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5. Let AA and BB be n×nn \times n matrices and let C=ABC=A B. Prove that if BB ...
Sep 29, 2024
Solution by Steps
step 1
Let BB be a singular matrix. By definition, this means that the determinant of BB is zero, i.e., det⁡(B)=0\det(B) = 0
step 2
The product of two matrices AA and BB is given by C=ABC = AB. The determinant of the product of two matrices is the product of their determinants: det⁡(C)=det⁡(A)⋅det⁡(B)\det(C) = \det(A) \cdot \det(B)
step 3
Since det⁡(B)=0\det(B) = 0, we have det⁡(C)=det⁡(A)⋅0=0\det(C) = \det(A) \cdot 0 = 0
step 4
Therefore, det⁡(C)=0\det(C) = 0 implies that CC is also a singular matrix
Answer
If BB is singular, then CC must also be singular.
Key Concept
The determinant of a product of matrices is the product of their determinants.
Explanation
A singular matrix has a determinant of zero, which leads to the product matrix also being singular.
Solution by Steps
step 1
Given the equation 2a1+a2=4a32 \boldsymbol{a}_{1} + \boldsymbol{a}_{2} = 4 \boldsymbol{a}_{3}, we can express one column in terms of the others. This indicates a linear dependence among the columns of matrix AA
step 2
Since the columns of AA are linearly dependent, the rank of AA is less than 3 (the number of columns). Therefore, the nullity of AA (the dimension of the solution space of Ax=0A \mathbf{x} = \mathbf{0}) is greater than 0
step 3
The system Ax=0A \mathbf{x} = \mathbf{0} will have infinitely many solutions because the nullity is greater than 0
step 4
Since AA has linearly dependent columns, it is not nonsingular. A matrix is nonsingular if and only if its columns are linearly independent
Answer
The system Ax=0A \mathbf{x} = \mathbf{0} has infinitely many solutions, and AA is not nonsingular.
Key Concept
Linear dependence and the implications for the null space of a matrix.
Explanation
The presence of linear dependence among the columns of matrix AA indicates that there are infinitely many solutions to the homogeneous system, and thus AA cannot be nonsingular.
Solution by Steps
step 1
To find the row echelon form of matrix A A , we perform Gaussian elimination, which involves using row operations to create zeros below the leading coefficients (pivots) in each column
step 2
The row echelon form will have the following characteristics: all non-zero rows are above any rows of all zeros, and the leading coefficient of a non-zero row is always to the right of the leading coefficient of the previous row
step 3
After applying the necessary row operations, we will arrive at the row echelon form R R of matrix A A
step 4
To find the reduced row echelon form, we continue the process by making the leading coefficients equal to 1 and ensuring that all entries in the column above and below each leading 1 are zeros
step 5
The final result will be the reduced row echelon form R′ R' of matrix A A
Answer
The row echelon form and reduced row echelon form of matrix A A can be determined through Gaussian elimination and further row operations.
Key Concept
Row echelon form and reduced row echelon form are used to simplify matrices for solving systems of linear equations.
Explanation
The row echelon form allows us to easily identify solutions to the system, while the reduced row echelon form provides a unique representation of the solution set.
Solution by Steps
step 1
To find the row echelon form of the matrix [2amp;3amp;6amp;44amp;1amp;2amp;40amp;3amp;8amp;7] \begin{bmatrix} 2 & 3 & 6 & 4 \\ 4 & 1 & 2 & 4 \\ 0 & 3 & 8 & 7 \end{bmatrix} , we perform row operations to create zeros below the leading coefficients
step 2
After applying the row operations, the row echelon form is [1amp;0amp;0amp;450amp;1amp;0amp;−1950amp;0amp;1amp;2310] \begin{bmatrix} 1 & 0 & 0 & \frac{4}{5} \\ 0 & 1 & 0 & -\frac{19}{5} \\ 0 & 0 & 1 & \frac{23}{10} \end{bmatrix}
step 3
The reduced row echelon form is obtained by further simplifying the row echelon form, resulting in [1amp;0amp;0amp;450amp;1amp;0amp;−1950amp;0amp;1amp;2310] \begin{bmatrix} 1 & 0 & 0 & \frac{4}{5} \\ 0 & 1 & 0 & -\frac{19}{5} \\ 0 & 0 & 1 & \frac{23}{10} \end{bmatrix}
Answer
The row echelon form and reduced row echelon form of the matrix are both [1amp;0amp;0amp;450amp;1amp;0amp;−1950amp;0amp;1amp;2310] \begin{bmatrix} 1 & 0 & 0 & \frac{4}{5} \\ 0 & 1 & 0 & -\frac{19}{5} \\ 0 & 0 & 1 & \frac{23}{10} \end{bmatrix}
Key Concept
Row Echelon Form and Reduced Row Echelon Form are used to simplify matrices for solving systems of equations.
Explanation
The row echelon form allows us to easily identify leading coefficients, while the reduced row echelon form provides a clearer solution set for the system of equations represented by the matrix.
Solution by Steps
step 1
An upper triangular matrix U U has all its entries below the main diagonal equal to zero. The determinant of U U is the product of its diagonal entries, which are all nonzero. Therefore, det⁡(U)≠0 \det(U) \neq 0 , indicating that U U is nonsingular
step 2
The inverse of a matrix U U , denoted U−1 U^{-1} , can be computed using the formula for the inverse of an upper triangular matrix. The entries of U−1 U^{-1} are determined by the entries of U U and will also maintain the upper triangular form. Thus, U−1 U^{-1} is upper triangular
Answer
U U is nonsingular and U−1 U^{-1} is upper triangular.
Key Concept
An upper triangular matrix is nonsingular if its diagonal entries are nonzero. The inverse of an upper triangular matrix is also upper triangular.
Explanation
Since the determinant of U U is nonzero, it confirms that U U is nonsingular. The structure of U U ensures that its inverse retains the upper triangular form.
Solution by Steps
step 1
A matrix B B is singular if its determinant is zero, meaning it does not have an inverse
step 2
A matrix A A is nonsingular if its determinant is non-zero, meaning it has an inverse
step 3
If B B is row equivalent to A A , then they have the same rank. Since B B is singular, its rank is less than n n
step 4
Therefore, A A cannot be nonsingular because it would imply that B B has full rank, which contradicts the fact that B B is singular
Answer
No, a singular matrix B B cannot be row equivalent to a nonsingular matrix A A .
Key Concept
Row equivalence and matrix singularity
Explanation
A singular matrix cannot be row equivalent to a nonsingular matrix because they would have different ranks.
Solution by Steps
step 1
Let AB AB be invertible. By the property of determinants, we have det⁡(AB)=det⁡(A)⋅det⁡(B) \det(AB) = \det(A) \cdot \det(B) . Since AB AB is invertible, it follows that det⁡(AB)≠0 \det(AB) \neq 0
step 2
Therefore, det⁡(A)⋅det⁡(B)≠0 \det(A) \cdot \det(B) \neq 0 . This implies that both det⁡(A)≠0 \det(A) \neq 0 and det⁡(B)≠0 \det(B) \neq 0
step 3
Since the determinants of both A A and B B are non-zero, it follows that both matrices A A and B B are invertible
Answer
Both A A and B B are invertible.
Key Concept
The product of two matrices is invertible if and only if both matrices are invertible.
Explanation
Since AB AB is invertible, it implies that both A A and B B must also be invertible due to the non-zero determinant condition.
Solution by Steps
step 1
Given that A A is an n×n n \times n matrix satisfying the condition |a_{ii}| > \sum_{j=1, j \neq i}^{n} |a_{ij}| for i=1,2,…,n i=1,2,\ldots,n , we can conclude that the diagonal entries dominate the sum of the absolute values of the other entries in their respective rows
step 2
This strict diagonal dominance implies that the matrix A A cannot have a zero determinant. If A A were singular, then there would exist a nontrivial solution to Ax=0 A\mathbf{x} = \mathbf{0} , which would contradict the strict diagonal dominance condition
step 3
Therefore, since the determinant of A A is non-zero, it follows that A A is nonsingular
Answer
A A is nonsingular.
Key Concept
Strictly diagonally dominant matrices are nonsingular.
Explanation
The condition of strict diagonal dominance ensures that the determinant of the matrix is non-zero, confirming that the matrix is nonsingular.
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