Solve the following pair of linear equations: 62x + 37y = 13 37x + 62y = -112 ✅ Chi Tiết
Kinh Nghiệm Hướng dẫn Solve the following pair of linear equations: 62x + 37y = 13 37x + 62y = -112 Mới Nhất
Họ và tên học viên đang tìm kiếm từ khóa Solve the following pair of linear equations: 62x + 37y = 13 37x + 62y = -112 được Update vào lúc : 2022-12-19 09:45:11 . Với phương châm chia sẻ Thủ Thuật về trong nội dung bài viết một cách Chi Tiết 2022. Nếu sau khi Read nội dung bài viết vẫn ko hiểu thì hoàn toàn có thể lại Comments ở cuối bài để Tác giả lý giải và hướng dẫn lại nha.The problem with the row picture of a $2times 2$ systems of linear equations is that it does not extend very well to larger $mtimes n$ systems of linear equations. To extend our ideas of consistent and independent to larger $mtimes n$ systems of linear equations, we need to view the same three systems using the column picture . Recall our three $2times 2$ systems of linear equations
$$beginarraylrcl text1. & x_1 + x_2 & = & 3 \ & x_1 - x_2 & = & -1 endarrayqquad beginarraylrcl 2. & x_1 - x_2 & = & -1 \ & x_1 - x_2 & = & 1 endarrayqquad beginarraylrcl 3. & x_1 + x_2 & = & 3 \ & 3x_1 + 3x_2 & = & 9 endarray$$
Let us re-write the first linear system in vector form .
$$1. beginbmatrix x_1 + x_2 \ x_1 - x_2 endbmatrix = beginbmatrix x_1 \ x_1 endbmatrix + beginbmatrix x_2 \ -x_2 endbmatrix = x_1beginbmatrix 1 \ 1 endbmatrix + x_2beginbmatrix 1 \ -1 endbmatrix = beginbmatrix 3 \ -1 endbmatrix$$
Notice that we start out with a column vector on the left side of the equation
$$beginbmatrix x_1 + x_2 \ x_1 - x_2 endbmatrix$$
This vector equals the vector on the right side of the equation
$$beginbmatrix 3 \ -1 endbmatrix$$
However, we use vector arithmetic, vector addition and scalar multiplication , to write the algebraic expression on the left-hand side of the equation as a linear combination .
$$x_1beginbmatrix 1 \ 1 endbmatrix + x_2beginbmatrix 1 \ -1 endbmatrix$$
A linear combination of vectors, multiplying vectors by scalars and adding them together, is one of the fundamental ideas of this course. You should take the time to recognize a linear combination each time it appears in your notes, the videos, the lectures, or these notes.
This gives us the vector equation
$$x_1beginbmatrix 1 \ 1 endbmatrix + x_2beginbmatrix 1 \ -1 endbmatrix = beginbmatrix 3 \ -1 endbmatrix$$
In English this equation says that some unknown linear combination of vectors $beginbmatrix 1 \ 1 endbmatrix$ and $beginbmatrix 1 \ -1 endbmatrix$ gives us the vector $beginbmatrix 3 \ -1 endbmatrix$. If we use variables for our vectors, we need to decorate them with a bar above or below the letter, or we type them in bold to indicate that the variable is a vector and not a scalar.
Bold face will be preferred in these notes, while arrows over or underlining will be the preferred method for handwritten work. The hatted vectors are typically reserved for special cases, such as the canonical basis vectors $hatimath$, $hatjmath$, and $hatk$ or a "changed" or shifted vector $mathbfxmapstohatmathbfx$. These will be discussed in more detail when they come up.
If we name our vectors $mathbfv_1$ and $mathbfv_2$, then we have
$$ mathbfv_1 = beginbmatrix 1 \ 1 endbmatrix,qquad mathbfv_2 = beginbmatrix 1 \ -1 endbmatrix$$
We typically name the vector on the right-hand side of the equation $mathbfb$. Our system of equations can now be written
$$ x_1mathbfv_1 + x_2mathbfv_2 = mathbfb. $$
Since $1,mathbfv_1 + 2,mathbfv_2 = mathbfb$, we have $x_1=1$ and $x_2=2$, so our unique solution is the vector
$$beginbmatrix x_1 \ x_2 endbmatrix = beginbmatrix 1 \ 2 endbmatrix$$
When there is any linear combination of the two vectors $mathbfv_1$ and $mathbfv_2$ that equals $mathbfb$, the linear system is called consistent .
When the vectors $mathbfv_1$ and $mathbfv_2$ point in different directions; that is they have different slopes, the linear system is called linearly independent .
Let us write the second system of linear equations in vector form.
$$beginbmatrix x_1-x_2 \ x_1-x_2 endbmatrix = beginbmatrix x_1 \ x_1 endbmatrix + beginbmatrix -x_2 \ -x_2 endbmatrix = x_1beginbmatrix 1 \ 1 endbmatrix + x_2beginbmatrix -1 \ -1 endbmatrix = beginbmatrix 1 \ -1 endbmatrix$$
In this case, $mathbfv_1 = beginbmatrix 1 \ 1 endbmatrix$ and $mathbfv_2 = beginbmatrix -1 \ -1 endbmatrix = -mathbfv_1$. In the language of our linear algebra videos, $mathbfv_1$ and $mathbfv_2$ lie on the same span . They point opposite directions but they lie on the same line so that any linear combination of vectors $mathbfv_1$ and $mathbfv_2$ will be on that line with slope $1$ and $y$-intercept $0$.
These vectors are linear dependent because they are colinear. Since $mathbfb$ is not a vector in their span , there is no linear combination of vectors $mathbfv_1$ and $mathbfv_1$ that will equal $mathbfb$. This system of linear equations has no solution.
When there is no linear combination of our vectors $mathbfv_1$ and $mathbfv_1$ that equals $mathbfb$, the linear system is called inconsistent .
When the vectors $mathbfv_1$ and $mathbfv_1$ belong to the same span , the linear system is called linearly dependent .
Finally, let us write the last linear system in vector form.
$$beginbmatrix x_1 + x_2 \ 3x_1 + 3x_2 endbmatrix = x_1beginbmatrix 1 \ 3 endbmatrix + x_2beginbmatrix 1 \ 3 endbmatrix = beginbmatrix 3 \ 9 endbmatrix = 3beginbmatrix 1 \ 3 endbmatrix $$
Notice that $mathbfv_1 = mathbfv_2 = beginbmatrix 1 \ 3 endbmatrix$, so they are certainly linearly dependent. Notice also that $mathbfb = 3beginbmatrix 1 \ 3 endbmatrix$. Our vector equation becomes
$$ left(x_1 + x_2right)beginbmatrix 1 \ 3 endbmatrix = 3beginbmatrix 1 \ 3 endbmatrix$$
Thus we have
$$x_1 + x_2 = 3.$$
We obtain an equation for $x_1$ and $x_2$ that has infinitely many solutions, $x_1 + x_2 = 3$. We could plot this equation on a Cartesian plane, and the set of solutions is a line with slope $-1$ and $y$-intercept $3$. All of the points $(x_1, x_2)$ on this line gives us a solution to the linear system. This linear system has infinitely many solutions.
When there are infinitely many solutions to a system of linear equations, the system is still called consistent .
When the vectors $mathbfv_1$ and $mathbfv_1$ belong to the same span , the linear system is called linearly dependent .
The column picture is more useful to use because we can apply the same ideas to larger systems of $m$ equations and $n$ unknowns.
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