Corrected exercises: linear systems, matrices and rank (MATH 133)

a) Subtract $2R_1$ from $R_2$ and $3R_1$ from $R_3$: the rows become $(0, 0, 3, 3)$ and $(0, 0, 5, 5)$. Divide the second by $3$ to get $(0, 0, 1, 1)$, then $R_3 - 5R_2$ kills the last row entirely. Finally $R_1 + R_2$ clears the third column above the pivot. The reduced form is $\begin{pmatrix} 1 & 2 & 0 & 4 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 \end{pmatrix}$.

b) Two pivots, in columns $1$ and $3$, so the rank is $2$. Column $2$ carries no pivot, so $y$ is free. The equations read $x + 2y = 4$ and $z = 1$, giving $x = 4 - 2y$ and $z = 1$. In parametric form, $(x, y, z) = (4, 0, 1) + t(-2, 1, 0)$ with $t$ any real. That is a LINE in space: one particular solution plus the whole null space direction. The zero row is not a nuisance, it is the third plane being a combination of the first two, which is the same statement as the rank being $2$ and not $3$.

c) Reducing gives $R_2 - 2R_1 = (0, 0 \mid 3)$, that is the equation $0 = 3$. The reduced form has a pivot in the AUGMENTED column, and that single feature is the whole test: a system is inconsistent if and only if the last column carries a pivot. Nothing else about the numbers matters.

d) With $m$ equations and $n$ unknowns and $m < n$, the rank is at most $m$, so it is strictly less than $n$. There are therefore at least $n - m$ free variables. Every choice of a non-zero value for a free variable produces a solution, and a homogeneous system is never inconsistent because the last column is all zeros. So the solution set contains a line at the very least, never just the origin.

e) Swapping two rows only reorders the equations. Multiplying a row by a non-zero constant $c$ replaces an equation by an equivalent one, and non-zero is essential: multiplying by $0$ would delete information and cannot be undone. Adding a multiple of one row to another produces a consequence of the two, and it is reversible by subtracting the same multiple, so nothing is lost. Reversibility is the reason the solution set survives: each operation is a bijection between systems.

a) Suppose $Au = b$ and $Av = b$ with $u \neq v$. Set $w = u - v$, so $Aw = b - b = 0$ and $w \neq 0$. Then for any real $t$, $A(u + tw) = Au + tAw = b + 0 = b$. So $u + tw$ is a solution for every $t$, and since $w \neq 0$ these are all distinct. Two solutions force infinitely many. This is exactly why the answer set of a linear system is either empty, a point, or an unbounded flat object, and never a finite scatter.

b) Reduce: $R_2 - 2R_1$ gives $(0, k - 4 \mid 0)$. If $k \neq 4$ then $y = 0$, then $x = 3$: one solution. If $k = 4$ the second row is all zeros, the rank drops to $1$, $y$ is free and the solutions are $(3 - 2t, t)$: infinitely many. There is no value of $k$ giving no solution here, because the right-hand side $6$ is exactly $2 \times 3$, so the second equation is consistent with the first whatever happens. Changing the $6$ to a $7$ would make $k = 4$ inconsistent, and that is the whole content of the third case.

c) One solution: the two lines cross at a point. No solution: they are parallel and distinct. Infinitely many: they are the same line written twice. The middle case is the only one that cannot happen for a homogeneous system, because both lines then pass through the origin and can never be parallel and distinct.

d) Exactly one. The rank is $3$ and there are $3$ unknowns, so there are $3$ pivots and no free variable. Having four equations rather than three is irrelevant: consistency says the fourth row reduced to zeros rather than to a contradiction, so it carried no information the first three did not already have. Counting equations is never the right count; counting pivots is.

e) The system is consistent if and only if rank$(A) =$ rank$(A \mid b)$. Adding one column to a matrix can add at most one pivot, so rank$(A \mid b)$ is either rank$(A)$ or rank$(A) + 1$, never more. The case rank$(A) + 1$ is precisely a pivot appearing in the augmented column, that is a row reading $0 =$ something non-zero. So the two formulations of inconsistency are the same statement.

a) $AB = \begin{pmatrix} 7 & 2 \\ 3 & 1 \end{pmatrix}$ and $BA = \begin{pmatrix} 1 & 2 \\ 3 & 7 \end{pmatrix}$. They differ, so matrix multiplication is not commutative, and this is not an accident of these entries: $A$ is a shear along $x$ and $B$ a shear along $y$, and shearing then shearing the other way is genuinely a different transformation from doing it in the other order. Note both products have the same trace $8$ and the same determinant $1$, which is a general fact about $AB$ and $BA$ and a useful sanity check.

b) $(A+B)^2 = (A+B)(A+B) = A^2 + AB + BA + B^2$. The middle terms cannot be combined into $2AB$ unless $AB = BA$. Here $AB \neq BA$, so the familiar identity is false. Every binomial-style formula from real algebra has to be re-derived for matrices, keeping the order of every factor, and the difference $AB - BA$ has a name for this reason: the commutator.

c) Take $A = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$ and $B = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}$. Then $AB = 0$ although neither factor is zero. This breaks cancellation: from $AC = AD$ one may NOT conclude $C = D$, because $A(C - D) = 0$ does not force $C = D$. Cancellation returns only when $A$ is invertible, and that is one more entry in the long list of things invertibility buys.

d) $(AB)(B^{-1}A^{-1}) = A(BB^{-1})A^{-1} = AIA^{-1} = AA^{-1} = I$, and the same computation on the other side gives $I$, so $B^{-1}A^{-1}$ is the inverse of $AB$. The order reverses for the reason socks and shoes reverse: to undo putting on socks then shoes, take off the shoes first.

e) Multiply $A^2 = A$ on the left by $A^{-1}$: $A^{-1}A^2 = A^{-1}A$, that is $A = I$. The hypothesis of invertibility is doing all the work, since without it $A^2 = A$ has many solutions, for instance any projection such as $\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$. A projection is idempotent precisely because projecting twice is projecting once, and it is never invertible unless it is the identity.

a) $3 \times 2.5 - 1 \times 1 = 7.5 - 1 = 6.5$, which is the area written on the figure. The sign is positive, meaning the pair $(u, v)$ turns counterclockwise, in the same sense as the standard basis. Swapping the two columns would give $-6.5$: the area is unchanged but the orientation is reversed, and that is what the word signed is doing.

b) Expand along the first column, which has a zero in it: $2 \times \det \begin{pmatrix} 3 & -1 \\ 2 & 4 \end{pmatrix} - 1 \times \det \begin{pmatrix} 0 & 1 \\ 2 & 4 \end{pmatrix} + 0$. That is $2(12 + 2) - 1(0 - 2) = 28 + 2 = 30$. Choosing the row or column with the most zeros is not a stylistic preference, it removes an entire $3 \times 3$ of work, and on a midterm it is the difference between finishing and not.

c) Swapping two rows multiplies the determinant by $-1$. Multiplying a row by $c$ multiplies it by $c$. Adding a multiple of one row to another leaves it UNCHANGED. Geometrically the third one slides the parallelogram along its own base: the base and the height are untouched, so the area is too. That is why elimination is the practical way to compute a large determinant: the expensive operation is free.

d) If $A$ is invertible then $AA^{-1} = I$, so $\det A \cdot \det(A^{-1}) = \det I = 1$. A product of two numbers equal to $1$ forces both to be non-zero, so $\det A \neq 0$, and $\det(A^{-1}) = 1/\det A$. The converse holds too, which is why $\det A \neq 0$ is one line of the invertible matrix theorem rather than a separate fact to memorise.

e) The determinant is $0$. Subtracting one of the identical columns from the other leaves a zero column and does not change the determinant, and a matrix with a zero column has determinant zero. Since the four columns are dependent, they span at most three dimensions, so the rank is at most $3$. In one glance the matrix is not invertible, the homogeneous system has a non-trivial solution, and the map is not onto.

a) Row reduce the matrix with these as columns. $R_2 - 2R_1$ and $R_3 - 3R_1$ give rows $(0, 0, 0)$ and $(0, 1, 1)$, so after a swap the pivots sit in columns $1$ and $2$ and the third column carries none. The rank is $2$, so the three are DEPENDENT. Solving gives $(2,4,7) = (1,2,3) + (1,2,4)$, a dependency you can check entry by entry: $1+1 = 2$, $2+2 = 4$, $3+4 = 7$.

b) No. Three vectors span $\mathbb{R}^3$ only when the rank is $3$, and here it is $2$. They span a plane through the origin. This is the point of the exercise: independence and spanning are not two computations, they are two readings of the same pivot count. For $n$ vectors in $\mathbb{R}^n$, rank $n$ gives both at once and any smaller rank gives neither.

c) The matrix holding them is $3 \times 4$. Its rank is at most $3$, being bounded by the number of rows, so with $4$ columns there is at least one free variable and therefore a non-trivial combination giving zero. More generally, more than $n$ vectors in $\mathbb{R}^n$ are always dependent, and fewer than $n$ can never span it.

d) rank$(A) +$ nullity$(A) = n$, the number of COLUMNS, never the number of rows. On the figure the $7$ columns split into $4$ pivot columns and $3$ free ones, and each free column contributes exactly one basis vector of the null space. The theorem is not a coincidence, it is that split written as an equation.

e) Nullity $= 7 - 4 = 3$. The image lives in $\mathbb{R}^5$ and has dimension equal to the rank, so $4 < 5$: the map is NOT onto. Consequently there are right-hand sides $b$ outside the image, and yes, $Ax = b$ can be inconsistent. When it is consistent, the solution set is a $3$-dimensional flat, never a single point.

a) The cheapest route is the determinant, since the matrix is small. Expanding gives $1(55 - 49) - 2(22 - 21) + 3(14 - 15) = 6 - 2 - 3 = 1$. Non-zero, so the matrix IS invertible, the rank is $3$, the columns are independent and $Ax = b$ has exactly one solution for every $b$. All of that came from one number. Had the determinant been $0$, the row reduction would then have been needed anyway to find the rank, which is why elimination is the more informative route on anything larger than $3 \times 3$.

b) The determinant is $t^2 - 4$, zero when $t = 2$ or $t = -2$. At $t = 2$ the matrix is $\begin{pmatrix} 2 & 1 \\ 4 & 2 \end{pmatrix}$, whose rows are proportional, so the rank is $1$ and the null space is the line spanned by $(1, -2)$. At $t = -2$ the null space is the line spanned by $(1, 2)$. In both cases the solution set is a line through the origin, never a single point: for a singular square matrix the homogeneous system always has room to move.

c) Suppose $A$ were invertible. Then $A^{-1}$ exists and multiplying $A^3 = 0$ by $(A^{-1})^3$ gives $I = 0$, impossible for $n \ge 1$. So $A$ is singular. Directly: $\det(A)^3 = \det(A^3) = 0$, so $\det A = 0$. Such a matrix is called nilpotent, and the argument shows nilpotent and invertible are mutually exclusive, with $\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ as the standard example.

d) From $AB = I$ take determinants: $\det A \det B = 1$, so $\det A \neq 0$ and $A$ is invertible. Multiply $AB = I$ on the left by $A^{-1}$ to get $B = A^{-1}$, hence $BA = A^{-1}A = I$. The argument uses squareness twice, once for the determinant to exist and once for $A^{-1}$ to be two-sided. For non-square matrices it genuinely fails: a $2 \times 3$ matrix can have a right inverse and no left inverse, because it can be onto without being one-to-one.

e) Take $A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix}$. If $AC = I$ then for every $b$ the vector $x = Cb$ solves $Ax = b$, so the map would be onto and the rank would be $3$. It is $2$. The image is a plane and no choice of $C$ can send anything outside it, so $AC = I$ is impossible. This is the argument in its useful form: rank is a property of $A$ alone and no companion matrix can raise it.

a) Eliminating $x$ gives $-2y + z = -1$ and $-y - 3z = -11$. From the first, $z = 2y - 1$; substituting, $-y - 6y + 3 = -11$, so $7y = 14$ and $y = 2$, then $z = 3$ and $x = 1$. The solution is $(1, 2, 3)$, and every equation checks: $1+2+3 = 6$, $1-2+6 = 5$, $2+2-3 = 1$. Rank $3$, three pivots, no free variable: the three planes meet in a single POINT, the first picture on the figure.

b) The second and third equations are $2$ and $3$ times the first, so after reduction only one row survives: rank $1$. Two free variables, so the solution set is the plane $x + y + z = 1$ itself, of dimension $2$. Parametrically $(1 - s - t, s, t)$. Three equations were written and one plane was described: the count of equations again says nothing, the rank says everything.

c) The third row reduces to $0 = 1$, a pivot in the augmented column, so the system becomes inconsistent. Geometrically $3x + 3y + 3z = 4$ is the plane $x + y + z = 4/3$, parallel to the first and distinct from it. Two parallel distinct planes never meet, so the whole system is empty however the second equation behaves.

d) The second row minus twice the first gives $(0, 0, a - 2 \mid 0)$. If $a \neq 2$ there is a pivot in the third column, the rank is $2$ and with three unknowns exactly one variable is free: a line. If $a = 2$ the second equation is a copy of the first, the rank drops to $1$, two variables are free and the solution set is the plane $x + 2y + z = 1$. So $a = 2$ is the answer.

e) Take three planes whose normals are independent in pairs but which meet pairwise in three DIFFERENT parallel lines, the way the three side faces of a triangular prism do. No two planes are parallel, each pair meets in a line, and yet the three lines never coincide, so no point lies on all three. In the algebra, rank$(A) = 2$ while rank$(A \mid b) = 3$: the coefficient matrix says the normals span only a plane, and the augmented column adds the pivot that makes it inconsistent.

a) FALSE. What decides is the rank, not the counts. The system $x = 1$, $2x = 2$, $3x = 3$ has three equations, one unknown, and exactly one solution. Correct statement: more equations than unknowns makes inconsistency LIKELY for a random right-hand side, because the rank is at most the number of unknowns while the augmented rank could exceed it, but it guarantees nothing.

b) FALSE, and backwards. $A^{-1} = \frac{1}{\det A}\,\text{adj}(A)$, so a small determinant makes the inverse LARGE, not small. Take $\begin{pmatrix} 1 & 0 \\ 0 & 0.001 \end{pmatrix}$: the determinant is $0.001$ and the inverse has an entry of $1000$. That blow-up is the whole practical meaning of near-singularity, and it is why a nearly singular system amplifies any error in $b$.

c) FALSE. Row operations preserve the row space and the null space, and they change the COLUMN space. In $\begin{pmatrix} 1 & 2 \\ 2 & 4 \end{pmatrix}$, reducing to $\begin{pmatrix} 1 & 2 \\ 0 & 0 \end{pmatrix}$ turns a column space spanned by $(1,2)$ into one spanned by $(1,0)$: different subspaces, same dimension. Correct: row reduction preserves the DIMENSION of the column space, which is why the rank can be read off the reduced form, but a basis of the column space must be taken from the ORIGINAL pivot columns.

d) TRUE only for a SQUARE matrix, and the statement omits that. For an $n \times n$ matrix, spanning and independence are equivalent, both meaning rank $n$. For a $2 \times 3$ matrix the three columns can span $\mathbb{R}^2$ while being dependent, for instance $(1,0)$, $(0,1)$, $(1,1)$. Correct statement: for $n$ vectors in $\mathbb{R}^n$, spanning and independence are equivalent; with more vectors than the dimension, spanning never implies independence.

e) FALSE unless $A$ is invertible. $AB = AC$ gives $A(B - C) = 0$, which forces $B = C$ only when the null space of $A$ is trivial. With $A = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$, $B = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}$ and $C$ the zero matrix, $AB = AC = 0$ while $B \neq C$. Correct: cancellation on the left requires $A$ to have independent columns.

a) Rearranging each row of the table: $x_1 - x_4 = 100$, $x_2 - x_1 = 100$, $x_3 - x_2 = -100$, $x_4 - x_3 = -100$. Note the four right-hand sides add to $0$, which is the total entering minus the total leaving: $800 - 800 = 0$. That balance is not an accident of the numbers, it is forced by counting every car twice, once in and once out.

b) Adding all four equations gives $0 = 0$: the fourth row is minus the sum of the first three, so it carries no information. The rank is $3$, not $4$, and it CANNOT be $4$ for the reason just given, because conservation at the last corner follows from conservation at the other three plus the global balance. Any closed network has this dependency, and it is why a traffic system always has at least one free flow.

c) With rank $3$ and four unknowns, one variable is free. Take $x_4 = t$. Then $x_1 = t + 100$, $x_2 = t + 200$, $x_3 = t + 100$. Every flow is pinned to every other, and only the overall level of circulation around the block is undetermined. That circulating amount is the null space direction $(1,1,1,1)$: adding the same number to every street changes nothing at any corner.

d) All four must be non-negative: $t \ge 0$, $t + 100 \ge 0$, $t + 200 \ge 0$, $t + 100 \ge 0$. The binding constraint is $t \ge 0$, so $t \in [0, \infty)$ and the smallest possible $x_2$ is $200$, reached when $t = 0$, that is when the street carrying $x_4$ is empty. Physical constraints do not enter the linear algebra, they select a piece of the solution line afterwards, and that two-step reading is what the question is testing.

e) Closing that street sets $x_3 = 0$, which forces $t = -100$ and therefore $x_4 = -100$, a negative flow. So the network is algebraically consistent but physically impossible: the equations have a solution, the one-way constraints do not. Either a street must be reversed or the corner flows must change. The system says exactly which: with $x_3 = 0$ imposed, conservation at $C$ demands $x_2 = 100$, and every other value follows.

a) $\det(A - \lambda I) = (4-\lambda)(3-\lambda) - 2 = \lambda^2 - 7\lambda + 10 = (\lambda - 2)(\lambda - 5)$, so the eigenvalues are $2$ and $5$. For $\lambda = 5$, $(A - 5I)v = 0$ reads $-v_1 + v_2 = 0$, giving $v = (1, 1)$. For $\lambda = 2$, $2v_1 + v_2 = 0$, giving $v = (1, -2)$. Both checks are one multiplication: $A(1,1) = (5,5)$ and $A(1,-2) = (2,-4)$.

b) $\lambda = 0$ is an eigenvalue exactly when $(A - 0I)v = Av = 0$ has a non-zero solution, that is when the null space is non-trivial, that is when $A$ is singular. So the invertible matrix theorem gains one more line for free, and the determinant of $A$ being the product of the eigenvalues makes the same statement a second time.

c) They are the diagonal entries. $A - \lambda I$ is still triangular, and the determinant of a triangular matrix is the product of its diagonal, so $\det(A - \lambda I) = \prod (a_{ii} - \lambda)$, whose roots are the $a_{ii}$. This is worth noticing on a midterm: a triangular matrix hands over its eigenvalues with no computation at all.

d) For a $2 \times 2$, $\det(A - \lambda I) = \lambda^2 - (\text{tr}A)\lambda + \det A$. Comparing with $(\lambda - \lambda_1)(\lambda - \lambda_2) = \lambda^2 - (\lambda_1 + \lambda_2)\lambda + \lambda_1\lambda_2$ gives the two identities. Here the trace is $4 + 3 = 7 = 2 + 5$ and the determinant is $12 - 2 = 10 = 2 \times 5$. Both hold, which is the fastest available check on a computed pair of eigenvalues.

e) NOT diagonalizable. The eigenvalue $2$ has algebraic multiplicity $2$ but geometric multiplicity $1$, so the eigenvectors span only $1 + 1 = 2$ dimensions and there is no basis of $\mathbb{R}^3$ made of them. The geometric multiplicity is the nullity of $A - 2I$, which is $1$, so by rank-nullity rank$(A - 2I) = 3 - 1 = 2$. The whole answer came from counting pivots once more, which is the thread of the set arriving at its last chapter.