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Below two problems are stated consecutively in book titled: Linear algebra, by: David cherney, Tom denton, Rohit thomas and Andrew waldron; at pg. #51, in review sec. 2.2.
It seems to me that they are linked, so stated together.

Q.8. Show that this pair of augmented matrices are row equivalent assuming $ad - bc \ne 0$:

$x = \begin{pmatrix}a & b & | & e\\c & d & | & f \end{pmatrix} \approx \begin{pmatrix}1 & 0 & | & \frac{dc-bf}{ad-bc}\\0 & 1 & | & \frac{af-ce}{ad-bc} \end{pmatrix}$

Q.9. Consider the augmented matrix:

$x = \begin{pmatrix}2 & -1 & | & 3\\-6 & 3 & | & 1 \end{pmatrix}$

Give a geometric reason why the associated system of equations has no solution. (Hint, plot the three vectors given by the columns of the augmented matrix in the plane.) Given a general augmented matrix

$x = \begin{pmatrix}a & b & | & e\\c & d & | & f \end{pmatrix}$

can you find a condition on the numbers a, b, c and d that corresponds to the geometric condition you found?

Solution attempts:

Q.8. I am unable to see any way to reach the second matrix from the first, so want to attempt derivation of the first from the second one.
As given $ad - bc \ne 0$, so:

$x = \begin{pmatrix}1 & 0 & | & \frac{dc-bf}{ad-bc}\\0 & 1 & | & \frac{af-ce}{ad-bc} \end{pmatrix} \rightarrow \begin{pmatrix}(ad-bc) & 0 & | & (dc-bf)\\0 & (ad-bc) & | & (af-ce) \end{pmatrix}$

But, unable to make any progress still.

Q.9. Assuming that the vectors (or matrix elements) are drawn from the field of reals, the three column vectors are:

$x_1 = \begin{bmatrix}2\\-6 \end{bmatrix}, x_2 = \begin{bmatrix}-1\\3 \end{bmatrix}, x_3 = \begin{bmatrix}3\\1 \end{bmatrix}$ in $\mathbb{R}^2$,

Due to the domain $\mathbb{R}^2$, the matrix's l.h.s. can be interpreted as dot product; by taking the first component as in $x$-coordinate, while the second component as $y$-coordinate:

$\begin{bmatrix}2\\-6 \end{bmatrix}.\begin{bmatrix}x\\y \end{bmatrix} + \begin{bmatrix}-1\\3 \end{bmatrix} .\begin{bmatrix}x\\y \end{bmatrix}= \begin{bmatrix}3\\1 \end{bmatrix}$ in $\mathbb{R}^2$

As per the question asking geometrical reason: the first two vectors have slope: $-3$, while the third vector has slope $\frac{1}{3}$; so third vector is perpendicular to the first two. But, still geometrically the third vector intersects the first two vectors. So, not clear why there is no solution associated with the three column vectors. I am not clear as why origin is not a solution, as here all three vectors intersect.

However, if take clue (without any understanding) from the earlier question, get the condition $ad-bc\ne 0$; that implies that satisfying $ad \ne bc\implies \frac ca \ne \frac db$.

jiten
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1 Answers1

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For your first questions, if you are familiar with the result that pre-multiplying the augmented system with an invertible matrix preserve the row equivalence, just pre-multiply with the inverse and you are done.

Alternatively, use the property they share the same unique solution, just directly verify.

Or you can consider $2$ cases, of which I will let you complete the details.

  • If $a \ne 0$,

$$\left( \begin{array}{cc|c} a & b & e \\ c & d & f\end{array}\right) \to \left( \begin{array}{cc|c} 1 & \frac{b}a & \frac{e}a \\ c & d & f\end{array}\right) \to \left( \begin{array}{cc|c} 1 & \frac{b}a & \frac{e}a \\ 0 & d-\frac{bc}a & f-\frac{ce}{a}\end{array}\right) \to \ldots $$

  • If $a=0$,

$$\left( \begin{array}{cc|c} 0 & b & e \\ c & d & f\end{array}\right) \to \left( \begin{array}{cc|c} c & d & f\\ 0 & 1 & \frac{e}b \end{array}\right)\to \left( \begin{array}{cc|c} c & 0 & f-\frac{de}{b}\\ 0 & 1 & \frac{e}b \end{array}\right)\to \ldots$$

For your second question, notice that $ad-bc=0$, also your first two columns are colinear but the third vector, they span the direction of $\begin{pmatrix} 1 \\ 3\end{pmatrix}$ but the vector $\begin{pmatrix} 3 \\ 1\end{pmatrix}$ does not lie on the line $y=3x$.

Siong Thye Goh
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  • For first question, for the first case; I can only proceed one additional step: $$\to \left( \begin{array}{cc|c} 1 & \frac{b}a & \frac{e}a \ 0 & \frac{ad-bc}a & \frac{af-ce}a \end{array}\right) $$. Request help, but even more important is that by some approach (algebraic/geometrical), it should be visible here the final answer. As only then the steps will go in that direction, in absence of any question. I mean the intuition should also be obvious. – jiten May 19 '20 at 13:46
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    The next move is to divide the second row by the $(2,2)$ entry. After that zero out the $(1,2)$-th entry and you are done. For an algebraic approach, just premultiply by matrix inverse. – Siong Thye Goh May 19 '20 at 13:57
  • Still am not clear why he geometrical approach does not yield the origin as the intersection point of all three column vectors. – jiten May 20 '20 at 00:27
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    you might like to view it as a point. – Siong Thye Goh May 20 '20 at 01:46
  • but viewing as points, would not solve even if the condition $ad - bc\ne 0$ is met. – jiten May 20 '20 at 02:00
  • if $ad-bc \ne 0$, you can walk through the whole $\mathbb{R}^2$ space. – Siong Thye Goh May 20 '20 at 02:14
  • Please elaborate your last comment, as I thought that problem is about finding the unique intersection point of three vectors. I stated in my last comment that even if no two vectors lie on the same slope line; still viewing them as points wouldn't help in finding intersection. But, you last comment seems about spanning the entire two dimensional space. – jiten May 20 '20 at 02:27
  • Please respond to my last comment. – jiten May 20 '20 at 03:24
  • Let us continue this discussion in chat. – Siong Thye Goh May 20 '20 at 03:25