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Problems 15-19 are hard as ever​

Problems 15-19 are hard as ever​-example-1
User Wyz
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2 Answers

4 votes

15. B: If
\det B is non-zero, then
\det B^(-1)=\frac1{\det B}. We have


\det B=4\cdot0-1\cdot2=-2

so


\det B^(-1)=-\frac12

16. C: In general,
AB\\eq BA for two matrices
A,B, but equality holds if
B=A^(-1). D is incorrect because
A^(-1)I=A^(-1)\\eq A.

17. C: This follows from a property of the determinant:


\det(kB)=k^n\det B

where
n is the size of the (square) matrix
B (and "size" refers to the number of rows or columns, both of which are the same).

18. A: A matrix has no inverse if its determinant is 0. The determinant of (A) is 0 because it contains a row made up entirely of 0s.

19. B: The matrix product
AB only exists if the number of columns of
A is equal to the number of rows of
B. In (a), the first matrix has 1 column and the other has 2 rows, so multiplication is invalid. In (c), the product on the left side would produce


\begin{bmatrix}1&-3\\-5&1\end{bmatrix}\begin{bmatrix}-3\\14\end{bmatrix}=\begin{bmatrix}-45\\29\end{bmatrix}=\begin{bmatrix}x\\y\end{bmatrix}

so that
x=-45 and
y=29, but these values don't work with the given equations because -45 - 3(29) = -132, not -3. In (d), the product on the left side is


\begin{bmatrix}-3\\14\end{bmatrix}\begin{bmatrix}x&y\end{bmatrix}=\begin{bmatrix}-3x&-3y\\14x&14y\end{bmatrix}=\begin{bmatrix}1&-3\\-5&1\end{bmatrix}

but then this would mean both
-3x=1 and
14x=-5, which are not consistent and have no solution.

User Arash Afsharpour
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8.1k points
2 votes

Answer:

15. b. -1/2

16. c.

17. c. 125

18. a.

19. b.

Explanation:

15. The determinant of the inverse of a matrix is the inverse of the determinant of the matrix. detB = -2, so the determinant of its inverse is ...

1/(-2) = -1/2 . . . . matches choice B

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16. A matrix and its inverse can be multiplied in either order: choice C.

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17. The determinant of a diagonal matrix is the product of the diagonal elements: 5×5×5 = 125, choice C.

__

18. As in 17, the determinant of a diagonal matrix is the product of the diagonal elements. The determinant of choice A is zero, so the matrix has no inverse.

__

19. The standard form of a linear matrix equation is ...

AX = B

where A is the matrix of coefficients, X is the column vector of variables, and B is the column vector of constants. This form matches choice B.

Choice C matches the form the solution might take if the square matrix were the inverse of the coefficient matrix, which it is not. The other choices cannot work because the matrix dimensions are wrong.

User Rocking
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8.5k points