
Concept explainers
a.
To find the determinant of the matrix A using the matrix capabilities of a graphing utility.
a.

Answer to Problem 44E
The value of the determinant A using the graphing utility is obtained as −46
Explanation of Solution
Given:
The matrices are given as
A=(−152 000 1 1 3−3−1 04 2 4 −1)B=(150 010−12 4 200 1−3 2 5 0)
Calculation:
The determinant of the matrix using the matrix capabilities of a graphing utility can be derived as,
- Enter the matrix by using the combination of keystrokes 2nd+x−1
- Use cursor key to select the edit option and then select row 1 of matrix A and press Enter .
- Enter the dimension of the matrix (4 rows by 4 columns).
- Enter the values of the matrix.
- Go to Matrix again by pressing 2nd+x−1
- Select the MATH option.
- Select the function det.
- Go to matrix again, choose [A]
- Press Enter to see the evaluation.
Conclusion:
Therefore,
The value of the determinant using the graphing utility is obtained as −46
b.
To find the determinant of the matrix B using the matrix capabilities of a graphing utility.
b.

Answer to Problem 44E
The value of the determinant B using the graphing utility is obtained as 89
Explanation of Solution
Given:
The matrices are given as,
A=(−152 000 1 1 3−3−1 04 2 4 −1)B=(150 010−12 4 200 1−3 2 5 0)
Calculation:
The determinant of the matrix using the matrix capabilities of a graphing utility can be derived as,
Enter the matrix by using the combination of keystrokes 2nd+x−1
Use cursor key to select the edit option and then select row 1 of matrix A and press Enter .
- Enter the dimension of the matrix (4 rows by 4 columns).
- Enter the values of the matrix.
- Select the MATH option.
- Select the function det.
- Go to matrix again, choose [A] Press Enter
- to see the evaluation.
Go to Matrix again by pressing 2nd+x−1
Conclusion:
Therefore,
The value of the determinant using the graphing utility is obtained as 89
c.
To find the matrix AB using the matrix capabilities of a graphing utility.
c.

Answer to Problem 44E
The product of the matrices using the graphing utility is obtained as
AB=(53−1010 22−12 5 1 −2918−6 −1335 16 −1 12)
Explanation of Solution
Given:
The matrices are given as,
A=(−152 000 1 1 3−3−1 04 2 4 −1)B=(150 010−12 4 200 1−3 2 5 0)
Calculation:
Enter the matrix by using the combination of keystrokes 2nd+x−1 .
Use the right arrow key to select the edit option and press Enter .
- Enter the dimension of the matrix (4 rows by 4 columns).
- Enter the values of the matrix.
Enter the second matrix again by pressing 2nd+x−1 .
Use the right arrow key to select the edit option, press [2] or highlight 2 and press Enter .
- Enter the dimension of the matrix (4 rows by 4 columns).
- Enter the values of the matrix.
- Press [2ND] and [MODE] to exit the matrix screen.
- Press [ENTER] to select matrix [A].
Press 2nd+x−1 from the blank screen and stay at the NAMES menu
Select the second matrix again by pressing 2nd+x−1 .
- Press [2] or highlight 2. [B] and press [ENTER].
- To multiply the matrices, press [ENTER].
Conclusion:
Therefore,
The product of the matrices using the graphing utility is obtained as
AB=(53−1010 22−12 5 1 −2918−6 −1335 16 −1 12)
d.
To find the determinant of the matrix AB using the matrix capabilities of a graphing utility.
d.

Answer to Problem 44E
The value of the determinant using the graphing utility is obtained as −4094 and it s noticed that, |AB|=|A|×|B|
Explanation of Solution
Given:
The matrices are given as,
A=(−152 000 1 1 3−3−1 04 2 4 −1)B=(150 010−12 4 200 1−3 2 5 0)
Calculation:
The determinant of the matrix using the matrix capabilities of a graphing utility can be derived as,
Enter the matrix by using the combination of keystrokes 2nd+x−1
Use cursor key to select the edit option and then select row 1 of matrix A and press Enter .
- Enter the dimension of the matrix (4 rows by 4 columns).
- Enter the values of the matrix.
- Select the MATH option.
- Select the function det.
- Go to matrix again, choose [A] Press Enter
- to see the evaluation.
Go to Matrix again by pressing 2nd+x−1
Conclusion:
Therefore,
The value of the determinant of the using the graphing utility is obtained as −4094 and it s noticed that, |AB|=|A|×|B|
Chapter 7 Solutions
Precalculus with Limits: A Graphing Approach
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