Let $$ A $$ and $$ B $$ be $$ 3 imes 3 $$ matrices, with $$ \operatorname{det} A=5 $$ and $$ \operatorname{det} B=4 $$. Use properties of determinants to complete parts (a) through (e) below. a. Compute det $$ A B $$. $$ \operatorname{det} A B=20 $$ (Type an integer or a fraction.) b. Compute det $$ 5 A $$. $$ \operatorname{det} 5 A=625 $$ (Type an integer or a fraction.) c. Compute $$ \operatorname{det} B^{T} $$. det $$ B^{ op}=4 $$ (Type an integer or a fraction.) d. Compute det $$ A^{-1} $$. det $$ A^{-1}=\square $$ (Type an integer or a simplified fraction.)

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Let $$ A $$ and $$ B $$ be $$ 3 	imes 3 $$ matrices, with $$ \operatorname{det} A=5 $$ and $$ \operatorname{det} B=4 $$. Use properties of determinants to complete parts (a) through (e) below. a. Compute det $$ A B $$. $$ \operatorname{det} A B=20 $$ (Type an integer or a fraction.) b. Compute det $$ 5 A $$. $$ \operatorname{det} 5 A=625 $$ (Type an integer or a fraction.) c. Compute $$ \operatorname{det} B^{T} $$. det $$ B^{	op}=4 $$ (Type an integer or a fraction.) d. Compute det $$ A^{-1} $$. det $$ A^{-1}=\square $$ (Type an integer or a simplified fraction.)

UpStudy AI · Accepted Answer

**a. Compute $$\det (AB)$$:**  
Using the property of determinants for the product of matrices, we have  
$$
\det (AB) = \det A \cdot \det B.
$$  
Given $$\det A = 5$$ and $$\det B = 4$$,  
$$
\det (AB) = 5 \cdot 4 = 20.
$$

**b. Compute $$\det (5A)$$:**  
When multiplying a matrix by a scalar, the determinant is multiplied by the scalar raised to the power of the size of the matrix. Since $$A$$ is a $$3 \times 3$$ matrix,  
$$
\det (5A) = 5^3 \cdot \det A.
$$  
Substitute the given value $$\det A = 5$$,  
$$
\det (5A) = 125 \cdot 5 = 625.
$$

**c. Compute $$\det (B^T)$$:**  
For any square matrix, the determinant of the transpose is equal to the determinant of the original matrix, so  
$$
\det (B^T) = \det B.
$$  
Thus, given $$\det B = 4$$,  
$$
\det (B^T) = 4.
$$

**d. Compute $$\det (A^{-1})$$:**  
The determinant of the inverse of a matrix is the reciprocal of the determinant of the matrix. Therefore,  
$$
\det (A^{-1}) = \frac{1}{\det A}.
$$  
Substitute the given value $$\det A = 5$$,  
$$
\det (A^{-1}) = \frac{1}{5}.
$$
a. $$\det (AB) = 20$$ b. $$\det (5A) = 625$$ c. $$\det (B^T) = 4$$ d. $$\det (A^{-1}) = \frac{1}{5}$$

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