MATH 133 practice midterm with full solutions (McGill)

The unknowns are $D$, $E$ and $F$, not $x$ and $y$. Once a point is fixed, $x^2 + y^2$ is a NUMBER, and it moves to the right-hand side: $Dx + Ey + F = -(x^2 + y^2)$, which is linear in $D$, $E$, $F$. For $P(5, 5)$: $5D + 5E + F = -50$. For $Q(-1, 5)$: $-D + 5E + F = -26$. For $R(6, -2)$: $6D - 2E + F = -40$. The sign is the first trap: $x^2 + y^2$ changes side, so the constants are negative.

The augmented matrix is $\left(\begin{array}{ccc|c} 5 & 5 & 1 & -50 \\ -1 & 5 & 1 & -26 \\ 6 & -2 & 1 & -40 \end{array}\right)$. A pivot of $5$ would bring fifths into every row, so start with $R_1 \leftrightarrow R_2$ and $R_1 \to -R_1$, which gives the row $(1, -5, -1 \mid 26)$. Then $R_2 \to R_2 - 5R_1$ gives $(0, 30, 6 \mid -180)$ and $R_3 \to R_3 - 6R_1$ gives $(0, 28, 7 \mid -196)$. Both rows have a common factor: $R_2 \to \frac{1}{6}R_2$ gives $(0, 5, 1 \mid -30)$ and $R_3 \to \frac{1}{7}R_3$ gives $(0, 4, 1 \mid -28)$. Now $R_2 \to R_2 - R_3$ gives $(0, 1, 0 \mid -2)$, and $R_3 \to R_3 - 4R_2$ gives $(0, 0, 1 \mid -20)$. Finally $R_1 \to R_1 + 5R_2 + R_3$ gives $(1, 0, 0 \mid 26 - 10 - 20) = (1, 0, 0 \mid -4)$. So $D = -4$, $E = -2$, $F = -20$.

Complete the squares in $x^2 + y^2 - 4x - 2y - 20 = 0$: $(x - 2)^2 - 4 + (y - 1)^2 - 1 = 20$, that is $(x - 2)^2 + (y - 1)^2 = 25$. Centre $(2, 1)$, radius $5$. For $(0, -3)$: $(0 - 2)^2 + (-3 - 1)^2 = 4 + 16 = 20 < 25$, so the point is inside the circle, at distance $\sqrt{20}$ from the centre. The corrected figure shows the circle through the three points.

The three points give $F = 0$, $D + E + F = -2$ and $2D + 2E + F = -8$. Order them as $\left(\begin{array}{ccc|c} 1 & 1 & 1 & -2 \\ 2 & 2 & 1 & -8 \\ 0 & 0 & 1 & 0 \end{array}\right)$. $R_2 \to R_2 - 2R_1$ gives $(0, 0, -1 \mid -4)$, then $R_3 \to R_3 + R_2$ gives $(0, 0, 0 \mid -4)$: the row $0 = -4$ makes the system inconsistent. Geometrically, the three points lie on the line $y = x$, and a line meets a circle in at most two points, so no circle can pass through all three. The algebra and the picture say the same thing, which is the check.

$R_2 \to R_2 - 2R_1$ gives $(0, 0, 1, -1)$, and $R_3 \to R_3 + R_1$ gives $(0, 0, 2, -2)$. Then $R_3 \to R_3 - 2R_2$ gives a zero row. The result $\begin{pmatrix} 1 & -1 & 0 & 2 \\ 0 & 0 & 1 & -1 \\ 0 & 0 & 0 & 0 \end{pmatrix}$ is already reduced: the pivots sit in columns $1$ and $3$, and the entry above the second pivot is already $0$. So $\text{rank}\,A = 2$, the leading variables are $x_1$ and $x_3$, and the free variables are $x_2 = s$ and $x_4 = t$. The rows read $x_3 = t$ and $x_1 = s - 2t$, so $x = s(1, 1, 0, 0) + t(-2, 0, 1, 1)$: the basic solutions are $(1, 1, 0, 0)$ and $(-2, 0, 1, 1)$. Check the second: $A(-2, 0, 1, 1) = (-2 + 2, -4 + 1 + 3, 2 + 2 - 4) = (0, 0, 0)$.

Carry $b$ through the same operations. After $R_2 \to R_2 - 2R_1$ and $R_3 \to R_3 + R_1$, the last column holds $b_1$, $b_2 - 2b_1$ and $b_3 + b_1$. After $R_3 \to R_3 - 2R_2$, the zero row carries $b_3 + b_1 - 2(b_2 - 2b_1) = 5b_1 - 2b_2 + b_3$. The system is consistent exactly when this number is $0$. For $(1, 3, 0)$: $5 - 6 + 0 = -1 \neq 0$, inconsistent. For $(1, 2, -1)$: $5 - 4 - 1 = 0$, consistent. In other words, the columns of $A$ span only the plane $5b_1 - 2b_2 + b_3 = 0$ of $\mathbb{R}^3$, and a right-hand side off that plane is not a combination of the columns. Every column must satisfy the condition, which is a free check: column $2$, $(-1, -2, 1)$, gives $-5 + 4 + 1 = 0$.

With $b = (1, 2, -1)$ the last column becomes $(1, 0, 0)$, so the reduced system is $x_1 - x_2 + 2x_4 = 1$ and $x_3 - x_4 = 0$. Setting $s = t = 0$ gives $p = (1, 0, 0, 0)$, and $x = (1, 0, 0, 0) + s(1, 1, 0, 0) + t(-2, 0, 1, 1)$. The particular solution could have been seen at once: $b$ is exactly the first column of $A$, so $Ae_1 = b$. The two basic solutions are the same as in a), because the homogeneous part does not depend on $b$.

$A$ is $3 \times 2$ and $B$ is $2 \times 3$. $AB$: inner sizes $2$ and $2$ agree, result $3 \times 3$. $BA$: inner sizes $3$ and $3$, result $2 \times 2$. $B^T$ is $3 \times 2$, the same size as $A$, so $A + B^T$ exists and is $3 \times 2$. $AB^T$ is $(3 \times 2)(3 \times 2)$: inner sizes $2$ and $3$ disagree, it does not exist. $A^TB^T$ is $(2 \times 3)(3 \times 2)$, a $2 \times 2$ matrix, and it is $(BA)^T$ by the reversal rule, not $(AB)^T$, which is $3 \times 3$.

Row $i$ of $B$ times column $j$ of $A$. Row $1$, $(1, 2, 0)$: with column $1$, $(2, 0, 1)$, it gives $2$; with column $2$, $(-1, 3, 1)$, it gives $-1 + 6 = 5$. Row $2$, $(-1, 1, 4)$: $-2 + 0 + 4 = 2$ and $1 + 3 + 4 = 8$. So $C = \begin{pmatrix} 2 & 5 \\ 2 & 8 \end{pmatrix}$.

The trace only uses the diagonal, so compute three entries and nothing else. $(AB)_{11}$ = row $1$ of $A$ times column $1$ of $B$ $= (2, -1) \cdot (1, -1) = 3$; $(AB)_{22} = (0, 3) \cdot (2, 1) = 3$; $(AB)_{33} = (1, 1) \cdot (0, 4) = 4$. So $\text{tr}(AB) = 10$, and $\text{tr}(C) = 2 + 8 = 10$ as well. This is the rule $\text{tr}(AB) = \text{tr}(BA)$, which holds even here where $AB$ and $BA$ do not have the same size; the matrices themselves are about as different as two matrices can be.

$C^2 = \begin{pmatrix} 2 & 5 \\ 2 & 8 \end{pmatrix}\begin{pmatrix} 2 & 5 \\ 2 & 8 \end{pmatrix} = \begin{pmatrix} 14 & 50 \\ 20 & 74 \end{pmatrix} = \begin{pmatrix} 14 & 50 \\ 20 & 74 \end{pmatrix}$, and $10C - 6I = \begin{pmatrix} 14 & 50 \\ 20 & 74 \end{pmatrix}$, the same matrix. The two numbers are not a coincidence: $10$ is the trace of $C$ and $6 = 16 - 10$ is $\det C$, and every $2 \times 2$ matrix satisfies $C^2 = (\text{tr}\,C)\,C - (\det C)\,I$.

TRUE. Row $i$ of $A$ times the vector $u = (1, 1, 1)$ is the sum of row $i$, which is $0$, so $Au = 0$ with $u \neq 0$. A matrix with a nontrivial null vector is not invertible, so $\det A = 0$. Second proof, with the determinant rules: the column operation $C_1 \to C_1 + C_2 + C_3$ does not change the determinant and produces a zero column. The size $3$ plays no role: the same holds for every $n$.

TRUE. $\det(A^TA) = \det(A^T)\det A = (\det A)^2$, which is non-zero when $\det A \neq 0$. One can even name the inverse: $(A^TA)^{-1} = A^{-1}(A^T)^{-1} = A^{-1}(A^{-1})^T$, by the reversal rule. The hypothesis matters for rectangular matrices: for a $3 \times 2$ matrix $A$, the $3 \times 3$ matrix $AA^T$ is never invertible, since its rank is at most $2$.

FALSE. The operation $R_2 \to 5R_2$ applied to $I_2$ gives $E = \begin{pmatrix} 1 & 0 \\ 0 & 5 \end{pmatrix}$, an elementary matrix with $\det E = 5$. Correct statement: a swap has determinant $-1$, a scaling by $c \neq 0$ has determinant $c$, a replacement $R_i \to R_i + cR_j$ has determinant $1$. What IS true of every elementary matrix is that its determinant is non-zero, which is why every elementary matrix is invertible.

TRUE. Write the system $Ax = b$. If $\det A \neq 0$, then $A$ is invertible and $x = A^{-1}b$ is the ONLY solution. Two different solutions therefore force $\det A = 0$. In fact the system then has infinitely many solutions: if $p \neq q$ both work, so does $p + t(q - p)$ for every $t$.

FALSE. Take $A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$. Each cofactor is plus or minus a $2 \times 2$ minor, and every $2 \times 2$ submatrix of $A$ contains a zero row or a zero column, so all nine cofactors vanish and $\text{adj}(A) = 0$, while $A \neq 0$. Correct statement: for $n \ge 3$, $\text{adj}(A) = 0$ happens exactly when every $(n-1) \times (n-1)$ minor is zero, which a non-zero matrix of small rank can do. For $2 \times 2$ matrices the statement is true, since $\text{adj}\begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$ contains every entry of $A$ up to sign.

The augmented matrix is $\left(\begin{array}{ccc|c} 1 & 1 & -1 & 1 \\ 2 & 3 & a & 3 \\ 1 & a & 3 & 2 \end{array}\right)$. $R_2 \to R_2 - 2R_1$ gives $(0, 1, a + 2 \mid 1)$, and $R_3 \to R_3 - R_1$ gives $(0, a - 1, 4 \mid 1)$. The second pivot is the $1$ of row $2$, a number, so the parameter in row $3$ is cleared with $R_3 \to R_3 - (a - 1)R_2$. That operation MULTIPLIES by $a - 1$, it never divides, so it is legal for every $a$, including $a = 1$, where it simply does nothing. It gives $(0, 0, 4 - (a - 1)(a + 2) \mid 1 - (a - 1))$.

If $a \neq 2$ and $a \neq -3$, the third pivot $-(a + 3)(a - 2)$ is non-zero: three pivots for three unknowns, exactly one solution. If $a = 2$, the last row is $(0, 0, 0 \mid 0)$: the system is consistent with rank $2$, so it has $3 - 2 = 1$ parameter and infinitely many solutions. If $a = -3$, the last row is $(0, 0, 0 \mid 5)$, the equation $0 = 5$: no solution. Every real $a$ lands in exactly one of the three cases, which is what the marker checks.

For $a = 2$ the system reduces to $x + y - z = 1$ and $y + 4z = 1$. With $z = t$ free: $y = 1 - 4t$ and $x = 1 - y + z = 1 - (1 - 4t) + t = 5t$. So $(x, y, z) = (5t, 1 - 4t, t) = (0, 1, 0) + t(5, -4, 1)$, a line of solutions. Check in the third equation with $a = 2$: $x + 2y + 3z = 5t + 2 - 8t + 3t = 2$ for every $t$.

For $a \neq 2, -3$, dividing by $-(a + 3)(a - 2)$ is now legal: $z = \frac{a - 2}{(a + 3)(a - 2)} = \frac{1}{a + 3}$. Back substitution: $y = 1 - (a + 2)z = \frac{(a + 3) - (a + 2)}{a + 3} = \frac{1}{a + 3}$, and $x = 1 - y + z = 1$. For $a = 0$: $(1, \frac{1}{3}, \frac{1}{3})$, and in the original system $1 + \frac{1}{3} - \frac{1}{3} = 1$, $2 + 1 + 0 = 3$, $1 + 0 + 1 = 2$. All three hold.

After dividing by $a - 2$ the last row becomes $-(a + 3)z = -1$, so the student announces $z = \frac{1}{a + 3}$ and the formulas of d) for every $a \neq -3$. At $a = 2$ that claims the single solution $(1, \frac{1}{5}, \frac{1}{5})$. It IS a solution: it is the point $t = \frac{1}{5}$ of the line of c). But it is not the only one, and the answer loses every other point of that line. The division was a division by zero exactly at $a = 2$, which is the one value where something interesting happens; on the exam this costs the whole infinitely many case.