) We need to determine the transpose of 5B, i.e., (5B)t. First, we'll handle the expression for Bt. From the first equation, we can isolate Bt. 3A+4Bt=(
7
−10
17
0
6
31
) Rearranging to solve for Bt. 4Bt=(
7
−10
17
0
6
31
)−3A Next, from the second equation, we solve for B. 2B−3At=(
−1
18
4
−6
−5
−7
) Rearranging to solve for B : ‌2B=(
−1
18
4
−6
−5
−7
)+3At ‌B=‌
1
2
((
−1
18
4
−6
−5
−7
)+3At) Now, let's find (5B)t. ‌5B=‌
5
2
((
−1
18
4
−6
−5
−7
)+3At) ‌(5B)t=‌
5
2
((
−1
4
−5
18
−6
−7
)+3A) Given that 3A+4Bt=(
7
−10
17
0
6
31
), let's substitute this value in the transpose equation: ‌(5B)t=‌
5
2
((
−1
4
−5
18
−6
−7
)+3A) ‌‌ Since ‌3A+4Bt=(
7
−10
17
0
6
31
) ‌3A=(
7
−10
17
0
6
31
)−4Bt
(5B)t=‌
5
2
((
−1
4
−5
18
−6
−7
)+((
7
−10
17
0
6
31
)−4Bt))
Therefore, upon solving the above expressions you will get the following matrix: (5B)t=(