Concordia MATH 204 Midterm Practice Mock

MATH 204 assessment practice solutions

How to use these solutions

Try the question paper before reading the solutions. Use the suggested marking guide out of 100 to check your work. Give the same points for any equivalent valid method. If one arithmetic error carries through later steps, still give credit for correct reasoning that follows consistently from it. Check any invalid division or reasoning step separately.

The eight short diagnostic retries appear after all solutions. The English and French versions contain identical mathematical tasks, so switching languages does not provide a fresh attempt.

Solution 1

The fourth left-hand side is the sum of the first three:

Let $b_i$ be the right-hand side of claim $i$, for $i=1,2,3,4$. For the four equations to agree, these numbers must satisfy $b_4=b_1+b_2+b_3$. The recorded numbers give $1+2+7=10\ne13$, so the four recorded equations are inconsistent.

If claim 4 is wrong, the first three are exact. From $x+y=1$ and $y+z=2$, obtain $y=1-x$ and $z=1+x$. Then

giving $(x,y,z)=(2,-1,3)$. Claim 4 must have right-hand side 10. The checks are

However, the faulty claim cannot be identified from the record alone. For example, suppose claim 1 is wrong instead. Compatibility requires

The triple $(3,1,1)$ gives $x+y=4$, $y+z=2$, $2x+z=7$ and $3x+2y+2z=13$. This changes exactly one recorded right-hand side and preserves the other three. These two distinct explanations are sufficient to disprove unique identification. There is no need to enumerate all possible repairs.

Suggested marking guide, 14 marks:

• Part 1: identify the left-hand-side relation (2); detect the incompatible constants (1)
• Part 2: solve the three retained equations coherently (3); correct claim 4 to 10 (1); verify (1)
• Part 3: construct a different valid correction (2); find a compatible triple (2); check the three unchanged claims (1); conclude that the data do not identify the error uniquely (1)

Diagnostic D1: A failed consistency check does not identify which input is wrong. If you named one faulty claim without ruling out alternatives, distinguish “one possible explanation” from “the only explanation.”

Solution 2

Since $A$ is $2\times2$ and $B$ is $3\times3$, $X$ must be $2\times3$. The product then has the same size as $C$.

Both outside factors are invertible:

Multiplying $AXB=C$ on the left by $A^{-1}$ and on the right by $B^{-1}$ gives

Each inverse must be adjacent to the factor it cancels; matrix factors cannot be rearranged as if they were scalars. Here

Check directly:

Every solution must equal the displayed expression $A^{-1}CB^{-1}$, and that expression satisfies the equation. The solution is unique. Solving the associated scalar equations is also valid, provided dimensions, factor order and uniqueness are justified.

Suggested marking guide, 12 marks:

• Part 1: dimensions $2\times3$, with justification (2)
• Part 2: valid algebra or equivalent equations, including order (3); correct inverses or equivalent elimination (2); correct $X$ (2)
• Part 3: original-equation check (2); uniqueness justification (1)

Diagnostic D2: Undo a left factor on the left and a right factor on the right. Check dimensions before trying to move an inverse. The expression $B^{-1}CA^{-1}$ is not even defined here.

Solution 3

Undo the operations in reverse order. Starting with $U$, negate row 3, then add twice the current row 1 to row 2:

Finally interchange rows 1 and 3:

The elementary matrices are

Set $E=E_3E_2E_1$. The original log says $EA=U$, not $EA=I_3$, and

All three elementary matrices are invertible, and $U$ is triangular with determinant 1. Thus $A=E^{-1}U$ is invertible. Alternatively, the two determinant sign changes in the log cancel, so $\det(A)=\det(U)=1$.

Solution 3 continued

Because

the inverse is

Direct multiplication gives $AA^{-1}=A^{-1}A=I_3$. A full row reduction of $[A\mid I_3]$ is an equally valid way to find this inverse.

The learner’s claim is false because $U\ne I_3$. Correct relationships include

Suggested marking guide, 14 marks:

• Part 1: reverse order and operations correctly (2); obtain $A$ (2)
• Part 2: justify invertibility (1); valid inverse method (2); correct inverse (2); check (1)
• Part 3: correctly interpret or compute the elementary product (2); reject the claim and give the corrected inverse relationship (2)

Diagnostic D3: An operation sequence gives an inverse directly only when it sends the original matrix to the identity. Also, the operation done first appears rightmost in a product of left-acting elementary matrices.

Solution 4

The row combinations can be written as $B=TA$, where

Since $T$ is triangular, $\det(T)=2\cdot1\cdot4=8$. Therefore

This is nonzero, so $B^{-1}$ exists. Using the determinant rules for a scalar multiple of a $3\times3$ matrix, a product, an inverse and a transpose gives

A row-operation derivation is also valid. For example, begin with rows $r_1,r_2,r_3$, scale row 1 by 2 and row 3 by 4, add row 2 to row 1, and then subtract $3/4$ of the current row 3 from row 2. This produces $B$ with determinant multiplier 8. The order matters when current rows are used.

Suggested marking guide, 10 marks:

• Valid row-combination or determinant argument (3); $\det(B)=-16$ (2)
• Establish that the inverse exists (1); apply the four determinant rules with the factor $3^3$ (3); obtain $27/8$ (1)

Diagnostic D4: Multiplying the whole $3\times3$ matrix by 3 contributes $3^3$. Multiplying one row by 3 would contribute only 3. Transposing does not change a determinant.

Solution 5

The coefficient matrix is

Expansion along the first row gives

It is invertible for every $t$; only the right-hand side depends on $t$. Every value of $t$ therefore gives exactly one solution.

For part 1, imposing $x=z$ in $2x+z=3$ gives $x=z=1$. Then $y+z=1$ gives $y=0$, and $x+2y=t$ forces $t=1$. Conversely, $t=1$ gives the solution $(1,0,1)$, so this value works and is the only one.

Another valid choice is to use Cramer’s rule for the two needed coordinates. The replacement determinants are

Then $x=z$ is equivalent to $(t+4)/5=(7-2t)/5$, which again gives $t=1$. The direct method is shorter here; the determinant method is still valid.

For part 2, $\det(H_x)=0$ precisely when $t=-4$. This means $x=0/5=0$, not that $H$ is singular. The relevant denominator remains $\det(H)=5$. At $t=-4$ the unique solution is $(0,-2,3)$, and its original left-hand sides are $-4,1,3$.

Suggested marking guide, 12 marks:

• Part 1: valid deduction of $x=z=1$ and $y=0$, or equivalent method (3); $t=1$ and a converse check (1); justify existence and uniqueness (2)
• Part 2: correct replacement determinant (2); $t=-4$ (1); distinguish it from the coefficient determinant and reject the singularity claim (2); conclude that the system still has a unique solution (1)

Diagnostic D5: Cramer’s rule applies only when the coefficient determinant is nonzero. A zero replacement determinant is permitted when the coefficient determinant is nonzero.

Solution 6

One example is

it is nonzero, and direct multiplication gives $PQ=0$.

For the complete description, write

Then

This is zero if and only if $c=-a$ and $d=-b$. Hence every allowed matrix, with no omitted cases, is

Every such matrix has determinant $-ab+ab=0$, so none is invertible. Equivalently, if $Q^{-1}$ existed, multiplying $PQ=0$ on the right by $Q^{-1}$ would imply $P=0$, contrary to the given matrix.

The proposed cancellation is invalid. A nonzero matrix need not be invertible; here $\det(P)=0$. The displayed nonzero example also directly disproves the conclusion $Q=0$.

Suggested marking guide, 10 marks:

• Part 1: nonzero example and product check (2)
• Part 2: obtain the two necessary equations (2); give the full family and its sufficiency (2)
• Part 3: justify that all such $Q$ are singular (2); explain why $P\ne0$ does not permit cancellation (2)

Diagnostic D6: Scalar zero-product and cancellation rules cannot be transferred to matrices without checking invertibility.

Solution 7

The data give $u\cdot u=4$, $v\cdot v=9$ and $u\cdot v=v\cdot u=3$. Expand each dot product by distributing over sums and taking numerical factors outside. This reduces every calculation to these three numbers.

First,

Thus $\|2u-v\|=\sqrt{13}$, a nonnegative scalar.

Next,

Both vectors are nonzero, so the angle formula is defined:

The cosine is negative, so the angle is obtuse.

The displacement between the two specified points is

Their distance is $\|2v\|=2\|v\|=6$.

Finally,

The coefficient 4 is nonzero, so this is the only value. Coordinates were unnecessary because every calculation depends only on the given norms and dot product.

Suggested marking guide, 14 marks:

• Part 1: correct expansion including both cross terms (3); positive square root (1)
• Part 2: numerator (1); valid nonzero denominator (1); cosine (1); obtuse interpretation (1)
• Part 3: displacement (2); distance (1)
• Part 4: orthogonality equation (2); unique parameter (1)

Diagnostic D7: A norm is not linear. Expand its square through the dot product; do not replace $\|2u-v\|$ by $2\|u\|-\|v\|$.

Solution 8

Write $P=(x,y,z)$. Equal distances to $A$ and $B$ give equal squared distances:

After cancellation, $-4x=-4y$, so $x=y$.

Orthogonality gives $x+y+z=0$, hence $z=-2x$. The norm condition now becomes

Therefore $x=1$ or $x=-1$, and the only candidates are

Every original constraint was transformed equivalently: distances are nonnegative, so equality is equivalent to equality of their squares; the orthogonality and norm substitutions did not divide by an unknown. Both square-root signs have been retained. Substitution confirms norm squared 6 and dot product 0 for each candidate, and $x=y$ confirms equal distances to $A$ and $B$. The list is complete.

The distances to $A$ are

Thus $P_+$ is nearer to $A$. Equal distance to $A$ and $B$ for each point does not mean that different possible points must be the same distance from $A$.

Suggested marking guide, 14 marks:

• Equal-distance equation and $x=y$ (2); orthogonality and $z=-2x$ (2); norm equation (2); both candidates and completeness (2)
• Two correct distance calculations (4)
• Correct nearer point with a comparison or check (2)

Diagnostic D8: A norm condition can leave two signs. Do not discard a negative coordinate just because a norm is nonnegative. Distances use displacements from the named point, not automatically from the origin.

Total marks

Use your mistakes to choose what to practise next. The suggested marks help you check your work.

Diagnose and repair

Record the first step you could not justify, rather than simply labelling a whole question “wrong.” Choose one or two categories first.

• D1, Q1: revisit whole-row operations and consistency; distinguish inconsistency from finding which recorded value is wrong
• D2, Q2: revisit matrix multiplication and inverses and guided question 1
• D3, Q3: revisit guided question 4; track the effect of the complete operation sequence
• D4, Q4: revisit guided question 3; keep a determinant multiplier beside every change
• D5, Q5: revisit guided question 5; identify the original coefficient determinant before the replacement determinant
• D6, Q6: revisit matrix multiplication and invertibility; test the proposed cancellation with an explicit product
• D7, Q7: revisit dot products and projection and guided question 7; use only the dot-product and norm material needed here
• D8, Q8: revisit guided question 7; write one equation for each condition and check every candidate in all of them

The later span/independence and plane/cross-product material in broader course resources is not required for this assessment.

Short fresh retries

Close the answer key below before attempting the items for your selected categories. These are repair checks, not another scored assessment.

1. D1. The equations $x-y=1$ and $2x+y=8$ are exact. A third record says $3x=k$. Find the only compatible $k$ and verify it with the solution.
2. D2. Let $A=\begin{pmatrix}1&1\\0&1\end{pmatrix}$ and $B=\begin{pmatrix}2&0\\0&1\end{pmatrix}$. Solve $AXB=I_2$. Check whether reversing the two inverse factors gives the same answer.
3. D3. Let $E_1=\begin{pmatrix}0&1\\1&0\end{pmatrix}$, $E_2=\begin{pmatrix}3&0\\0&1\end{pmatrix}$, and $E_2E_1K=I_2$. Find $K^{-1}$ and $K$, keeping the factor order visible.
4. D4. A $2\times2$ matrix $A$ has determinant 3. Find $\det(2(A^{-1})^{\mathsf T})$.
5. D5. For $x+y=t$, $x-y=2$, find the value of $t$ for which the Cramer numerator for $y$ is zero. Is the coefficient matrix singular at that value?
6. D6. For $P=\begin{pmatrix}1&0\\0&0\end{pmatrix}$, give one nonzero $Q$ with $PQ=0$. Explain why no such $Q$ can be invertible.
7. D7. If $\|u\|=\|v\|=1$ and $\|u-v\|=1$, find $u\cdot v$ and the angle between the vectors.
8. D8. Find all points in $\mathbb R^2$ that are equally distant from $(2,0)$ and $(0,2)$ and have distance 2 from the origin. Which is nearer to $(1,1)$?