Answer:
[tex]A^{-1}=\left[\begin{array}{ccc}2&0&-1\\3&-1&-1\\-4&1&2\end{array}\right][/tex]
X = 4
Step-by-step explanation:
* At first lets revise how to find the inverse of 3 × 3 matrix
- To find the inverse of a 3 x 3 matrix,
# first calculate the determinant of the matrix.
- If the determinant is 0, the matrix has no inverse.
# Second, transpose the matrix by rewriting the first row as the first
column, the middle row as the middle column, and the third row
as the third column.
# Third, Find the determinant of each of the 2 x 2 minor matrices,
- To find the right minor matrix for each term, first highlight the row
and column of the term you begin with. This should include five
terms of the matrix. The remaining four terms make up the
minor matrix.
- Find the determinant of each minor matrix by cross-multiplying
the diagonals and subtracting
# Fourth, create the matrix of cofactors.
- Place the results of the previous step into a new matrix of
co-factors by aligning each minor matrix determinant with the
corresponding position in the original matrix
- When assigning signs, the first element of the first row keeps its
original sign. The second element is reversed. The third element
keeps its original sign. Continue on with the rest of the matrix
- The final result of this step is called the Adj matrix of the original
# The inverse matrix = 1/determinant × Adj matrix
* Now lets find the matrix A from the story problem
∵ x, y, and z represent the cost of a burger, an order of fries, and
a soft drink
- Order of Winston: x + y + z = 8
- Order of Tia: 2x + z = 10.50
- Order of George: x + 2y + 2z = 12
* Lets make the matrix A
# [tex]A=\left[\begin{array}{ccc}1&1&1\\2&0&1\\1&2&2\end{array}\right][/tex]
# Determinant of A = 1(0×2 - 1×2) - 1(2×2 - 1×1) + 1(2×2 - 0×1)
∴ Determinant of A = 1(-2) - 1(3) + 1(4) = -2 - 3 + 4 = -1
* Lets transposed A
# [tex]A^{T}=\left[\begin{array}{ccc}1&2&1\\1&0&2\\1&1&2\end{array}\right][/tex]
* Lets find the minor matrix for each term
- The 1st row
# (0×2 - 2×1) = -2 , (1×2 - 2×1) = 0 , (1×1 - 0×1) = 1
- The 2nd row
# (2×2 - 1×1) = 3 , (1×2 - 1×1) = 1 , (1×1 - 2×1) = -1
- The 3rd row
# (2×2 - 1×0) = 4 , (1×2 - 1×1) = 1 , (1×0 - 2×1) = -2
* Lets Make Adj A
# [tex]AdjA=\left[\begin{array}{ccc}-2&0&1\\3&1&-1\\4&1&-2\end{array}\right] *\left[\begin{array}{ccc}+&-&+\\-&+&-\\+&-&+\end{array}\right]=\left[\begin{array}{ccc}-2&0&1\\-3&1&1\\4&-1&-2\end{array}\right][/tex]
* Lets write inverse of A
# [tex]A^{-1}=\frac{1}{-1}\left[\begin{array}{ccc}-2&0&1\\-3&1&1\\4&-1&-2\end{array}\right]=\left[\begin{array}{ccc}2&0&-1\\3&-1&-1\\-4&1&2\end{array}\right][/tex]
* Now lets find the solution matrix for X
# [tex]A_{x}=\left[\begin{array}{ccc}8&1&1\\10.5&0&1\\12&2&2\end{array}\right][/tex]
∴ Ax = 8(0×2 - 1×2) - 1(10.5×2 - 1×12) + 1(10.5×2 - 0×12)
∴ Ax = -16 - 9 + 21 = -4
* Lets find the value of X
∵ X = Ax/A
∴ X = -4/-1 = 4
Answer: [tex]\left[\begin{array}{ccc}2&0&-1\\3&-1&-1\\-4&1&2\end{array}\right] and\ \left[\begin{array}{ccc}4\\1.5\\2.5\end{array}\right][/tex]
Step-by-step explanation: I got this right on Edmentum.
