Topic 1 Systems of Linear Equations & Matrix Algebra

This topic covers the foundational mechanics of Linear Algebra: manipulating matrices, solving systems using row reduction, computing inverses, and utilizing matrix factorizations (like LU decomposition).

Question 1: Reduced Row Echelon Form (RREF)

Source: Practice Exam 4, Question 2
Exact Question: Give the reduced row echelon form for the matrix $\left[\begin{array}{ccc}1& 2& 3\\ 1& 0& 1\\ 0& 6& 6\end{array}\right]$.

Methodology:

1. Objective: Transform the matrix using elementary row operations into Reduced Row Echelon Form (RREF). A matrix is in RREF if:
   • All nonzero rows are above any rows of all zeros.
   • Each leading entry (pivot) of a row is in a column to the right of the leading entry of the row above it.
   • The leading entry in each nonzero row is exactly 1.
   • Each leading 1 is the only nonzero entry in its column (i.e., zeros above and below the pivot).
2. Forward Phase (Elimination): Create zeros below the pivots, working from left to right, top to bottom.
3. Backward Phase: Create zeros above the pivots, working from right to left, bottom to top.
4. Scaling: Multiply rows by scalars to ensure every pivot is exactly 1.

Solution:

• Step 1: Eliminate the entry in row 2, col 1. Subtract Row 1 from Row 2 ($R_2\to R_2-R_1$).$$\left[\begin{array}{ccc}1& 2& 3\\ 0& -2& -2\\ 0& 6& 6\end{array}\right]$$
• Step 2: Eliminate the entry in row 3, col 2 using the row 2 pivot. Add 3 times Row 2 to Row 3 ($R_3\to R_3+3R_2$).$$\left[\begin{array}{ccc}1& 2& 3\\ 0& -2& -2\\ 0& 0& 0\end{array}\right]$$
• Step 3: The matrix is now in standard echelon form. Now, we begin the backward phase to reach reduced echelon form. Eliminate the entry in row 1, col 2. Add Row 2 to Row 1 ($R_1\to R_1+R_2$).$$\left[\begin{array}{ccc}1& 0& 1\\ 0& -2& -2\\ 0& 0& 0\end{array}\right]$$
• Step 4: Scale Row 2 so its pivot is exactly 1. Multiply Row 2 by $-\frac{1}{2}$ ($R_2\to -\frac{1}{2}{R}_2$).
  Final Answer:$$\left[\begin{array}{ccc}1& 0& 1\\ 0& 1& 1\\ 0& 0& 0\end{array}\right]$$

Question 2: Finding a Matrix Inverse ($3\times 3$)

Source: Exam 2, Question 2
Exact Question: Find the inverse matrix of the matrix $A=\left[\begin{array}{ccc}3& 1& 1\\ 1& 0& -1\\ 2& 1& 1\end{array}\right]$.

Methodology:

1. Objective: Compute $A^{-1}$ using the augmented matrix algorithm.
2. Set Up: Form the augmented matrix $[A\mid I_3]$, where $I_3$ is the $3\times 3$ identity matrix.
3. Row Reduction: Apply elementary row operations to transform the left side $A$ into $I_3$.
4. Result: If $A$ is invertible, the augmented matrix will reduce to the form $[I_3\mid A^{-1}]$. The right side of the partition is your answer.

Solution:

• Set up the matrix:$$[A\mid I_3]=\left[\begin{array}{cccccc}3& 1& 1& 1& 0& 0\\ 1& 0& -1& 0& 1& 0\\ 2& 1& 1& 0& 0& 1\end{array}\right]$$
• Step 1: To make the math easier, swap Row 1 and Row 2 to get a 1 in the top-left pivot position ($R_1\leftrightarrow R_2$). (Note: The official solution skips the swap and does more complex row operations, but swapping is highly recommended to avoid fractions). Following the provided solution's path for exactness: Let's stick to the exam's approach which simply reduces the system:$$\left[\begin{array}{cccccc}1& 0& -1& 0& 1& 0\\ 3& 1& 1& 1& 0& 0\\ 2& 1& 1& 0& 0& 1\end{array}\right]\text{(after swap)}$$
• Step 2: Eliminate below pivot 1. $R_2\to R_2-3R_1$, and $R_3\to R_3-2R_1$.$$\left[\begin{array}{cccccc}1& 0& -1& 0& 1& 0\\ 0& 1& 4& 1& -3& 0\\ 0& 1& 3& 0& -2& 1\end{array}\right]$$
• Step 3: Eliminate below pivot 2. $R_3\to R_3-R_2$.$$\left[\begin{array}{cccccc}1& 0& -1& 0& 1& 0\\ 0& 1& 4& 1& -3& 0\\ 0& 0& -1& -1& 1& 1\end{array}\right]$$
• Step 4: Scale Row 3 to make the pivot positive 1 ($R_3\to -R_3$).$$\left[\begin{array}{cccccc}1& 0& -1& 0& 1& 0\\ 0& 1& 4& 1& -3& 0\\ 0& 0& 1& 1& -1& -1\end{array}\right]$$
• Step 5: Backward phase. Eliminate above pivot 3. $R_2\to R_2-4R_3$ and $R_1\to R_1+R_3$.$$\left[\begin{array}{cccccc}1& 0& 0& 1& 0& -1\\ 0& 1& 0& -3& 1& 4\\ 0& 0& 1& 1& -1& -1\end{array}\right]$$Final Answer:$$A^{-1}=\left[\begin{array}{ccc}1& 0& -1\\ -3& 1& 4\\ 1& -1& -1\end{array}\right]$$

Question 3: Matrix Inverse with Scaling

Source: Practice Exam 3, Question 11
Exact Question: Let $A=\left[\begin{array}{ccc}0& -1& 5\\ -3& 4& 3\\ -9& 12& 6\end{array}\right]$. Compute $A^{-1}$ if it exists.

Methodology:

1. Set Up: $[A\mid I_3]$.
2. Row Swaps: Because the $(1,1)$ entry is $0$, we must swap Row 1 with another row (e.g., Row 2) to get a non-zero pivot.
3. Forward/Backward Phase: Proceed as standard, being careful with arithmetic since scaling will involve fractions.

Solution:

• Set up:$$\left[\begin{array}{cccccc}0& -1& 5& 1& 0& 0\\ -3& 4& 3& 0& 1& 0\\ -9& 12& 6& 0& 0& 1\end{array}\right]$$
• Step 1: Swap $R_1$ and $R_2$.$$\left[\begin{array}{cccccc}-3& 4& 3& 0& 1& 0\\ 0& -1& 5& 1& 0& 0\\ -9& 12& 6& 0& 0& 1\end{array}\right]$$
• Step 2: Eliminate below pivot 1. $R_3\to R_3-3R_1$.$$\left[\begin{array}{cccccc}-3& 4& 3& 0& 1& 0\\ 0& -1& 5& 1& 0& 0\\ 0& 0& -3& 0& -3& 1\end{array}\right]$$
• Step 3: Backward phase. Eliminate above pivot 2. $R_1\to R_1+4R_2$.$$\left[\begin{array}{cccccc}-3& 0& 23& 4& 1& 0\\ 0& -1& 5& 1& 0& 0\\ 0& 0& -3& 0& -3& 1\end{array}\right]$$
• Step 4: Scale rows to make pivots equal to 1. $R_1\to -\frac{1}{3}{R}_1$, $R_2\to -R_2$, $R_3\to -\frac{1}{3}{R}_3$.$$\left[\begin{array}{cccccc}1& 0& -\frac{23}{3}& -\frac{4}{3}& -\frac{1}{3}& 0\\ 0& 1& -5& -1& 0& 0\\ 0& 0& 1& 0& 1& -\frac{1}{3}\end{array}\right]$$
• Step 5: Eliminate above pivot 3. $R_2\to R_2+5R_3$, and $R_1\to R_1+\frac{23}{3}{R}_3$.$$\left[\begin{array}{cccccc}1& 0& 0& -\frac{4}{3}& \frac{22}{3}& -\frac{23}{9}\\ 0& 1& 0& -1& 5& -\frac{5}{3}\\ 0& 0& 1& 0& 1& -\frac{1}{3}\end{array}\right]$$Final Answer:$$A^{-1}=\left[\begin{array}{ccc}-\frac{4}{3}& \frac{22}{3}& -\frac{23}{9}\\ -1& 5& -\frac{5}{3}\\ 0& 1& -\frac{1}{3}\end{array}\right]$$

Question 4: Determinants and Invertibility

Source: Practice Exam 4, Question 4
Exact Question: For which value(s) of $h$ is the matrix $\left[\begin{array}{cc}h& -3\\ -2& 2h\end{array}\right]$ invertible?

Methodology:

1. Objective: Find the condition for invertibility.
2. Formula: A $2\times 2$ matrix $A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$ is invertible if and only if its determinant is non-zero ($det(A)=ad-bc\ne 0$).
3. Process: Calculate the determinant in terms of $h$. Set it equal to zero to find the values that make the matrix singular (not invertible). The final answer will be all values except those.

Solution:

• Step 1: Calculate the determinant.$$det(A)=(h)(2h)-(-3)(-2)=2h^2-6$$
• Step 2: Set to zero to find critical values.$$2h^2-6=0⟹2h^2=6⟹h^2=3⟹h=\pm \sqrt{3}$$
• Step 3: Formulate conclusion.
  Final Answer: The matrix is invertible when $h\ne \pm \sqrt{3}$.

Question 5: Matrix Algebra and Dimensions

Source: Practice Exam 4, Question 9
Exact Question: Let $A=\left[\begin{array}{ccc}3& 0& 2\\ 1& 1& 1\end{array}\right]$ and $B=\left[\begin{array}{cc}1& 2\\ 2& 1\end{array}\right]$. Compute $A^T{B}^{-1}$, if it exists.

Methodology:

1. Check Dimensions: Matrix multiplication $CD$ is only defined if the number of columns of $C$ equals the number of rows of $D$. $A$ is $2\times 3$, so $A^T$ is $3\times 2$. $B$ is $2\times 2$, so $B^{-1}$ is $2\times 2$. The product $(3\times 2)\times (2\times 2)$ is valid and will yield a $3\times 2$ matrix.
2. Transpose: Find $A^T$ by turning the rows of $A$ into columns.
3. Inverse Formula: Find $B^{-1}$ using the $2\times 2$ rule: $B^{-1}=\frac{1}{det(B)}\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]$.
4. Multiply: Compute the dot products of the rows of $A^T$ with the columns of $B^{-1}$.

Solution:

• Step 1: Find $A^T$.$$A^T=\left[\begin{array}{cc}3& 1\\ 0& 1\\ 2& 1\end{array}\right]$$
• Step 2: Find $B^{-1}$. $det(B)=(1)(1)-(2)(2)=-3$.$$B^{-1}=-\frac{1}{3}\left[\begin{array}{cc}1& -2\\ -2& 1\end{array}\right]=\left[\begin{array}{cc}-\frac{1}{3}& \frac{2}{3}\\ \frac{2}{3}& -\frac{1}{3}\end{array}\right]$$
• Step 3: Multiply $A^T{B}^{-1}$.$$\left[\begin{array}{cc}3& 1\\ 0& 1\\ 2& 1\end{array}\right]\left[\begin{array}{cc}-\frac{1}{3}& \frac{2}{3}\\ \frac{2}{3}& -\frac{1}{3}\end{array}\right]=\left[\begin{array}{cc}3(-\frac{1}{3})+1(\frac{2}{3})& 3(\frac{2}{3})+1(-\frac{1}{3})\\ 0(-\frac{1}{3})+1(\frac{2}{3})& 0(\frac{2}{3})+1(-\frac{1}{3})\\ 2(-\frac{1}{3})+1(\frac{2}{3})& 2(\frac{2}{3})+1(-\frac{1}{3})\end{array}\right]$$Final Answer:$$A^T{B}^{-1}=\left[\begin{array}{cc}-\frac{1}{3}& \frac{5}{3}\\ \frac{2}{3}& -\frac{1}{3}\\ 0& 1\end{array}\right]$$

Question 6: LU Decomposition & Solving $Ax=b$

Source: Exam 2, Question 7
Exact Question: Consider the matrix $A=\left[\begin{array}{ccc}1& 2& 2\\ 3& 1& -2\\ 2& 9& 5\end{array}\right]$.
a. Find an LU decomposition of $A$.
b. Use this LU decomposition to solve the system $Ax=\left[\begin{array}{c}6\\ -1\\ 10\end{array}\right]$.

Methodology:

1. Part a (Find LU):
   • Reduce $A$ to an upper triangular matrix $U$ using only row replacements (subtracting a multiple of a row from a row below it). Never swap rows or scale rows!
   • Keep track of the multipliers. The scalar $m$ used to eliminate the entry in row $i$, column $j$ becomes the $(i,j)$ entry in the lower triangular matrix $L$.
   • $L$ will have 1s on its main diagonal.
2. Part b (Solve $Ax=b$):
   • Since $A=LU$, the equation $Ax=b$ becomes $LUx=b$.
   • Let $Ux=y$. Now the system becomes $Ly=b$.
   • Solve $Ly=b$ for $y$ using forward substitution (starting from $y_1$ and working down).
   • Once $y$ is found, solve $Ux=y$ for $x$ using backward substitution (starting from $x_3$ and working up).

Solution:
Part a:

• Find $U$: Let's eliminate below the first pivot (column 1).
  $R_2\to R_2-3R_1$ (Multiplier $L_{21}=3$)
  $R_3\to R_3-2R_1$ (Multiplier $L_{31}=2$)$$\left[\begin{array}{ccc}1& 2& 2\\ 0& -5& -8\\ 0& 5& 1\end{array}\right]$$
• Now eliminate below the second pivot (column 2).
  $R_3\to R_3-(-1)R_2$ (Wait, $R_3+R_2$ is identical to $R_3-(-1)R_2$. The multiplier is $L_{32}=-1$).$$U=\left[\begin{array}{ccc}1& 2& 2\\ 0& -5& -8\\ 0& 0& -7\end{array}\right]$$
• Construct $L$: Place the multipliers into a lower triangular matrix with 1s on the diagonal.$$L=\left[\begin{array}{ccc}1& 0& 0\\ 3& 1& 0\\ 2& -1& 1\end{array}\right]$$

Part b:

• Step 1: Solve $Ly=b⟹\left[\begin{array}{ccc}1& 0& 0\\ 3& 1& 0\\ 2& -1& 1\end{array}\right]\left[\begin{array}{c}{y}_1\\ y_2\\ y_3\end{array}\right]=\left[\begin{array}{c}6\\ -1\\ 10\end{array}\right]$
  Row 1: $y_1=6$
  Row 2: $3(6)+y_2=-1⟹y_2=-19$
  Row 3: $2(6)-(-19)+y_3=10⟹12+19+y_3=10⟹y_3=-21$
  So, $y=\left[\begin{array}{c}6\\ -19\\ -21\end{array}\right]$.

• Step 2: Solve $Ux=y⟹\left[\begin{array}{ccc}1& 2& 2\\ 0& -5& -8\\ 0& 0& -7\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=\left[\begin{array}{c}6\\ -19\\ -21\end{array}\right]$
  Row 3: $-7x_3=-21⟹x_3=3$
  Row 2: $-5x_2-8(3)=-19⟹-5x_2=5⟹x_2=-1$
  Row 1: $x_1+2(-1)+2(3)=6⟹x_1-2+6=6⟹x_1=2$

  Final Answer:

  $$x=\left[\begin{array}{c}2\\ -1\\ 3\end{array}\right]$$

Question 7: LU Factorization Only

Source: Practice Exam 4, Question 8
Exact Question: Find an LU-factorization of the matrix $A=\left[\begin{array}{ccc}1& 2& 4\\ 2& 5& 9\\ -2& 7& 1\end{array}\right]$.

Methodology:
This is identically structured to Part A of Question 6. Reduce $A$ to $U$ while tracking the exact multipliers $m$ used to subtract rows. Remember: the formula is $Ro{w}_{target}=Ro{w}_{target}-m\times Ro{w}_{source}$. That $m$ goes into $L$.

Solution:

• Step 1: Target column 1.
  $R_2\to R_2-2R_1⟹m_{21}=2$.
  $R_3\to R_3-(-2)R_1⟹m_{31}=-2$.$$\left[\begin{array}{ccc}1& 2& 4\\ 0& 1& 1\\ 0& 11& 9\end{array}\right]$$
• Step 2: Target column 2.
  $R_3\to R_3-11R_2⟹m_{32}=11$.$$U=\left[\begin{array}{ccc}1& 2& 4\\ 0& 1& 1\\ 0& 0& -2\end{array}\right]$$
• Step 3: Build $L$ using the multipliers.
  Final Answer:$$L=\left[\begin{array}{ccc}1& 0& 0\\ 2& 1& 0\\ -2& 11& 1\end{array}\right],U=\left[\begin{array}{ccc}1& 2& 4\\ 0& 1& 1\\ 0& 0& -2\end{array}\right]$$