1. a)
i. Which of the following is true for the system?
x + y = 14
x-y-14=0
a) The system has a unique solution
b) The system has no solution
c) The system has an infinite number of solutions
d) All of the above
Answers
Step-by-step explanation:
a)
i. Which of the following is true for the system?
x + y = 14
x-y-14=0
a) The system has a unique solution
b) The system has no solution
c) The system has an infinite number of solutions
d) All of the above
answer) a)The system has a unique solution
Step-by-step explanation:
Solving a system of linear equations using the inverse of a matrix requires the definition of two new matrices: \displaystyle XX is the matrix representing the variables of the system, and \displaystyle BB is the matrix representing the constants. Using matrix multiplication, we may define a system of equations with the same number of equations as variables as
\displaystyle AX=BAX=B
To solve a system of linear equations using an inverse matrix, let \displaystyle AA be the coefficient matrix, let \displaystyle XX be the variable matrix, and let \displaystyle BB be the constant matrix. Thus, we want to solve a system \displaystyle AX=BAX=B. For example, look at the following system of equations.
\displaystyle \begin{array}{c}{a}_{1}x+{b}_{1}y={c}_{1}\\ {a}_{2}x+{b}_{2}y={c}_{2}\end{array}
a
1
x+b
1
y=c
1
a
2
x+b
2
y=c
2
From this system, the coefficient matrix is
\displaystyle A=\left[\begin{array}{cc}{a}_{1}& {b}_{1}\\ {a}_{2}& {b}_{2}\end{array}\right]A=[
a
1
a
2
b
1
b
2
]
The variable matrix is
\displaystyle X=\left[\begin{array}{c}x\\ y\end{array}\right]X=[
x
y
]
And the constant matrix is
\displaystyle B=\left[\begin{array}{c}{c}_{1}\\ {c}_{2}\end{array}\right]B=[
c
1
c
2
]
Then \displaystyle AX=BAX=B looks like
\displaystyle \left[\begin{array}{cc}{a}_{1}& {b}_{1}\\ {a}_{2}& {b}_{2}\end{array}\right]\text{ }\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}{c}_{1}\\ {c}_{2}\end{array}\right][
a
1
a
2
b
1
b
2
] [
x
y
]=[
c
1
c
2
]
Recall the discussion earlier in this section regarding multiplying a real number by its inverse, \displaystyle \left({2}^{-1}\right)2=\left(\frac{1}{2}\right)2=1(2
−1
)2=(
2
1
)2=1. To solve a single linear equation \displaystyle ax=bax=b for \displaystyle xx, we would simply multiply both sides of the equation by the multiplicative inverse (reciprocal) of \displaystyle aa. Thus,
\displaystyle \begin{array}{c}\text{ }ax=b\\ \text{ }\left(\frac{1}{a}\right)ax=\left(\frac{1}{a}\right)b\\ \left({a}^{-1}\text{ }\right)ax=\left({a}^{-1}\right)b\\ \left[\left({a}^{-1}\right)a\right]x=\left({a}^{-1}\right)b\\ \text{ }1x=\left({a}^{-1}\right)b\\ \text{ }x=\left({a}^{-1}\right)b\end{array}
ax=b
(
a
1
)ax=(
a
1
)b
(a
−1
)ax=(a
−1
)b
[(a
−1
)a]x=(a
−1
)b
1x=(a
−1
)b
x=(a
−1
)b
The only difference between a solving a linear equation and a system of equations written in matrix form is that finding the inverse of a matrix is more complicated, and matrix multiplication is a longer process. However, the goal is the same—to isolate the variable.
We will investigate this idea in detail, but it is helpful to begin with a \displaystyle 2\times 22×2 system and then move on to a \displaystyle 3\times 33×3 system.