The interpretation of the given question is as follows:
Use the given inverse to solve the system of equations

The inverse of ![\left[\begin{array}{ccc}1&-1&1\\0&2&1\\3&-8&0\end{array}\right] is \left[\begin{array}{ccc}-8&8&3\\-3&3&1\\6&-5&-2\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%26-1%261%5C%5C0%262%261%5C%5C3%26-8%260%5Cend%7Barray%7D%5Cright%5D%20%20%20is%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-8%268%263%5C%5C-3%263%261%5C%5C6%26-5%26-2%5Cend%7Barray%7D%5Cright%5D)
x =
y =
z =
Answer:
x = - 1.5
y = - 0.5
z = - 5
Step-by-step explanation:
Using the correlation of inverse of matrix AX = B to solve the question above;
AX = B
⇒ A⁻¹(AX) = A⁻¹ B
X = A⁻¹ B
So ;
X = A⁻¹ B
![\left[\begin{array}{ccc}-6\\ -6\\- \dfrac{1}{2}\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-6%5C%5C%20-6%5C%5C-%20%5Cdfrac%7B1%7D%7B2%7D%5Cend%7Barray%7D%5Cright%5D)
![\left[\begin{array}{c}x\\y\\z\end{array}\right] =\left[\begin{array}{ccc}(-8*-6)+(8*-6)+(3*-\dfrac{1}{2})\\(-3*-6)+(3*-6)+(1*-\dfrac{1}{2})\\(6*-6)+(5*-6)+(-2* - \dfrac{1}{2})\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%5C%5Cy%5C%5Cz%5Cend%7Barray%7D%5Cright%5D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%28-8%2A-6%29%2B%288%2A-6%29%2B%283%2A-%5Cdfrac%7B1%7D%7B2%7D%29%5C%5C%28-3%2A-6%29%2B%283%2A-6%29%2B%281%2A-%5Cdfrac%7B1%7D%7B2%7D%29%5C%5C%286%2A-6%29%2B%285%2A-6%29%2B%28-2%2A%20-%20%5Cdfrac%7B1%7D%7B2%7D%29%5Cend%7Barray%7D%5Cright%5D)
![\left[\begin{array}{c}x\\y\\z\end{array}\right] =\left[\begin{array}{ccc}(48)+(-48)+(\dfrac{-3}{2})\\(18)+(-18)+(\dfrac{-1}{2})\\(-36)+(30)+(1)\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%5C%5Cy%5C%5Cz%5Cend%7Barray%7D%5Cright%5D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%2848%29%2B%28-48%29%2B%28%5Cdfrac%7B-3%7D%7B2%7D%29%5C%5C%2818%29%2B%28-18%29%2B%28%5Cdfrac%7B-1%7D%7B2%7D%29%5C%5C%28-36%29%2B%2830%29%2B%281%29%5Cend%7Barray%7D%5Cright%5D)
![\left[\begin{array}{c}x\\y\\z\end{array}\right] =\left[\begin{array}{ccc}(\dfrac{-3}{2})\\(\dfrac{-1}{2})\\(-5)\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%5C%5Cy%5C%5Cz%5Cend%7Barray%7D%5Cright%5D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%28%5Cdfrac%7B-3%7D%7B2%7D%29%5C%5C%28%5Cdfrac%7B-1%7D%7B2%7D%29%5C%5C%28-5%29%5Cend%7Barray%7D%5Cright%5D)
![\left[\begin{array}{c}x\\y\\z\end{array}\right] =\left[\begin{array}{ccc}-1.5\\-0.5\\ -5\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%5C%5Cy%5C%5Cz%5Cend%7Barray%7D%5Cright%5D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-1.5%5C%5C-0.5%5C%5C%20-5%5Cend%7Barray%7D%5Cright%5D)
The solutions to the given system of equations is (0, -6) and (1, -5)
<h3>Simultaneous equations</h3>
From the question, we are to determine the solutions to the given system of equations
The equations are
x − y = 6 --------- (1)
y = x² −6 ---------- (2)
From equation (1)
x - y = 6
∴ x = 6 + y ------- (3)
Substitute into equation (2)
y = x² −6
y = (6+y)² −6
y = (6+y)(6+y) -6
y = 36 + 6y + 6y +y² -6
y = 36 + 12y + y² - 6
Simplifying
y² + 12y - y + 30 = 0
y² + 11y + 30 = 0
Solve quadratically
y² + 11y + 30 = 0
y² + 6y + 5y + 30 = 0
y(y +6) +5 (y +6) = 0
(y + 5)(y + 6) = 0
y + 5 = 0 OR y + 6 = 0
y = -5 OR y = -6
Substitute the values of y into equation (3)
x = 6 + y
When y = -5
x = 6 + (-5)
x = 6 -5
x = 1
When y = -6
x = 6 + (-6)
x = 6 -6
x = 0
∴ When x = 0, y = -6 and when x = 1, y = -5
Hence, the solutions to the given system of equations is (0, -6) and (1, -5)
Learn more on Solving simultaneous equations here: brainly.com/question/16863577
#SPJ!
The simplified form is 20x^2. Because you 2x*2, you get 4x. You multiply that by 5x, and then you get your answer.
Answer:
t=12
Step-by-step explanation:
Step 1: Subtract t from both sides.
−t+5−t=t−19−t
−2t+5=−19
Step 2: Subtract 5 from both sides.
−2t+5−5=−19−5
−2t=−24
Step 3: Divide both sides by -2.
−2t−2=−24−2