MATH 2010 Exam 3 Review Problems

We are given the system:

Substitute $x_1 = 1$ into the second equation:

Case Analysis:

(i) Infinitely many solutions:
This occurs when both coefficients and constants are zero:

Solving $a^2+a-2=0$: $(a+2)(a-1)=0 \implies a = -2, 1$
Solving $a^2-a-6=0$: $(a-3)(a+2)=0 \implies a = -2, 3$
The common value is $a = -2$.

When $a = -2$, the equation becomes $0 \cdot x_2 = 0$, which is always true.
Solution: $x_1 = 1$, $x_2$ is free.
Vector form: $\vec{x} = \begin{bmatrix} 1 \\ 0 \end{bmatrix} + x_2 \begin{bmatrix} 0 \\ 1 \end{bmatrix}$

(ii) No solution: This occurs when the coefficient is zero but the constant is non-zero.
From above, $a^2+a-2=0$ gives $a = -2, 1$

• For $a = -2$: $0 \cdot x_2 = 0$ (infinitely many solutions, not no solution)
• For $a = 1$: $0 \cdot x_2 = 1^2-1-6 = -6 \neq 0$ (no solution)

Answer: $a = 1$

(iii) Unique solution with $x_2 = 0$:
For unique solution, we need $a^2+a-2 \neq 0$, so $a \neq -2, 1$.
For $x_2 = 0$, we need $a^2-a-6 = 0$, so $a = -2, 3$.
Since $a \neq -2$, we must have $a = 3$.

When $a = 3$: $x_2 = 0$ and $x_1 = 1$.
Vector form: $\vec{x} = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$

Final Answer

• (i) Infinitely many solutions: $a = -2$, $\vec{x} = \begin{bmatrix} 1 \\ 0 \end{bmatrix} + x_2 \begin{bmatrix} 0 \\ 1 \end{bmatrix}$
• (ii) No solution: $a = 1$
• (iii) Unique solution with $x_2=0$: $a = 3$, $\vec{x} = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$

We form the augmented matrix and perform row reduction:

Step 0: Write the augmented matrix (already done above)

Step 1: Begin with the leftmost nonzero column. Make the leading entry 1 by $R_1 \leftarrow \frac{1}{2}R_1$:

Step 2: Use elimination to create zeros below the leading 1.
$R_2 \leftarrow R_2 - 4R_1$, $R_3 \leftarrow R_3 - 7R_1$:

Step 3: Repeat steps 1 and 2 (ignoring Row 1). Make the leading entry in Row 2 equal to 1 by $R_2 \leftarrow -\frac{1}{3}R_2$:

Step 4: Create zeros below the leading 1 in Row 2.
$R_3 \leftarrow R_3 + 6R_2$:

The matrix is now in echelon form. The last row is all zeros, so the system is consistent. Since Column 3 has no leading 1, $x_3$ is a free variable, and the system has infinitely many solutions.

Step 5: Create zeros above each leading 1 (working upward from right to left) to get Reduced Row Echelon Form (RREF).
$R_1 \leftarrow R_1 - 2R_2$:

Step 6: Write the system of equations corresponding to the RREF:

Solve for pivot variables:

Final Answer The system has infinitely many solutions. In vector form:

We form the augmented matrix and perform Gaussian elimination:

Step 1: Eliminate $x_1$ from rows 2 and 3

• $R_2 \leftarrow R_2 - 2R_1$: $[4, 5, 6, 3] - [4, 8, 12, 0] = [0, -3, -6, 3]$
• $R_3 \leftarrow R_3 - \frac{7}{2}R_1$: $[7, 8, 9, 0] - [7, 14, 21, 0] = [0, -6, -12, 0]$

Step 2: Simplify row 2

• $R_2 \leftarrow -\frac{1}{3}R_2$: $[0, 1, 2, -1]$

Step 3: Eliminate $x_2$ from row 3

• $R_3 \leftarrow R_3 + 6R_2$: $[0, -6, -12, 0] + [0, 6, 12, -6] = [0, 0, 0, -6]$

Analysis: The last row represents the equation:

which simplifies to $0 = -6$, a contradiction.

This indicates that the system is inconsistent and has no solution.

Verification: Let’s check if the equations are compatible by examining the relationships:

• From equations 1 and 2, we found $x_2 = -1 - 2x_3$ and $x_1 = 2 + x_3$ (as in Q2)
• Substituting into equation 3: $7(2+x_3) + 8(-1-2x_3) + 9x_3 = 14 + 7x_3 - 8 - 16x_3 + 9x_3 = 6 \neq 0$

This confirms the inconsistency.

Final Answer The system has no solutions (inconsistent).

(a) The matrix is:

This represents:

• $x_1 = 7$
• $x_2 = -2$
• $x_3 = 3$
• $0 = 0$ (consistent)

Answer: Consistent. Solution: $\vec{x} = \begin{bmatrix} 7 \\ -2 \\ 3 \end{bmatrix}$

(b) The matrix is:

This represents:

• $x_1 - 2x_2 + 7x_4 = 5$
• $x_3 - 4x_4 = 3$
• $0 = 0$ (consistent)

Free variables: $x_2, x_4$

From equation 2: $x_3 = 3 + 4x_4$ From equation 1: $x_1 = 5 + 2x_2 - 7x_4$

Answer: Consistent. Solution: $\vec{x} = \begin{bmatrix} 5 \\ 0 \\ 3 \\ 0 \end{bmatrix} + x_2 \begin{bmatrix} 2 \\ 1 \\ 0 \\ 0 \end{bmatrix} + x_4 \begin{bmatrix} -7 \\ 0 \\ 4 \\ 1 \end{bmatrix}$

(c) The matrix is:

The last row gives $0 = 1$, which is a contradiction.

Answer: Inconsistent.

Final Answer

• (a) Consistent: $\vec{x} = \begin{bmatrix} 7 \\ -2 \\ 3 \end{bmatrix}$
• (b) Consistent: $\vec{x} = \begin{bmatrix} 5 \\ 0 \\ 3 \\ 0 \end{bmatrix} + x_2 \begin{bmatrix} 2 \\ 1 \\ 0 \\ 0 \end{bmatrix} + x_4 \begin{bmatrix} -7 \\ 0 \\ 4 \\ 1 \end{bmatrix}$
• (c) Inconsistent

Part (a): Find consistency conditions

Form the augmented matrix:

Row reduction:
$R_2 \leftarrow R_2 + R_1$: $[0, 0, 1, -2 | b_1 + b_2]$
$R_3 \leftarrow R_3 - R_1$: $[0, 0, 3, -6 | b_3 - b_1]$

$R_3 \leftarrow R_3 - 3R_2$: $[0, 0, 0, 0 | b_3 - b_1 - 3(b_1 + b_2)] = [0, 0, 0, 0 | b_3 - 4b_1 - 3b_2]$

For consistency, we need:

Answer (a): The system is consistent if and only if $b_3 = 4b_1 + 3b_2$.

Part (b): Check specific cases

(I) $b_1 = -1, b_2 = 2, b_3 = 1$
Check: $4b_1 + 3b_2 = 4(-1) + 3(2) = -4 + 6 = 2 \neq 1 = b_3$
Answer: Inconsistent.

(II) $b_1 = 1, b_2 = 1, b_3 = 7$
Check: $4b_1 + 3b_2 = 4(1) + 3(1) = 7 = b_3$
Answer: Consistent.

Find the solution:
From row reduction:

• $x_3 - 2x_4 = b_1 + b_2 = 2 \implies x_3 = 2 + 2x_4$
• $x_1 + 5x_2 + 3x_4 = b_1 = 1 \implies x_1 = 1 - 5x_2 - 3x_4$

Free variables: $x_2, x_4$

Vector form: $\vec{x} = \begin{bmatrix} 1 \\ 0 \\ 2 \\ 0 \end{bmatrix} + x_2 \begin{bmatrix} -5 \\ 1 \\ 0 \\ 0 \end{bmatrix} + x_4 \begin{bmatrix} -3 \\ 0 \\ 2 \\ 1 \end{bmatrix}$

Final Answer

• (a) Consistency condition: $b_3 = 4b_1 + 3b_2$
• (b)(I) Inconsistent
• (b)(II) Consistent: $\vec{x} = \begin{bmatrix} 1 \\ 0 \\ 2 \\ 0 \end{bmatrix} + x_2 \begin{bmatrix} -5 \\ 1 \\ 0 \\ 0 \end{bmatrix} + x_4 \begin{bmatrix} -3 \\ 0 \\ 2 \\ 1 \end{bmatrix}$

(a) $CA$
$C$ is $3 \times 2$, $A$ is $2 \times 2$. Product is defined ($3 \times 2$).

(b) $AC$
$A$ is $2 \times 2$, $C$ is $3 \times 2$. Product is NOT defined (columns of A = 2, rows of C = 3).
Answer: Not defined.

(c) $(A-B)D$
$A-B = \begin{bmatrix} -3 & 1 \\ 2 & -1 \end{bmatrix} - \begin{bmatrix} 0 & 4 \\ -2 & 5 \end{bmatrix} = \begin{bmatrix} -3 & -3 \\ 4 & -6 \end{bmatrix}$
$(A-B)$ is $2 \times 2$, $D$ is $2 \times 3$. Product is defined ($2 \times 3$).

(d) $B(C^T + D)$
$C^T = \begin{bmatrix} 5 & -1 & 3 \\ 0 & 4 & 3 \end{bmatrix}$, $D = \begin{bmatrix} 1 & 0 & -3 \\ -2 & 5 & -1 \end{bmatrix}$
$C^T + D = \begin{bmatrix} 6 & -1 & 0 \\ -2 & 9 & 2 \end{bmatrix}$
$B$ is $2 \times 2$, $C^T + D$ is $2 \times 3$. Product is defined ($2 \times 3$).

(e) $CE$
$C$ is $3 \times 2$, $E$ is $3 \times 3$. Product is NOT defined (columns of C = 2, rows of E = 3).
Answer: Not defined.

(f) $C^T B$
$C^T$ is $2 \times 3$, $B$ is $2 \times 2$. Product is NOT defined (columns of C^T = 3, rows of B = 2).
Answer: Not defined.

Final Answer

• (a) $CA = \begin{bmatrix} -15 & 5 \\ 11 & -5 \\ -3 & 0 \end{bmatrix}$
• (b) Not defined (dimension mismatch)
• (c) $(A-B)D = \begin{bmatrix} 3 & -15 & 12 \\ 16 & -30 & -6 \end{bmatrix}$
• (d) $B(C^T + D) = \begin{bmatrix} -8 & 36 & 8 \\ -22 & 47 & 10 \end{bmatrix}$
• (e) Not defined (dimension mismatch)
• (f) Not defined (dimension mismatch)

Let $\vec{v}_1 = \begin{bmatrix} 1 \\ 0 \\ 2 \end{bmatrix}, \vec{v}_2 = \begin{bmatrix} 2 \\ -3 \\ 1 \end{bmatrix}, \vec{v}_3 = \begin{bmatrix} 1 \\ 3 \\ 5 \end{bmatrix}$

Step 1: Let $a_1\vec{v}_1 + a_2\vec{v}_2 + a_3\vec{v}_3 = \vec{0}$, is $a_1 = a_2 = a_3 = 0$ the only solution?

Step 2: Perform EROs to reduce to RREF:

Step 3: Analyze the RREF:

• Leading 1 columns: Column 1 and Column 2
• Free column: Column 3 $\Rightarrow$ $a_3$ is a free variable

Since there is a free variable, the system has infinitely many (nontrivial) solutions.

Step 4: To write one vector in terms of the others, choose a nonzero value for the free variable.

Let $a_3 = 1 \Rightarrow a_1 = -3(1) = -3, a_2 = 1$

Check: $3\vec{v}_1 - \vec{v}_2 = 3\begin{bmatrix} 1 \\ 0 \\ 2 \end{bmatrix} - \begin{bmatrix} 2 \\ -3 \\ 1 \end{bmatrix} = \begin{bmatrix} 3 \\ 0 \\ 6 \end{bmatrix} - \begin{bmatrix} 2 \\ -3 \\ 1 \end{bmatrix} = \begin{bmatrix} 1 \\ 3 \\ 5 \end{bmatrix} = \vec{v}_3$ ✓

Conclusion:

Since the system has nontrivial solution, $\{\vec{v}_1, \vec{v}_2, \vec{v}_3\}$ is linearly dependent.

One vector can be expressed as:

Step-by-Step Solution

(a) $\left\{ \begin{bmatrix} 0 \\ -2 \\ 8 \end{bmatrix}, \begin{bmatrix} 4 \\ 4 \\ 9 \end{bmatrix} \right\}$

Observation:

• This is a set of 2 vectors in $\mathbb{R}^3$.
• Check for proportionality: The first component of the first vector is $0$, while the second is $4$.
• There is no scalar $c$ such that $\begin{bmatrix} 0 \\ -2 \\ 8 \end{bmatrix} = c \begin{bmatrix} 4 \\ 4 \\ 9 \end{bmatrix}$.

Conclusion: ✅ Linearly Independent

Reason: The two vectors are not scalar multiples of each other.

(b) $\left\{ \begin{bmatrix} 3 \\ 2 \\ -4 \end{bmatrix}, \begin{bmatrix} -6 \\ 1 \\ 7 \end{bmatrix}, \begin{bmatrix} 6 \\ -5 \\ 2 \end{bmatrix}, \begin{bmatrix} 3 \\ 7 \\ -5 \end{bmatrix} \right\}$

Observation:

• This is a set of 4 vectors in $\mathbb{R}^3$.
• Number of vectors $p = 4$, dimension $m = 3$.
• Condition $p > m$ (4 > 3) is satisfied.

Conclusion: ❌ Linearly Dependent

Reason: In $\mathbb{R}^m$, any set with more than $m$ vectors must be linearly dependent (Pigeonhole Principle).

(c) $\left\{ \begin{bmatrix} 3 \\ 1 \\ 5 \end{bmatrix}, \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 7 \\ 8 \end{bmatrix} \right\}$

Observation:

• The set contains the zero vector $\vec{0} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$.

Conclusion: ❌ Linearly Dependent

Reason: Any set containing the zero vector is linearly dependent.

• Proof: $1 \cdot \vec{0} + 0 \cdot \vec{v}_1 + 0 \cdot \vec{v}_3 = \vec{0}$ is a nontrivial solution.

(d) $\left\{ \begin{bmatrix} 3 \\ 1 \\ -2 \end{bmatrix}, \begin{bmatrix} 2 \\ -1 \\ 5 \end{bmatrix}, \begin{bmatrix} 12 \\ 4 \\ -8 \end{bmatrix} \right\}$

Observation:

• Check the 1st and 3rd vectors: $$\begin{bmatrix} 12 \\ 4 \\ -8 \end{bmatrix} = 4 \times \begin{bmatrix} 3 \\ 1 \\ -2 \end{bmatrix}$$
• The 3rd vector is exactly 4 times the 1st vector.

Conclusion: ❌ Linearly Dependent

Reason: There is a scalar multiple relationship, $\vec{v}_3 = 4\vec{v}_1$.

Summary Table

Final Answer

• (a) Linearly Independent
• (b) Linearly Dependent (4 vectors in $\mathbb{R}^3$)
• (c) Linearly Dependent (contains zero vector)
• (d) Linearly Dependent ($\vec{v}_3 = 4\vec{v}_1$)

Part (a): Find values of $\lambda$ for which $A$ is nonsingular

A matrix is nonsingular if and only if its determinant is non-zero.

For $A$ to be nonsingular: $\det(A) \neq 0$

Answer (a): $A$ is nonsingular for all $\lambda \neq 4$ and $\lambda \neq -1$.

Part (b): Find $A^{-1}$

For a $2 \times 2$ matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$, the inverse is:

So:

Final Answer

• (a) $A$ is nonsingular for $\lambda \neq 4$ and $\lambda \neq -1$
• (b) $A^{-1} = \frac{1}{(\lambda-4)(\lambda+1)} \begin{bmatrix} \lambda-3 & -2 \\ -2 & \lambda \end{bmatrix}$

Part (a): Find $A^{-1}$ using Gaussian Elimination

Following the augmented matrix method, we construct $[A | I_3]$ and perform elementary row operations (EROs) to transform the left side into the identity matrix $I_3$. If successful, the right side will become $A^{-1}$.

Initial Augmented Matrix:

Step 1: Create zeros in the first column below the first pivot.
Apply $R_2 \leftarrow R_2 - R_1$ and $R_3 \leftarrow R_3 + R_1$:

Step 2: Create a zero in the second column below the second pivot.
Apply $R_3 \leftarrow R_3 - R_2$:

Step 3: Create zeros in the third column above the third pivot (back-substitution phase).
Apply $R_2 \leftarrow R_2 - R_3$ and $R_1 \leftarrow R_1 + 3R_3$: