Practice Set-1 — Solutions

0. Basic Operations

1. Problem: Add the matrices \[ A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}, \quad B = \begin{bmatrix} 5 & 6 \\ 7 & 8 \end{bmatrix} \]
   \[ A+B=\begin{bmatrix}1+5 & 2+6\\[4pt] 3+7 & 4+8\end{bmatrix} =\begin{bmatrix}6 & 8 \\ 10 & 12\end{bmatrix} \]
2. Problem: Subtract the matrices \[ \begin{bmatrix} 4 & -2 \\ 1 & 0 \end{bmatrix} - \begin{bmatrix} 2 & 3 \\ -1 & 5 \end{bmatrix} \]
   \[ \begin{bmatrix}4-2 & -2-3 \\[4pt] 1-(-1) & 0-5 \end{bmatrix} =\begin{bmatrix}2 & -5 \\ 2 & -5\end{bmatrix} \]
3. Problem: Multiply the matrices \[ \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix} \cdot \begin{bmatrix} 2 & 0 \\ 3 & 4 \end{bmatrix} \]
   Multiply row by column: \[ \begin{bmatrix} 1\cdot2+2\cdot3 & 1\cdot0+2\cdot4\\[6pt] 0\cdot2+1\cdot3 & 0\cdot0+1\cdot4 \end{bmatrix} =\begin{bmatrix}8 & 8 \\ 3 & 4\end{bmatrix} \]

4. Problem: Find the determinant of: \[ \begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix} \]
   \[ \det = 2\cdot4 - 3\cdot1 = 8-3=5. \]
5. Problem: Evaluate the determinant of: \[ \begin{bmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 1 & 0 & 6 \end{bmatrix} \]
   Expand along the first row: \[ \det = 1\cdot\begin{vmatrix}4&5\\0&6\end{vmatrix} -2\cdot\begin{vmatrix}0&5\\1&6\end{vmatrix} +3\cdot\begin{vmatrix}0&4\\1&0\end{vmatrix} \] \[ =1(24-0)-2(0-5)+3(0-4)=24+10-12=22. \]

6. Problem: Find the inverse of: \[ \begin{bmatrix} 4 & 7 \\ 2 & 6 \end{bmatrix} \]
   Determinant \(=4\cdot6-7\cdot2=24-14=10\). Inverse: \[ A^{-1}=\frac{1}{10}\begin{bmatrix}6 & -7\\ -2 & 4\end{bmatrix} =\begin{bmatrix}0.6 & -0.7\\ -0.2 & 0.4\end{bmatrix}. \] (Or keep as fractions: \(\frac{1}{10}\begin{bmatrix}6 & -7\\ -2 & 4\end{bmatrix}\).)
7. Problem: Show that the matrix is not invertible: \[ \begin{bmatrix} 2 & 4 \\ 1 & 2 \end{bmatrix} \]
   Determinant \(=2\cdot2-4\cdot1=4-4=0\). Since \(\det=0\), the matrix is singular (not invertible).

8. Problem: Identify if the following is symmetric, skew-symmetric, or neither: \[ \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix} \]
   Transpose equals \(\begin{bmatrix}0 & -2\\ 2 & 0\end{bmatrix} = -\begin{bmatrix}0&2\\-2&0\end{bmatrix}\). So the matrix is skew-symmetric.
9. Problem: Write a \(3\times3\) diagonal matrix and a scalar matrix with scalar \(5\).
   Example diagonal matrix: \[ \begin{bmatrix}1&0&0\\0&2&0\\0&0&3\end{bmatrix}. \quad\text{Scalar matrix }5I_3=\begin{bmatrix}5&0&0\\0&5&0\\0&0&5\end{bmatrix}. \]
10. Problem: Verify if the identity matrix \[ I = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \] satisfies \(AI = IA = A\) for \[ A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}. \]
    Multiplying any matrix by the identity leaves it unchanged, so \[ AI = A,\qquad IA = A. \] (One can multiply to check: rows of \(A\) times columns of \(I\) reproduce the rows of \(A\).)

A. Row Reduced Echelon Form (RREF)

1. Problem: Reduce to RREF: \[ A = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 3 & 6 & 9 \end{bmatrix} \]
   Rows 2 and 3 are multiples of row 1. The RREF is \[ \operatorname{RREF}(A)= \begin{bmatrix}1 & 2 & 3\\[4pt] 0 & 0 & 0\\[4pt] 0 & 0 & 0\end{bmatrix}. \] (Rank = 1.)
2. Problem: Find the RREF of: \[ B = \begin{bmatrix} 2 & 1 & -1 \\ -1 & 0 & 3 \\ 4 & 1 & 2 \end{bmatrix} \]
   After row operations the RREF is the \(3\times3\) identity matrix: \[ \operatorname{RREF}(B)=\begin{bmatrix}1&0&0\\[3pt]0&1&0\\[3pt]0&0&1\end{bmatrix}. \] So \(B\) is invertible (rank = 3).
3. Problem: Bring to RREF: \[ C = \begin{bmatrix} 0 & 2 & 4 & 6 \\ 1 & 3 & 5 & 7 \\ 2 & 4 & 6 & 8 \end{bmatrix} \]
   RREF is \[ \operatorname{RREF}(C)= \begin{bmatrix} 1 & 0 & -1 & -2\\[4pt] 0 & 1 & 2 & 3\\[4pt] 0 & 0 & 0 & 0 \end{bmatrix}. \] (Two leading variables — rank = 2.)

B. Rank of a Matrix

4. Problem: Find the rank of: \[ D = \begin{bmatrix} 1 & 0 & -1 & 2 \\ 2 & 1 & 0 & 3 \\ 3 & 1 & -1 & 5 \end{bmatrix} \]
   \(\operatorname{rank}(D)=2.\)
5. Problem: Compute the rank of: \[ E = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix} \]
   \(\operatorname{rank}(E)=2\) (rows/columns are linearly dependent).
6. Problem: Find the rank of: \[ F = \begin{bmatrix} 2 & 4 & 6 & 8 \\ 1 & 2 & 3 & 4 \\ 0 & 0 & 0 & 0 \end{bmatrix} \]
   \(\operatorname{rank}(F)=1\) (row 1 is twice row 2; only one independent row).

C. System of Linear Equations

7. Problem: Solve using row-reduction: \[ \begin{cases} x + 2y + 3z = 9 \\ 2x + 3y + z = 8 \\ 3x + y + 2z = 7 \end{cases} \]
   Unique solution: \[ (x,y,z)=\left(\tfrac{2}{3},\,\tfrac{5}{3},\,\tfrac{5}{3}\right). \]
   (You can check by substitution; the augmented matrix reduces to a unique solution.)
8. Problem: Determine if the system has unique, infinite, or no solution: \[ \begin{cases} x + y + z = 6 \\ 2x + 2y + 2z = 12 \\ 3x + 3y + 3z = 18 \end{cases} \]
   All three equations are multiples of the first → they are dependent. So the system has infinitely many solutions. General form: \[ x=6-y-z \quad\text{(two free variables, e.g. set }y=s,z=t\text{).} \]
9. Problem: Solve the system: \[ \begin{cases} x + y - z = 1 \\ 2x - y + z = 3 \\ 3x + 2y = 4 \end{cases} \]
   Unique solution: \[ (x,y,z)=\left(\tfrac{4}{3},\,0,\,\tfrac{1}{3}\right). \]
10. Problem: A system of equations is given by: \[ \begin{cases} x_1 + x_2 + x_3 + x_4 = 5 \\ 2x_1 + 3x_2 + 5x_3 + 7x_4 = 15 \\ x_1 - x_2 + 2x_3 + 3x_4 = 4 \end{cases} \]

    • Write the augmented matrix.
    • Reduce to RREF.
    • Find the general solution.
    Augmented matrix: \[ \left[\begin{array}{cccc|c} 1 & 1 & 1 & 1 & 5\\[4pt] 2 & 3 & 5 & 7 & 15\\[4pt] 1 & -1 & 2 & 3 & 4 \end{array}\right] \] RREF: \[ \left[\begin{array}{cccc|c} 1 & 0 & 0 & -\tfrac{4}{7} & \tfrac{18}{7}\\[6pt] 0 & 1 & 0 & -\tfrac{1}{7} & \tfrac{8}{7}\\[6pt] 0 & 0 & 1 & \tfrac{12}{7} & \tfrac{9}{7} \end{array}\right] \] Let \(x_4=t\) (free parameter). Then the general solution is \[ \begin{aligned} x_1 &= \tfrac{18}{7} + \tfrac{4}{7}t,\\[4pt] x_2 &= \tfrac{8}{7} + \tfrac{1}{7}t,\\[4pt] x_3 &= \tfrac{9}{7} - \tfrac{12}{7}t,\\[4pt] x_4 &= t\quad (t\in\mathbb{R}). \end{aligned} \]