Problems 21-28
21 The left hand-side matrix is
$$ \ \phantom{,} \begin{bmatrix} 2x & + & y & & & & \\ x & + & 2y & + & z & & \\ & & y & + & 2z & + & t \\ & & & & z & + & 2t \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 5 \end{bmatrix} \ , $$so the first pivot is $\frac{l_{21}}{l_{11}} = \frac{1}{2}$. Multiplying the first row by the pivot and subtracting the given product from the second row leads to matrix
$$ \phantom{.} \begin{bmatrix} 2x & + & y & & & & \\ & & \frac{3}{2}y & + & z & & \\ & & y & + & 2z & + & t \\ & & & & z & + & 2t \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 5 \end{bmatrix} \ . $$The next pivot is $\frac{2}{3}$, so subtracting the second row multiplied by the pivot from the third row results in matrix
$$ \phantom{.} \begin{bmatrix} 2x & + & y & & & & \\ & & \frac{3}{2}y & + & z & & \\ & & & & \frac{4}{3}z & + & t \\ & & & & z & + & 2t \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 5 \end{bmatrix} \ . $$The last pivot is $\frac{3}{4}$ and continuing the elimination, that is, subtracting the third row multiplied by the pivot from the fourth is
$$ \phantom{.} \begin{bmatrix} 2x & + & y & & & & \\ & & \frac{3}{2}y & + & z & & \\ & & & & \frac{4}{3}z & + & t \\ & & & & & & \frac{5}{4}t \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 5 \end{bmatrix} \ . $$from where it can be seen that
$$ \begin{array}{l c r c r} t & = & 5 \left( \dfrac{4}{5} \right) & = & 4 \ , \\[0.5em] z & = & -4 \left( \dfrac{3}{4} \right) & = & -3 \ , \\[0.5em] y & = & 3 \left( \dfrac{2}{3} \right) & = & 2 \ \phantom{,} \end{array} $$and
$$ \begin{array}{l c r c r} x & = & -2 \left( \dfrac{1}{2} \right) & = & -1 \ . \end{array} $$The first pivot of the right hand-side equation $\frac{l_{21}}{l_{11}}$ $=$ $-\frac{1}{2}$, so multiplying the first row by the pivot and subtracting the row from the second means that
$$ \ \phantom{,} \begin{bmatrix} 2x & - & y & & & & \\ & - & \frac{3}{2}y & + & z & & \\ & - & y & + & 2z & + & t \\ & & & - & z & + & 2t \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 5 \end{bmatrix} \ . $$The second pivot is $\frac{l_{32}}{l_{22}}$ $=$ $\frac{-1}{-(3/2)}$ $=$ $\frac{2}{3}$. Multiplying the second row with the second pivot and subtracting the row from the third leads to
$$ \ \phantom{,} \begin{bmatrix} 2x & - & y & & & & \\ & - & \frac{3}{2}y & + & z & & \\ & & & & \frac{4}{3}z & + & t \\ & & & - & z & + & 2t \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 5 \end{bmatrix} \ . $$The last, i.e. the third pivot is $\frac{l_{43}}{l_{33}}$ $=$ $-\frac{1}{4/3}$ $=$ $-\frac{3}{4}$, so after multiplying the third row with the pivot and subtracting this product from the fourth row, the system of equations becomes
$$ \phantom{.} \begin{bmatrix} 2x & - & y & & & & \\ & - & \frac{3}{2}y & + & z & & \\ & & & & \frac{4}{3}z & + & t \\ & & & & & - & \frac{5}{4}t \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 5 \end{bmatrix} \ . $$As with the left hand-side, applying the back substitution means that
$$ \begin{array}{l c r c r} t & = & 5 \left( -\dfrac{4}{5} \right) & = & -4 \ , \\[0.5em] z & = & -4 \left( \dfrac{3}{4} \right) & = & -3 \ , \\[0.5em] y & = & -3 \left( \dfrac{2}{3} \right) & = & -2 \ \phantom{,} \end{array} $$and
$$ \begin{array}{l c r c r} x & = & -2 \left( \dfrac{1}{2} \right) & = & -1 \ . \end{array} $$22
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23 If the system of equation leads to
$$ \phantom{,} \ \left\{ \ \begin{array}{r r r r r} x & + & y & = & 1 \\ & & 2y & = & 3 \end{array} \ \right. \ , $$then the original system of equations could’ve been
$$ \phantom{,} \ \left\{ \ \begin{array}{r r r r r} x & + & y & = & 1 \\ 2x & + & 4y & = & 5 \end{array} \ \right. \ , $$because subtracting the first row multiplied by the pivot this setup gives, leads to the system of equations defined in the problem statement.
A original system of equations could have been also
$$ \phantom{,} \ \left\{ \ \begin{array}{r r r r r} x & + & y & = & 1 \\ 0x & + & 2y & = & 3 \end{array} \ \right. \ , $$because it leads to the given system of equations. In this form no elimination is needed.
Lastly, the original system of equations could have been
$$ \phantom{,} \ \left\{ \ \begin{array}{r r r r r} x & + & y & = & 1 \\ x & + & 3y & = & 4 \end{array} \ \right. \ , $$because after finding out the pivot, multiplying the second row with it and subtracting the product leads the given systems of equations.
24 When $a = 2$ or $a = 0$, the elimination leads to failure with infinitely many solutions because these numbers lead to zeros in the pivot values.
25 – 26 Any number $a \in \mathbb{R}$ means that for matrix
$$ A = \left[ \ \begin{array}{c c c} a & 2 & 3 \\ a & a & 4 \\ a & a & a \end{array} \ \right] $$leads to zeros in the pivots ($a - a = 0$) and into situation where the elimination fails with permanent failure with no solution.
27 Solving the system of equations
$$ \left\{ \ \ \begin{array}{ c c c c c c } 3x & & & & & = & 3 \\ 6x & + & 2y & & & = & 8 \\ 9x & - & 2y & + & z & = & 9 \end{array} \right. $$by the suggested forward susbstitution, means that starting from the first row, on can see that $x = 1$. Substituting $x$ to the second equation means that $2y + 6 = 8$ so $y = 1$. Substituting $x$ and $y$ to the third equation leads to $9 - 2 + z = 9$, so $z = 2$. Substituting $x$, $y$ and $z$ to the given equation shows that forward substitution yielded right answers.
28 Let matrix
$$ \phantom{.} \ \mathbf{A} = \left[ \ \begin{array}{c c c} a & b \\ c & d \end{array} \ \right] \ . $$The transpose of $\mathbf{A}$ is
$$ \phantom{.} \ \mathbf{A}^\top = \left[ \ \begin{array}{c c c} a & c \\ b & d \end{array} \ \right] \ . $$Multiplying $\mathbf{A}^\top$ with partial identity matrix
$$ \mathbf{P} = \left[ \ \begin{array}{c c c} 0 & 0 \\ 0 & 1 \end{array} \ \right] $$means that
$$ \phantom{.} \ \mathbf{A}^\top \mathbf{P} = \left[ \ \begin{array}{c c c} a & c \\ b & d \end{array} \ \right] \left[ \ \begin{array}{c c c} 0 & 0 \\ 0 & 1 \end{array} \ \right] = \left[ \ \begin{array}{c c c} 0 & c \\ 0 & d \end{array} \ \right] \ , $$so
$$ \phantom{,} \ (\mathbf{A}^\top \mathbf{P})^\top = \left[ \ \begin{array}{c c c} 0 & 0 \\ c & d \end{array} \ \right] \ . $$Multiplying this transpose with exchange matrix
$$ \mathbf{E} = \left[ \ \begin{array}{c c c} 0 & 1 \\ 1 & 0 \end{array} \ \right] $$leads to
$$ \phantom{,} \ \mathbf{E}(\mathbf{A}^\top \mathbf{P})^\top = \left[ \ \begin{array}{c c c} c & d \\ 0 & 0 \end{array} \ \right] \ , $$which can be multiplied with $-3$ to get a subtraction expression that subtracts $\mathbf{A}$’s second row three times subtracted from $\mathbf{A}$, that is,
Encoding the subtraction expression into a simple MATLAB expression could be done by writing a
function result = subtract_second_row_thrice(A)
result = A - (3 * E * (A' * P)');
end
which is enough to solve the given problem.