Corrected exercises: matrix inverses and elementary matrices (MATH 133)

a) $AB = \begin{pmatrix} 3 & 5 \\ 1 & 2 \end{pmatrix}\begin{pmatrix} 2 & -5 \\ -1 & 3 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = I$, and $BA = \begin{pmatrix} 2 & -5 \\ -1 & 3 \end{pmatrix}\begin{pmatrix} 3 & 5 \\ 1 & 2 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = I$. Both products are the identity, so $B = A^{-1}$. For square matrices a theorem of the chapter says one product is enough, but on a first question the marker expects the definition, and the definition has two sides. Here $ad - bc = 6 - 5 = 1$, which is why the inverse has integer entries: the formula divides by $1$.

b) Suppose $AB = BA = I$ and $AC = CA = I$. Then $B = BI = B(AC) = (BA)C = IC = C$. The proof uses associativity and nothing else, and it uses $BA = I$ on one side and $AC = I$ on the other. This is what licenses the notation $A^{-1}$: the article THE inverse is only allowed once uniqueness is proved. It also means that any matrix you find by any method, formula, algorithm or guess, is THE inverse as soon as it passes the test.

c) $ad - bc = 2 \times 3 - 1 \times 4 = 2$, so $M^{-1} = \frac{1}{2}\begin{pmatrix} 3 & -1 \\ -4 & 2 \end{pmatrix} = \begin{pmatrix} \frac{3}{2} & -\frac{1}{2} \\ -2 & 1 \end{pmatrix}$. The gesture is: swap the diagonal entries, change the SIGN of the off-diagonal ones, divide by $ad - bc$. The system is $M\mathbf{x} = \mathbf{b}$ with $\mathbf{b} = (3, 7)$, and multiplying on the LEFT by $M^{-1}$ gives $\mathbf{x} = M^{-1}\mathbf{b} = \left(\frac{9}{2} - \frac{7}{2},\ -6 + 7\right) = (1, 1)$. Check: $2 + 1 = 3$ and $4 + 3 = 7$. This is the crossing point of the two lines on the left panel, and it exists for EVERY right-hand side because $M^{-1}$ exists: change $\mathbf{b}$, the lines slide parallel to themselves, and they still cross exactly once.

d) Take any $2 \times 2$ matrix $B$ with rows $\mathbf{r}_1$ and $\mathbf{r}_2$. The rows of $PB$ are $4\mathbf{r}_1 + 6\mathbf{r}_2$ and $2\mathbf{r}_1 + 3\mathbf{r}_2$, so the first row of $PB$ is ALWAYS twice the second. The identity has rows $(1, 0)$ and $(0, 1)$, and $(1, 0)$ is not twice $(0, 1)$. So no $B$ gives $PB = I$ and $P$ has no inverse. Consistently, $ad - bc = 12 - 12 = 0$ and the formula would divide by zero. On the figure the two equations $4x + 6y = 6$ and $2x + 3y = 6$ are parallel lines: the rows of $P$ are proportional, the lines have the same slope, and for this right-hand side there is no solution at all. An inverse would have produced one, so it cannot exist.

e) The $(1,1)$ entry of the student's matrix times $M$ is $\frac{1}{2} \times 2 + 1 \times 4 = 5$, not $1$, so the product is not $I$ and the matrix is not $M^{-1}$. Inverting a matrix is not inverting its entries: the inverse must undo the whole map, and the formula of part c) shows that every entry of $M^{-1}$ depends on all four entries of $M$. Cost on an exam: the whole question, and every part that uses the inverse afterwards.

a) Start from $\left(\begin{array}{ccc|ccc} 1 & 0 & 2 & 1 & 0 & 0 \\ 2 & 1 & 3 & 0 & 1 & 0 \\ 1 & 1 & 2 & 0 & 0 & 1 \end{array}\right)$. $R_2 \to R_2 - 2R_1$ and $R_3 \to R_3 - R_1$ give $\left(\begin{array}{ccc|ccc} 1 & 0 & 2 & 1 & 0 & 0 \\ 0 & 1 & -1 & -2 & 1 & 0 \\ 0 & 1 & 0 & -1 & 0 & 1 \end{array}\right)$. Then $R_3 \to R_3 - R_2$ gives the third row $(0, 0, 1 \mid 1, -1, 1)$: the left block is now upper triangular with pivots $1, 1, 1$, so the matrix is invertible and the rest is clearing upwards. $R_2 \to R_2 + R_3$ gives $(0, 1, 0 \mid -1, 0, 1)$ and $R_1 \to R_1 - 2R_3$ gives $(1, 0, 0 \mid -1, 2, -2)$. The block reads $\left(\begin{array}{ccc|ccc} 1 & 0 & 0 & -1 & 2 & -2 \\ 0 & 1 & 0 & -1 & 0 & 1 \\ 0 & 0 & 1 & 1 & -1 & 1 \end{array}\right)$, so $A^{-1} = \begin{pmatrix} -1 & 2 & -2 \\ -1 & 0 & 1 \\ 1 & -1 & 1 \end{pmatrix}$. Five named operations: that list is where the method marks are.

b) $AA^{-1} = \begin{pmatrix} 1 & 0 & 2 \\ 2 & 1 & 3 \\ 1 & 1 & 2 \end{pmatrix}\begin{pmatrix} -1 & 2 & -2 \\ -1 & 0 & 1 \\ 1 & -1 & 1 \end{pmatrix}$. Row $1$ times the columns: $-1 + 0 + 2 = 1$, $2 + 0 - 2 = 0$, $-2 + 0 + 2 = 0$. Row $2$: $-2 - 1 + 3 = 0$, $4 + 0 - 3 = 1$, $-4 + 1 + 3 = 0$. Row $3$: $-1 - 1 + 2 = 0$, $2 + 0 - 2 = 0$, $-2 + 1 + 2 = 1$. The product is $I$. Nine dot products cost about a minute, and a single sign error in the reduction shows up here as a wrong entry, never silently.

c) Multiply $A\mathbf{x} = \mathbf{b}$ on the LEFT by $A^{-1}$: $\mathbf{x} = A^{-1}\mathbf{b}$. For $\mathbf{b} = (3, 5, 4)$: $\mathbf{x} = (-3 + 10 - 8,\ -3 + 0 + 4,\ 3 - 5 + 4) = (-1, 1, 2)$, and indeed $A\mathbf{x} = (-1 + 4,\ -2 + 1 + 6,\ -1 + 1 + 4) = (3, 5, 4)$. For $\mathbf{b} = (1, 1, 1)$: $\mathbf{x} = (-1 + 2 - 2,\ -1 + 0 + 1,\ 1 - 1 + 1) = (-1, 0, 1)$. The inverse pays off when the same $A$ meets several right-hand sides; for a single system, reducing $[A \mid \mathbf{b}]$ is shorter. Note also that $A^{-1}\mathbf{b}$ and never $\mathbf{b}A^{-1}$: a column vector can only be multiplied by $A^{-1}$ on the left, the other product does not even exist.

d) Each operation of part a) is the multiplication on the left by an elementary matrix $E_i$, the matrix obtained by doing that operation to $I$. Doing all five to the block gives $E_5 E_4 E_3 E_2 E_1 [A \mid I] = [E_5 \cdots E_1 A \mid E_5 \cdots E_1]$, because multiplying a block matrix on the left multiplies each block. The left half is $I$, so $E_5 \cdots E_1 A = I$, which says that $E_5 \cdots E_1$ is the inverse of $A$ (for square matrices, one side suffices). The right half is exactly that product. The identity on the right was only a blank page on which the operations wrote their own product.

e) A column operation is a multiplication on the RIGHT: adding column $1$ to column $3$ replaces the left block by $(\text{current block})F$ for an elementary matrix $F$, while the right half is untouched. Suppose the student then finishes with row operations only. The left block reaches $I$, but what it holds is $E A F = I$, where $E$ is the product of the row operations, and the right half holds $E$ alone. From $EAF = I$ we get $A^{-1} = FE$, not $E$: the right half is wrong by a factor $F$ sitting on the LEFT of it, and the answer copied from the page is not the inverse. The algorithm is a statement about left multiplication only. Mixing sides is the thread of this whole chapter: an inverse undoes a matrix from one side, and a computation that switches sides halfway computes something else.

a) From $\left(\begin{array}{ccc|ccc} 1 & 2 & -1 & 1 & 0 & 0 \\ 2 & 5 & 1 & 0 & 1 & 0 \\ 3 & 7 & 0 & 0 & 0 & 1 \end{array}\right)$: $R_2 \to R_2 - 2R_1$ gives $(0, 1, 3 \mid -2, 1, 0)$ and $R_3 \to R_3 - 3R_1$ gives $(0, 1, 3 \mid -3, 0, 1)$. Then $R_3 \to R_3 - R_2$ gives $(0, 0, 0 \mid -1, -1, 1)$. The left block has a zero row and no further operation can create a pivot there, since every other row already has its pivot in an earlier column. The algorithm stops: $A$ is not invertible. Continuing to clear upwards would earn nothing, and dividing by a zero pivot is the one move that is never allowed.

b) The right half records which combination of the ORIGINAL rows produced the zero row: $-R_1 - R_2 + R_3 = \mathbf{0}$, that is $R_3 = R_1 + R_2$. Check: $(1, 2, -1) + (2, 5, 1) = (3, 7, 0)$. No two rows are proportional, yet the three rows are dependent: dependence is a relation among ALL the rows, and scanning pairs for multiples misses it. This is the use of the right half that students forget: even when the algorithm fails, it hands you the reason.

c) Let $E = E_3E_2E_1$ be the product of the three operations, so $EA$ has a zero third row. If some $B$ satisfied $AB = I$, then $(EA)B = E$. The third row of $(EA)B$ is (zero row)$\cdot B = \mathbf{0}$, while $E$ is invertible, a product of invertible elementary matrices, so it cannot have a zero row. Contradiction: no such $B$ exists. The argument is general: once a zero row appears on the left, no later operation can turn the left block into $I$.

d) Apply the same three operations to $[A \mid \mathbf{b}]$: the last row becomes $(0, 0, 0 \mid -b_1 - b_2 + b_3)$. The system is consistent exactly when $b_3 = b_1 + b_2$. So $\mathbf{b} = (1, 1, 1)$ gives $0 = -1$, no solution, while $\mathbf{b} = (1, 1, 2)$ gives $0 = 0$ and has solutions (infinitely many, one free variable). An invertible matrix would solve $A\mathbf{x} = \mathbf{b}$ for every $\mathbf{b}$, by $\mathbf{x} = A^{-1}\mathbf{b}$, so the first vector is already a proof that $A^{-1}$ does not exist.

e) Continue on the left block only: from $x_2 + 3x_3 = 0$ and $x_1 + 2x_2 - x_3 = 0$, take $x_3 = 1$, then $x_2 = -3$, $x_1 = 7$. So $\mathbf{x} = (7, -3, 1)$, and $A\mathbf{x} = (7 - 6 - 1,\ 14 - 15 + 1,\ 21 - 21 + 0) = \mathbf{0}$. If $A^{-1}$ existed, multiplying $A\mathbf{x} = \mathbf{0}$ on the left by $A^{-1}$ would give $\mathbf{x} = A^{-1}\mathbf{0} = \mathbf{0}$, which is false. A matrix that sends a non-zero vector to zero has lost information, and nothing can undo a loss.

a) $R_2 \to R_2 - R_1$ gives $(0, 1, 2 \mid -1, 1, 0)$ and $R_3 \to R_3 - R_1$ gives $(0, 2, k - 1 \mid -1, 0, 1)$. Then $R_3 \to R_3 - 2R_2$ gives $(0, 0, k - 5 \mid 1, -2, 1)$. The letter only ever sat in the last column, so it only reaches the last pivot, which is $k - 5$. The right half of that row, $(1, -2, 1)$, does not depend on $k$ at all: it records the operations, and the operations were the same for every $k$.

b) If $k \neq 5$, the pivot $k - 5$ is non-zero, $R_3 \to \frac{1}{k-5}R_3$ is a legal operation, and clearing upwards turns the left block into $I$: $A_k$ is invertible. If $k = 5$, the third row of the left block is zero and the algorithm stops, exactly as in the previous exercise: $A_5$ is not invertible. Both directions are needed, and the step that is lost most often is the sentence before the division: writing $R_3 \to \frac{1}{k-5}R_3$ without first setting $k = 5$ aside divides by zero for one value and costs the case analysis, usually half the question.

c) With $k = 6$ the third row is already $(0, 0, 1 \mid 1, -2, 1)$. $R_2 \to R_2 - 2R_3$ gives $(0, 1, 0 \mid -3, 5, -2)$. $R_1 \to R_1 - R_3$ gives $(1, 1, 0 \mid 0, 2, -1)$, then $R_1 \to R_1 - R_2$ gives $(1, 0, 0 \mid 3, -3, 1)$. So $A_6^{-1} = \begin{pmatrix} 3 & -3 & 1 \\ -3 & 5 & -2 \\ 1 & -2 & 1 \end{pmatrix}$. Check the third row of $A_6A_6^{-1}$: $(1, 3, 6)$ against the three columns gives $3 - 9 + 6 = 0$, $-3 + 15 - 12 = 0$, $1 - 6 + 6 = 1$, the third row of $I$. The inverse happens to be symmetric, like $A_6$ itself: the inverse of a symmetric matrix is always symmetric, because $(A^{-1})^T = (A^T)^{-1}$.

d) For $k \neq 5$, the operation $R_3 \to \frac{1}{k-5}R_3$ turns the third row into $\left(0, 0, 1 \mid \frac{1}{k-5}, \frac{-2}{k-5}, \frac{1}{k-5}\right)$, and the later operations only change rows $1$ and $2$. So the third row of $A_k^{-1}$ is $\frac{1}{k-5}(1, -2, 1)$. For $k = 6$ this is $(1, -2, 1)$, the third row found in c). The formula also shows what happens near the excluded value: for $k = 5.01$ the entries are $100, -200, 100$. The inverse does not fade out gracefully as $k \to 5$, it blows up.

e) At $k = 5$, $A_5\mathbf{x}$ for $\mathbf{x} = (1, -2, 1)$ is $(1 - 2 + 1,\ 1 - 4 + 3,\ 1 - 6 + 5) = (0, 0, 0)$. A non-zero vector sent to zero means $A_5$ cannot be undone, which confirms b) from the other side. For this symmetric matrix the same triple $(1, -2, 1)$ is also the relation between the rows recorded in a): $R_1 - 2R_2 + R_3 = \mathbf{0}$ when $k = 5$.

a) Both have $ad - bc = 1$: $A^{-1} = \begin{pmatrix} 1 & -1 \\ -1 & 2 \end{pmatrix}$ and $B^{-1} = \begin{pmatrix} 1 & -2 \\ 0 & 1 \end{pmatrix}$. Undoing $AB$ means undoing $A$ LAST, since it acted last: $(AB)^{-1} = B^{-1}A^{-1} = \begin{pmatrix} 1 & -2 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & -1 \\ -1 & 2 \end{pmatrix} = \begin{pmatrix} 3 & -5 \\ -1 & 2 \end{pmatrix}$. The other order gives $A^{-1}B^{-1} = \begin{pmatrix} 1 & -3 \\ -1 & 4 \end{pmatrix}$, a different matrix. Confirmation: $AB = \begin{pmatrix} 2 & 5 \\ 1 & 3 \end{pmatrix}$, with $ad - bc = 6 - 5 = 1$, so the formula gives $\begin{pmatrix} 3 & -5 \\ -1 & 2 \end{pmatrix}$, the first answer. The rule saves real work on larger matrices: two known inverses and one product replace a whole new reduction.

b) Candidate: $(A^{-1})^T$. Using $(XY)^T = Y^TX^T$, $A^T(A^{-1})^T = (A^{-1}A)^T = I^T = I$ and $(A^{-1})^TA^T = (AA^{-1})^T = I^T = I$. Both products are $I$, so $A^T$ is invertible and, by uniqueness of the inverse, $(A^T)^{-1} = (A^{-1})^T$. Note the pattern of the proof, which is the only one this chapter uses: name a candidate, multiply on both sides, conclude by uniqueness. Transposing also reverses order, which is why the two reversals cancel into a clean statement.

c) $(3A)\left(\frac{1}{3}A^{-1}\right) = \frac{3}{3}AA^{-1} = I$, and the same on the other side, so $(3A)^{-1} = \frac{1}{3}A^{-1} = \begin{pmatrix} \frac{1}{3} & -\frac{1}{3} \\ -\frac{1}{3} & \frac{2}{3} \end{pmatrix}$. The classic slip is $3A^{-1}$: the scalar must be inverted too. For the square, $A^2(A^{-1})^2 = A(AA^{-1})A^{-1} = AA^{-1} = I$, so $(A^2)^{-1} = (A^{-1})^2 = \begin{pmatrix} 1 & -1 \\ -1 & 2 \end{pmatrix}^2 = \begin{pmatrix} 2 & -3 \\ -3 & 5 \end{pmatrix}$. Check: $A^2 = \begin{pmatrix} 5 & 3 \\ 3 & 2 \end{pmatrix}$, $ad - bc = 10 - 9 = 1$, and the formula returns $\begin{pmatrix} 2 & -3 \\ -3 & 5 \end{pmatrix}$.

d) $A + B = \begin{pmatrix} 3 & 3 \\ 1 & 2 \end{pmatrix}$, with $ad - bc = 6 - 3 = 3$, so $(A + B)^{-1} = \frac{1}{3}\begin{pmatrix} 2 & -3 \\ -1 & 3 \end{pmatrix}$. But $A^{-1} + B^{-1} = \begin{pmatrix} 2 & -3 \\ -1 & 3 \end{pmatrix}$, three times too large: $(A + B)(A^{-1} + B^{-1}) = 3I$, not $I$. Worse, the sum of two invertible matrices need not be invertible at all: $I$ and $-I$ are both invertible and $I + (-I) = 0$. There is no formula for $(A + B)^{-1}$; the only method is to add first, then invert the sum.

e) Invert both sides: $C^T - 2I = \begin{pmatrix} 2 & 1 \\ -1 & 0 \end{pmatrix}^{-1}$. Here $ad - bc = 0 + 1 = 1$, so $C^T - 2I = \begin{pmatrix} 0 & -1 \\ 1 & 2 \end{pmatrix}$, hence $C^T = \begin{pmatrix} 2 & -1 \\ 1 & 4 \end{pmatrix}$ and $C = \begin{pmatrix} 2 & 1 \\ -1 & 4 \end{pmatrix}$. Check: $C^T - 2I = \begin{pmatrix} 0 & -1 \\ 1 & 2 \end{pmatrix}$ and $\begin{pmatrix} 0 & -1 \\ 1 & 2 \end{pmatrix}\begin{pmatrix} 2 & 1 \\ -1 & 0 \end{pmatrix} = I$. The order of the peeling matters: the inverse is the outermost operation, so it goes first, then the $2I$, then the transpose.