Compute the inverse of the following matrix with steps.
1 answer:
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Answer:
The base is 8 and the height is 3, so the area is 8 (3) = 24 square units.
Step-by-step explanation:
The area of the parallelogram is given by the following expression:
The vectors are, respectively:
The base of the parallelogram is 8 units.
The height of the parallelogram is 3 units.
The cross product of both vectors is:
The area of the parallelogram is given by the norm of the resulting vector:
Check the picture below.
you can pretty much just count off the grid the units for JK and MI.
now, let's check how long are KI and JM
![\bf \textit{distance between 2 points}\\ \quad \\ \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) K&({{ -4}}\quad ,&{{ 4}})\quad % (c,d) I&({{ -2}}\quad ,&{{ 3}}) \end{array}\qquad % distance value d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2} \\\\\\ KI=\sqrt{[-2-(-4)]^2+[3-4]^2}\implies KI=\sqrt{(-2+4)^2+(3-4)^2} \\\\\\ KI=\sqrt{2^2+(-1)^2}\implies KI=\sqrt{4+1}\implies \boxed{KI=\sqrt{5}}\\\\ -------------------------------](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bdistance%20between%202%20points%7D%5C%5C%20%5Cquad%20%5C%5C%0A%5Cbegin%7Barray%7D%7Blllll%7D%0A%26x_1%26y_1%26x_2%26y_2%5C%5C%0A%25%20%20%28a%2Cb%29%0AK%26%28%7B%7B%20-4%7D%7D%5Cquad%20%2C%26%7B%7B%204%7D%7D%29%5Cquad%20%0A%25%20%20%28c%2Cd%29%0AI%26%28%7B%7B%20-2%7D%7D%5Cquad%20%2C%26%7B%7B%203%7D%7D%29%0A%5Cend%7Barray%7D%5Cqquad%20%0A%25%20%20distance%20value%0Ad%20%3D%20%5Csqrt%7B%28%7B%7B%20x_2%7D%7D-%7B%7B%20x_1%7D%7D%29%5E2%20%2B%20%28%7B%7B%20y_2%7D%7D-%7B%7B%20y_1%7D%7D%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0AKI%3D%5Csqrt%7B%5B-2-%28-4%29%5D%5E2%2B%5B3-4%5D%5E2%7D%5Cimplies%20KI%3D%5Csqrt%7B%28-2%2B4%29%5E2%2B%283-4%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0AKI%3D%5Csqrt%7B2%5E2%2B%28-1%29%5E2%7D%5Cimplies%20KI%3D%5Csqrt%7B4%2B1%7D%5Cimplies%20%5Cboxed%7BKI%3D%5Csqrt%7B5%7D%7D%5C%5C%5C%5C%0A-------------------------------)
![\bf \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) J&({{ -7}}\quad ,&{{ 4}})\quad % (c,d) M&({{ -8}}\quad ,&{{ 3}}) \end{array}\qquad % distance value \\\\\\ JM=\sqrt{[-8-(-7)]^2+[3-4]^2}\implies JM=\sqrt{(-8+7)^2+(3-4)^2} \\\\\\ JM=\sqrt{(-1)^2+(-1)^2}\implies \boxed{JM=\sqrt{2}}](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Barray%7D%7Blllll%7D%0A%26x_1%26y_1%26x_2%26y_2%5C%5C%0A%25%20%20%28a%2Cb%29%0AJ%26%28%7B%7B%20-7%7D%7D%5Cquad%20%2C%26%7B%7B%204%7D%7D%29%5Cquad%20%0A%25%20%20%28c%2Cd%29%0AM%26%28%7B%7B%20-8%7D%7D%5Cquad%20%2C%26%7B%7B%203%7D%7D%29%0A%5Cend%7Barray%7D%5Cqquad%20%0A%25%20%20distance%20value%0A%5C%5C%5C%5C%5C%5C%0AJM%3D%5Csqrt%7B%5B-8-%28-7%29%5D%5E2%2B%5B3-4%5D%5E2%7D%5Cimplies%20JM%3D%5Csqrt%7B%28-8%2B7%29%5E2%2B%283-4%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0AJM%3D%5Csqrt%7B%28-1%29%5E2%2B%28-1%29%5E2%7D%5Cimplies%20%5Cboxed%7BJM%3D%5Csqrt%7B2%7D%7D)
so, add all sides, and that's the perimeter of the trapezoid.
you can pretty much just count off the grid the units for JK and MI.
now, let's check how long are KI and JM
so, add all sides, and that's the perimeter of the trapezoid.