Test #1 Practice Problems

Practice problems for Linear Algebra Test #1

Use this sheet to test your knowledge of the concepts covered in Test #1. These problems are drawn directly from the examples and exercises in your course materials.

Test #1 Topics

  1. Gauss-Jordan
  2. Parametric linear system (including systems with unknown constants kk)
  3. Solve linear system by inverting the coefficient matrix using the algorithm
  4. Write AA and A−1A^{-1} as product of elementary matrices
  5. Prove if XX is symmetric, skewed symmetric, or neither

1. Gauss-Jordan Method

Write the augmented matrix and solve the following linear systems using the Gauss-Jordan Method:

Problem 1.1

{x+y+2z=8−x−2y+3z=13x−7y+4z=10\begin{cases} x + y + 2z = 8 \\ -x - 2y + 3z = 1 \\ 3x - 7y + 4z = 10 \end{cases}

Solution: (x,y,z)=(3,1,2)(x, y, z) = (3, 1, 2)

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Problem 1.2

{2x+2y+2z=0−2x+5y+2z=18x+y+4z=−1\begin{cases} 2x + 2y + 2z = 0 \\ -2x + 5y + 2z = 1 \\ 8x + y + 4z = -1 \end{cases}

Solution: Let z=tz = t.

x=−17−37t,y=17−47t,z=tx = -\frac{1}{7} - \frac{3}{7}t, \quad y = \frac{1}{7} - \frac{4}{7}t, \quad z = t
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Problem 1.3

{x1−x2+2x3−x4=−12x1+x2−2x3−2x4=−2−x1+2x2−4x3+x4=13x1−3x4=−3\begin{cases} x_1 - x_2 + 2x_3 - x_4 = -1 \\ 2x_1 + x_2 - 2x_3 - 2x_4 = -2 \\ -x_1 + 2x_2 - 4x_3 + x_4 = 1 \\ 3x_1 - 3x_4 = -3 \end{cases}

Solution: Let x3=s, x4=tx_3 = s, \ x_4 = t.

x1=t−1,x2=2s,x3=s,x4=tx_1 = t - 1, \quad x_2 = 2s, \quad x_3 = s, \quad x_4 = t
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2. Parametric Linear Systems

Solve the following systems which require the use of parameters (free variables):

Problem 2.1

{x1+2x2+x3+3x4=42x1+3x2+3x3+4x4=7−x1+x2−4x3+3x4=−13x1+5x2+4x3+7x4=11\begin{cases} x_1 + 2x_2 + x_3 + 3x_4 = 4 \\ 2x_1 + 3x_2 + 3x_3 + 4x_4 = 7 \\ -x_1 + x_2 - 4x_3 + 3x_4 = -1 \\ 3x_1 + 5x_2 + 4x_3 + 7x_4 = 11 \end{cases}

Solution: Let x3=s, x4=tx_3 = s, \ x_4 = t.

x1=2−3s+t,x2=1+s−2t,x3=s,x4=tx_1 = 2 - 3s + t, \quad x_2 = 1 + s - 2t, \quad x_3 = s, \quad x_4 = t
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Problem 2.2

{x2−2x3−2x4−x5=7−3x1−2x2−x3−6x4=−5x1+x2+2x4−x5=2\begin{cases} x_2 - 2x_3 - 2x_4 - x_5 = 7 \\ -3x_1 - 2x_2 - x_3 - 6x_4 = -5 \\ x_1 + x_2 + 2x_4 - x_5 = 2 \end{cases}

Solution: Let x4=s, x5=tx_4 = s, \ x_5 = t.

x1=7−4t,x2=−5−2s+5t,x3=−6−2s+2t,x4=s,x5=tx_1 = 7 - 4t, \quad x_2 = -5 - 2s + 5t, \quad x_3 = -6 - 2s + 2t, \quad x_4 = s, \quad x_5 = t
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Problem 2.3: Systems with Unknown Constants (kk)

Determine the values of kk for which the system has (i) no solutions, (ii) exactly ONE solution, and (iii) infinitely many solutions.

Problem 2.3a:

{x+2y−3z=43x−y+5z=24x+y+(k2−14)z=k+2\begin{cases} x + 2y - 3z = 4 \\ 3x - y + 5z = 2 \\ 4x + y + (k^2 - 14)z = k + 2 \end{cases}

Solution using Row Reduction:

Starting with the augmented matrix:

[12−343−15241k2−14k+2]\left[\begin{array}{ccc|c} 1 & 2 & -3 & 4 \\ 3 & -1 & 5 & 2 \\ 4 & 1 & k^2-14 & k+2 \end{array}\right]

Step 1: R2←R2−3R1R_2 \leftarrow R_2 - 3R_1, R3←R3−4R1R_3 \leftarrow R_3 - 4R_1

[12−340−714−100−7k2−2k−14]\left[\begin{array}{ccc|c} 1 & 2 & -3 & 4 \\ 0 & -7 & 14 & -10 \\ 0 & -7 & k^2-2 & k-14 \end{array}\right]

Step 2: R3←R3−R2R_3 \leftarrow R_3 - R_2

[12−340−714−1000k2−16k−4]\left[\begin{array}{ccc|c} 1 & 2 & -3 & 4 \\ 0 & -7 & 14 & -10 \\ 0 & 0 & k^2-16 & k-4 \end{array}\right]

Analyzing the bottom row: (k2−16)z=k−4(k^2-16)z = k-4

Cases:

  • If k2−16≠0k^2 - 16 \neq 0 (i.e., k≠±4k \neq \pm 4): Pivot in zz exists ⇒\Rightarrow exactly one solution.
  • If k=4k = 4: Bottom row becomes 0=00 = 0 ⇒\Rightarrow one free variable ⇒\Rightarrow infinitely many solutions.
  • If k=−4k = -4: Bottom row becomes 0=−80 = -8 ⇒\Rightarrow contradiction ⇒\Rightarrow no solution.

Answers:

(i) No solution: k=−4\boxed{k = -4}

(ii) Exactly one solution: k∈R∖{−4,4}\boxed{k \in \mathbb{R} \setminus \{-4, 4\}}

(iii) Infinitely many solutions: k=4\boxed{k = 4}

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Problem 2.3b:

{x+y+3z=2−3x+(k−1)y−10z=−74x+(2−k)y+(k2−8)z=k−2\begin{cases} x + y + 3z = 2 \\ -3x + (k - 1)y - 10z = -7 \\ 4x + (2 - k)y + (k^2 - 8)z = k - 2 \end{cases}

Solution using Row Reduction:

Starting with the augmented matrix:

[1132−3k−1−10−742−kk2−8k−2]\left[\begin{array}{ccc|c} 1 & 1 & 3 & 2 \\ -3 & k-1 & -10 & -7 \\ 4 & 2-k & k^2-8 & k-2 \end{array}\right]

Step 1: R2←R2+3R1R_2 \leftarrow R_2 + 3R_1

R2:[0, (k−1)+3, −10+9, −7+6]=[0, k+2, −1, −1]R_2: [0, \ (k-1)+3, \ -10+9, \ -7+6] = [0, \ k+2, \ -1, \ -1]

Step 2: R3←R3−4R1R_3 \leftarrow R_3 - 4R_1

R3:[0, (2−k)−4, (k2−8)−12, (k−2)−8]=[0, −(k+2), k2−20, k−10]R_3: [0, \ (2-k)-4, \ (k^2-8)-12, \ (k-2)-8] = [0, \ -(k+2), \ k^2-20, \ k-10]

Result:

[11320k+2−1−10−(k+2)k2−20k−10]\left[\begin{array}{ccc|c} 1 & 1 & 3 & 2 \\ 0 & k+2 & -1 & -1 \\ 0 & -(k+2) & k^2-20 & k-10 \end{array}\right]

Step 3: R3←R3+R2R_3 \leftarrow R_3 + R_2

[11320k+2−1−100k2−21k−11]\left[\begin{array}{ccc|c} 1 & 1 & 3 & 2 \\ 0 & k+2 & -1 & -1 \\ 0 & 0 & k^2-21 & k-11 \end{array}\right]

Analyzing the bottom row: (k2−21)z=k−11(k^2-21)z = k-11

Cases:

  • If k2−21=0k^2 - 21 = 0 (i.e., k=±21k = \pm\sqrt{21}): Bottom row becomes 0=k−110 = k - 11. Since k≠11k \neq 11, this is a contradiction ⇒\Rightarrow no solution.
  • If k=−2k = -2: Second row becomes 0y−z=−1⇒z=10y - z = -1 \Rightarrow z = 1, but bottom row gives (−17)z=−13⇒z=1317(-17)z = -13 \Rightarrow z = \frac{13}{17} ⇒\Rightarrow contradiction ⇒\Rightarrow no solution.
  • Otherwise: All pivots exist ⇒\Rightarrow exactly one solution.
  • Infinite solutions would need k2−21=0k^2 - 21 = 0 AND k−11=0k - 11 = 0 together, which is impossible ⇒\Rightarrow none.

Answers:

(i) No solution: k∈{−2,21,−21}\boxed{k \in \{-2, \sqrt{21}, -\sqrt{21}\}}

(ii) Exactly one solution: k∈R∖{−2,21,−21}\boxed{k \in \mathbb{R} \setminus \{-2, \sqrt{21}, -\sqrt{21}\}}

(iii) Infinitely many solutions: None\boxed{\text{None}}

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3. Solving Linear Systems by Inverting the Coefficient Matrix

Solve the following systems by finding A−1A^{-1} using the inversion algorithm, then computing X=A−1BX = A^{-1}B:

Problem 3.1

{x+y+2z=8−x−2y+3z=13x−7y+4z=10\begin{cases} x + y + 2z = 8 \\ -x - 2y + 3z = 1 \\ 3x - 7y + 4z = 10 \end{cases}

Steps:

  1. Write the coefficient matrix AA
  2. Find A−1A^{-1} using the algorithm: [A∣I]→[I∣A−1]\left[A \mid \mathbb{I}\right] \rightarrow \left[\mathbb{I} \mid A^{-1}\right]
  3. Compute X=A−1BX = A^{-1}B where BB is the constants vector

Solution: (x,y,z)=(3,1,2)(x, y, z) = (3, 1, 2)

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Problem 3.2

{5x+y=33x+2y=7\begin{cases} 5x + y = 3 \\ 3x + 2y = 7 \end{cases}

Solution: (x,y)=(−17,267)(x, y) = \left(-\frac{1}{7}, \frac{26}{7}\right)

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Problem 3.3

{2x+5y+5z=1−x−y=22x+4y+3z=−1\begin{cases} 2x + 5y + 5z = 1 \\ -x - y = 2 \\ 2x + 4y + 3z = -1 \end{cases}

Solution: (x,y,z)=(−2,0,1)(x, y, z) = (-2, 0, 1)

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4. Writing Matrices as Products of Elementary Matrices

Problem 4.1

Write A=[5132]A = \begin{bmatrix} 5 & 1 \\ 3 & 2 \end{bmatrix} as a product of elementary matrices.

Hint: Row reduce AA to I\mathbb{I} and track each elementary row operation. Each operation corresponds to an elementary matrix.

Solution:

[5132]=[5001][1031][10075][11501]\begin{bmatrix}5 & 1 \\ 3 & 2\end{bmatrix} = \begin{bmatrix}5 & 0 \\ 0 & 1\end{bmatrix} \begin{bmatrix}1 & 0 \\ 3 & 1\end{bmatrix} \begin{bmatrix}1 & 0 \\ 0 & \tfrac{7}{5}\end{bmatrix} \begin{bmatrix}1 & \tfrac{1}{5} \\ 0 & 1\end{bmatrix}
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Problem 4.2

Write A=[123141219]A = \begin{bmatrix} 1 & 2 & 3 \\ 1 & 4 & 1 \\ 2 & 1 & 9 \end{bmatrix} and its inverse A−1A^{-1} as products of elementary matrices.

Solution: det⁡(A)=0⇒A\det(A) = 0 \Rightarrow A is singular, so AA cannot be written as a product of elementary matrices and A−1A^{-1} does not exist.

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Problem 4.3

Given B=[3122]B = \begin{bmatrix} 3 & 1 \\ 2 & 2 \end{bmatrix}, express both BB and B−1B^{-1} as products of elementary matrices.

Solution:

[3122]=[3001][1021][10043][11301]\begin{bmatrix}3 & 1 \\ 2 & 2\end{bmatrix} = \begin{bmatrix}3 & 0 \\ 0 & 1\end{bmatrix} \begin{bmatrix}1 & 0 \\ 2 & 1\end{bmatrix} \begin{bmatrix}1 & 0 \\ 0 & \tfrac{4}{3}\end{bmatrix} \begin{bmatrix}1 & \tfrac{1}{3} \\ 0 & 1\end{bmatrix}[3122]−1=[1−1301][10034][10−21][13001]=[12−14−1234]\begin{bmatrix}3 & 1 \\ 2 & 2\end{bmatrix}^{-1} = \begin{bmatrix}1 & -\tfrac{1}{3} \\ 0 & 1\end{bmatrix} \begin{bmatrix}1 & 0 \\ 0 & \tfrac{3}{4}\end{bmatrix} \begin{bmatrix}1 & 0 \\ -2 & 1\end{bmatrix} \begin{bmatrix}\tfrac{1}{3} & 0 \\ 0 & 1\end{bmatrix} = \begin{bmatrix}\tfrac{1}{2} & -\tfrac{1}{4} \\ -\tfrac{1}{2} & \tfrac{3}{4}\end{bmatrix}
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5. Symmetric and Skewed Symmetric Matrices

Prove whether each matrix XX is symmetric, skewed symmetric, or neither.

Definitions:

  • Symmetric: XT=XX^T = X
  • Skewed Symmetric: XT=−XX^T = -X
  • Neither: XT≠XX^T \neq X and XT≠−XX^T \neq -X

Problem 5.1

X=[3−25−201514]X = \begin{bmatrix} 3 & -2 & 5 \\ -2 & 0 & 1 \\ 5 & 1 & 4 \end{bmatrix}

Solution: Symmetric (since XT=XX^T = X)

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Problem 5.2

X=[03−5−3025−20]X = \begin{bmatrix} 0 & 3 & -5 \\ -3 & 0 & 2 \\ 5 & -2 & 0 \end{bmatrix}

Solution: Skew-symmetric (since XT=−XX^T = -X)

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Problem 5.3

X=[123456789]X = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}

Solution: Neither (since XT≠XX^T \neq X and XT≠−XX^T \neq -X)

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Problem 5.4

X=[2000−10005]X = \begin{bmatrix} 2 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 5 \end{bmatrix}

Solution: Symmetric (since XT=XX^T = X)

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Problem 5.5

X=[0−440]X = \begin{bmatrix} 0 & -4 \\ 4 & 0 \end{bmatrix}

Solution: Skew-symmetric (since XT=−XX^T = -X)

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6. Mixed Practice Problems

Problem 6.1: Matrix Inverses

Find the inverse of the following matrices using the Inversion Algorithm:

a) A=[5132]A = \begin{bmatrix} 5 & 1 \\ 3 & 2 \end{bmatrix}

b) A=[8−4−21]A = \begin{bmatrix} 8 & -4 \\ -2 & 1 \end{bmatrix}

c) A=[9321]A = \begin{bmatrix} 9 & 3 \\ 2 & 1 \end{bmatrix}

Solutions:

a) [5132]−1=17[2−1−35]\begin{bmatrix} 5 & 1 \\ 3 & 2 \end{bmatrix}^{-1} = \frac{1}{7}\begin{bmatrix} 2 & -1 \\ -3 & 5 \end{bmatrix}

b) det⁡=0⇒\det = 0 \Rightarrow no inverse.

c) [9321]−1=13[1−3−29]\begin{bmatrix} 9 & 3 \\ 2 & 1 \end{bmatrix}^{-1} = \frac{1}{3}\begin{bmatrix} 1 & -3 \\ -2 & 9 \end{bmatrix}

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Problem 6.2: Verifying Inverses

Determine whether the given two matrices are inverses of each other by computing ABAB and BABA:

a) A=[5312]A = \begin{bmatrix} 5 & 3 \\ 1 & 2 \end{bmatrix} and B=[2/7−3/7−1/75/7]B = \begin{bmatrix} 2/7 & -3/7 \\ -1/7 & 5/7 \end{bmatrix}

b) A=[−5178]A = \begin{bmatrix} -5 & 1 \\ 7 & 8 \end{bmatrix} and B=[−8/471/477/475/47]B = \begin{bmatrix} -8/47 & 1/47 \\ 7/47 & 5/47 \end{bmatrix}

Solutions:

a) Yes (both AB=IAB = \mathbb{I} and BA=IBA = \mathbb{I}).

b) Yes (both AB=IAB = \mathbb{I} and BA=IBA = \mathbb{I}).

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Problem 6.3: Row Equivalence

Show that A=[123141219]A = \begin{bmatrix} 1 & 2 & 3 \\ 1 & 4 & 1 \\ 2 & 1 & 9 \end{bmatrix} and B=[10502−2114]B = \begin{bmatrix} 1 & 0 & 5 \\ 0 & 2 & -2 \\ 1 & 1 & 4 \end{bmatrix} are row equivalent by finding a sequence of elementary row operations that transforms AA into BB.

Solution: One valid sequence of operations transforming A→BA \to B:

  1. R2←R2−R1R_2 \leftarrow R_2 - R_1
  2. R1←R1−R2R_1 \leftarrow R_1 - R_2
  3. R3←R3−R1R_3 \leftarrow R_3 - R_1
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Study Tips

  1. Gauss-Jordan: Practice getting matrices into RREF quickly and accurately. Watch for leading 1s and ensure zeros above and below them.

  2. Parametric Systems: When you have free variables, assign parameters (like s,ts, t) to the non-leading variables and express other variables in terms of these parameters.

  3. Systems with Unknown Constants (kk):

    • Use row reduction to get the augmented matrix into row echelon form.
    • Analyze the bottom row(s) which will contain expressions with kk.
    • For each critical value of kk (where coefficients become zero), check:
      • If 0=00 = 0: Free variable exists ⇒\Rightarrow infinitely many solutions
      • If 0=non-zero0 = \text{non-zero}: Contradiction ⇒\Rightarrow no solution
    • If no coefficients are zero: All pivots exist ⇒\Rightarrow exactly one solution

  4. Matrix Inversion Algorithm: Remember the setup [A∣I]→[I∣A−1]\left[A \mid \mathbb{I}\right] \rightarrow \left[\mathbb{I} \mid A^{-1}\right]. Whatever row operations you do to the left side, you must do to the right side.

  5. Elementary Matrices: Each elementary row operation corresponds to multiplying by an elementary matrix on the left. Track your operations carefully.

  6. Symmetric Matrices: Just compute XTX^T and compare it to XX and −X-X. Show your work step by step.