Question

Part 2: Use a decoding matrix.Raya sent an encoded message back to Darnell that provides the location of the next clue.To decode the message, Darnell will need to use the inverse of the 2 x 2 encoding matrix, A. Complete the following problems to determine the next clue.a) What is the determinant of A? Show your work or explain. (3 points) b) What is the inverse matrix of A? Show how to get A–1. (4 points) c) What is A–1 × C? (4 points) d) Describe how to use the resultant matrix from part c to decode the message. What does the decoded message say? (4 points)

Answer 1

a. The given A matrix is:

$A=\begin{bmatrix}{3} & {4} & {} \\ {1} & {-2} & {} \\ {} & {} & {}\end{bmatrix}$

The determinant of a 2x2 matrix can be found as:

$\begin{gathered} A=\begin{bmatrix}{a} & {b} & {} \\ {c} & {d} & {} \\ {} & {} & {}\end{bmatrix} \\ \det (A)=|A|=a\times d-b\times c \\ \therefore\det (A)=3\times(-2)-1\times4=-6-4=-10 \end{gathered}$

The determinant of matrix A=-10.

b. The inverse of a 2x2 matrix is given by:

$\begin{gathered} A^{-1}=\frac{1}{\det(A)}\begin{bmatrix}{d} & {-b} & {} \\ {-c} & {a} & {} \\ {} & {} & {}\end{bmatrix} \\ \therefore A^{-1}=\frac{1}{-10}\begin{bmatrix}{-2} & {-4} & {} \\ {-1} & {3} & {} \\ {} & {} & {}\end{bmatrix} \\ \therefore A^{-1}=\begin{bmatrix}{\frac{-2}{-10}} & {\frac{-4}{-10}} & {} \\ {\frac{-1}{-10}} & {\frac{3}{-10}} & {} \\ {} & {} & {}\end{bmatrix}=\begin{bmatrix}{\frac{2}{10}} & {\frac{4}{10}} & {} \\ {\frac{1}{10}} & {-\frac{3}{10}} & {} \\ {} & {} & {}\end{bmatrix} \end{gathered}$

c. A^-1xC is:

And the result can be found as:

This is the resulting matrix of multiplying A^-1xC.

d. In order to decode the message we need to replace the numbers in the matrix with their corresponding letters according to the alphabet order, then:

19->S, 5->E, 1->A, 18->R, 3->C, 8->H, 0->space, 20->T, 8->H, 5->E, 0->space, 7->G 25->Y, 13->M.

By replacing these letters into the matrix we obtain:

And reading in column order, the decoded message says:

"SEARCH_THE_GYM"