deduce that the determinant of matrix \(A\) is zero.
4Step 4: Determine if A is invertible
Since \(\operatorname{det}(A) = 0\), we can conclude that matrix \(A\) is not invertible, because a matrix is invertible only if its determinant is non-zero.
Key Concepts
Invertibility of MatricesPowers of a MatrixMatrix Inverses
Invertibility of Matrices
A matrix is invertible if it has a unique inverse matrix such that the product of the two matrices is the identity matrix. This is a crucial concept in linear algebra. An identity matrix is essentially like the number 1 in regular arithmetic – when you multiply anything by 1, the value stays the same. The key factor that determines if a matrix is invertible is its determinant.
- If the determinant of a matrix is zero, the matrix does not have an inverse, making it non-invertible.
- If the determinant is non-zero, the matrix can be inverted.
Powers of a Matrix
When dealing with powers of matrices, such as raising a matrix to a certain power, it is crucial to understand how determinants behave under these operations. There is a specific property regarding determinants and matrix powers:
This property helps to deduce critical information about the invertibility of a matrix. For instance, if \(\operatorname{det}(A^3) = 0\), we can derive that \(\operatorname{det}(A) = 0\). This directly implies that the matrix itself is non-invertible, as a zero determinant means no inverse exists.
- The determinant of a matrix raised to a power is the determinant of the matrix raised to that same power.
This property helps to deduce critical information about the invertibility of a matrix. For instance, if \(\operatorname{det}(A^3) = 0\), we can derive that \(\operatorname{det}(A) = 0\). This directly implies that the matrix itself is non-invertible, as a zero determinant means no inverse exists.
Matrix Inverses
Finding a matrix inverse is akin to division in regular arithmetic, but it holds much greater significance in the realm of linear algebra. The existence of an inverse directly depends on whether the matrix is invertible. Here's how you can identify and understand matrix inverses:
- A matrix \(A\) has an inverse \(A^{-1}\) if and only if the product \(A \times A^{-1} = I\), where \(I\) is the identity matrix.
- A non-square matrix cannot have an inverse, making it non-invertible by definition.
- To find an inverse, if it exists, one performs row operations to transform the matrix into its identity counterpart.
Other exercises in this chapter
Problem 55
Find \(A^{-1}.\) $$A=\left[\begin{array}{cc} 3 e^{t} & e^{2 t} \\ 2 e^{t} & 2 e^{2 t} \end{array}\right]$$
View solution Problem 55
The alternating symbol \(\epsilon_{i j k}\) is defined by \(\epsilon_{i j k}=\left\\{\begin{aligned} 1, & \text { if }(i j k) \text { is an even permutation of
View solution Problem 56
Find \(A^{-1}.\) $$A=\left[\begin{array}{ccc} e^{3 t} & 9 t e^{3 t} & -e^{-2 t} \\ -t e^{3 t} & e^{3 t} & e^{-2 t} \\ -t e^{3 t} & e^{3 t} & 0 \end{array}\right
View solution Problem 56
If \(A\) is the general \(n \times n\) matrix, determine the sign attached to the term $$a_{1 n} a_{2} n-1 a_{3} n-2 \cdots a_{n 1}$$ which arises in \(\operato
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