MHB Finding the difference of mixed measures

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To solve the problem of subtracting mixed measures, first, borrow one degree from the top line, converting it to minutes, resulting in 18 degrees 77 minutes 33 seconds. The next step is to subtract the second measure, 7 degrees 45 minutes 28 seconds, from this adjusted value. The subtraction can be performed by aligning the degrees, minutes, and seconds separately. After completing the subtraction, the final result will provide the difference in mixed measures. This method effectively simplifies the calculation process for mixed units.
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How would I work the following problem-19(degrees) 17' 33"
- 7(degrees) 45' 28"Please help asap
 
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We are given:

$$\begin{array}{r}19^{\circ}\,17^{\prime}\,33^{\prime\prime} \\ -\,\underline{\,\,7^{\circ}\,45^{\prime}\,28^{\prime\prime}}\end{array}$$

If, in the top line, we "borrow" a degree, and add it to the minutes, we have:

$$\begin{array}{r}18^{\circ}\,77^{\prime}\,33^{\prime\prime} \\ -\,\underline{\,\,7^{\circ}\,45^{\prime}\,28^{\prime\prime}}\end{array}$$

Now we can just subtract like terms...can you proceed?
 
Thread 'Erroneously finding discrepancy in transpose rule'

Thread 'Erroneously finding discrepancy in transpose rule'

Obviously, there is something elementary I am missing here. To form the transpose of a matrix, one exchanges rows and columns, so the transpose of a scalar, considered as (or isomorphic to) a one-entry matrix, should stay the same, including if the scalar is a complex number. On the other hand, in the isomorphism between the complex plane and the real plane, a complex number a+bi corresponds to a matrix in the real plane; taking the transpose we get which then corresponds to a-bi...
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