TU BCA Mathematics II Master Guide: Matrices, Gauss Elimination, Eigenvalues & Cramer’s Rule (Solved Board Problems)

Author: Bhuban Subedi | Subject: Mathematics II (CAMT154) | Semester: Second Semester

In the second semester of the Tribhuvan University BCA curriculum, Mathematics II (CAMT154) transitions students into Linear Algebra, Discrete Mathematics, and Vector Spaces. Linear algebra forms the mathematical foundation for computer graphics 3D rendering, machine learning tensor computations, and cryptography.

In the 60-mark external TU final examination, linear algebra numericals in Group B and Group C account for over 30% of the total score. Examiners consistently test candidates on finding the rank of matrices, solving 3-variable linear systems via Gauss-Jordan elimination, and computing Eigenvalues and Eigenvectors.

In this guide, I will solve standard TU board exam problems step-by-step.

1. System of Linear Equations: Cramer’s Rule

Consider a system of 3 linear equations:
$$\begin{aligned}
a_1 x + b_1 y + c_1 z &= d_1 \
a_2 x + b_2 y + c_2 z &= d_2 \
a_3 x + b_3 y + c_3 z &= d_3
\end{aligned}$$

The coefficient determinant $D$ is:
$$D = \begin{vmatrix} a_1 & b_1 & c_1 \ a_2 & b_2 & c_2 \ a_3 & b_3 & c_3 \end{vmatrix}$$

If $D \neq 0$, the unique solution is given by:
$$x = \frac{D_x}{D}, \quad y = \frac{D_y}{D}, \quad z = \frac{D_z}{D}$$

Solved TU Board Problem 1: Cramer’s Rule

Problem: Solve the following system using Cramer’s Rule:
$$\begin{aligned}
x + y + z &= 6 \
x – y + z &= 2 \
2x + y – z &= 1
\end{aligned}$$

Step 1: Compute Main Determinant $D$:
$$D = \begin{vmatrix} 1 & 1 & 1 \ 1 & -1 & 1 \ 2 & 1 & -1 \end{vmatrix}$$
$$D = 1((-1)(-1) – (1)(1)) – 1((1)(-1) – (1)(2)) + 1((1)(1) – (-1)(2))$$
$$D = 1(1 – 1) – 1(-1 – 2) + 1(1 + 2) = 0 – 1(-3) + 1(3) = 3 + 3 = \mathbf{6} \quad (D \neq 0)$$

Step 2: Compute $D_x$ (Replace Column 1 with Constant Vector $[6, 2, 1]^T$):
$$D_x = \begin{vmatrix} 6 & 1 & 1 \ 2 & -1 & 1 \ 1 & 1 & -1 \end{vmatrix} = 6(1 – 1) – 1(-2 – 1) + 1(2 – (-1)) = 0 + 3 + 3 = \mathbf{6}$$

Step 3: Compute $D_y$ (Replace Column 2 with Constant Vector):
$$D_y = \begin{vmatrix} 1 & 6 & 1 \ 1 & 2 & 1 \ 2 & 1 & -1 \end{vmatrix} = 1(-2 – 1) – 6(-1 – 2) + 1(1 – 4) = -3 + 18 – 3 = \mathbf{12}$$

Step 4: Compute $D_z$ (Replace Column 3 with Constant Vector):
$$D_z = \begin{vmatrix} 1 & 1 & 6 \ 1 & -1 & 2 \ 2 & 1 & 1 \end{vmatrix} = 1(-1 – 2) – 1(1 – 4) + 6(1 – (-2)) = -3 + 3 + 18 = \mathbf{18}$$

Step 5: Calculate Variables:
$$x = \frac{D_x}{D} = \frac{6}{6} = \mathbf{1}, \quad y = \frac{D_y}{D} = \frac{12}{6} = \mathbf{2}, \quad z = \frac{D_z}{D} = \frac{18}{6} = \mathbf{3}$$

Final Solution: $\mathbf{x = 1, \; y = 2, \; z = 3}$

2. Eigenvalues and Eigenvectors

For an $n \times n$ square matrix $A$, a scalar $\lambda$ is called an Eigenvalue if there exists a non-zero vector $X$ such that:
$$A X = \lambda X \implies (A – \lambda I) X = 0$$

The Eigenvalues are the roots of the Characteristic Equation:
$$\det(A – \lambda I) = 0$$

Solved TU Board Problem 2: Finding Eigenvalues & Eigenvectors

Problem: Find the Eigenvalues and corresponding Eigenvectors for matrix $A = \begin{pmatrix} 4 & 2 \ 1 & 3 \end{pmatrix}$.

Step 1: Form Characteristic Equation:
$$\det(A – \lambda I) = \begin{vmatrix} 4 – \lambda & 2 \ 1 & 3 – \lambda \end{vmatrix} = 0$$
$$(4 – \lambda)(3 – \lambda) – (2)(1) = 0$$
$$\lambda^2 – 7\lambda + 12 – 2 = 0 \implies \lambda^2 – 7\lambda + 10 = 0$$
$$(\lambda – 5)(\lambda – 2) = 0 \implies \mathbf{\lambda_1 = 5, \; \lambda_2 = 2}$$

Step 2: Find Eigenvector for $\lambda = 5$:
$$(A – 5I) X = 0 \implies \begin{pmatrix} 4-5 & 2 \ 1 & 3-5 \end{pmatrix} \begin{pmatrix} x_1 \ x_2 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \end{pmatrix}$$
$$\begin{pmatrix} -1 & 2 \ 1 & -2 \end{pmatrix} \begin{pmatrix} x_1 \ x_2 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \end{pmatrix} \implies -x_1 + 2x_2 = 0 \implies x_1 = 2x_2$$
Let $x_2 = k$, then $x_1 = 2k$.
$$\mathbf{X_1 = k \begin{pmatrix} 2 \ 1 \end{pmatrix}} \quad (\text{For } k = 1, \; X_1 = \begin{pmatrix} 2 \ 1 \end{pmatrix})$$

Step 3: Find Eigenvector for $\lambda = 2$:
$$(A – 2I) X = 0 \implies \begin{pmatrix} 4-2 & 2 \ 1 & 3-2 \end{pmatrix} \begin{pmatrix} x_1 \ x_2 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \end{pmatrix}$$
$$\begin{pmatrix} 2 & 2 \ 1 & 1 \end{pmatrix} \begin{pmatrix} x_1 \ x_2 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \end{pmatrix} \implies x_1 + x_2 = 0 \implies x_1 = -x_2$$
Let $x_2 = k$, then $x_1 = -k$.
$$\mathbf{X_2 = k \begin{pmatrix} -1 \ 1 \end{pmatrix}} \quad (\text{For } k = 1, \; X_2 = \begin{pmatrix} -1 \ 1 \end{pmatrix})$$

3. Rank of a Matrix & Row Echelon Form

The rank of a matrix is the maximum number of linearly independent row vectors in the matrix, which equals the number of non-zero rows in its Row Echelon Form (REF).

+-------------------------------------------------------------------------------+
|                       GAUSS ELIMINATION PROCEDURE                             |
| 1. Use row operations (R_i -> R_i - k * R_j) to create zeros below the pivot. |
| 2. Transform the matrix into Upper Triangular (Row Echelon) form.             |
| 3. Count non-zero rows = Rank(A).                                             |
| 4. Perform back-substitution to find variable values.                         |
+-------------------------------------------------------------------------------+

Frequently Asked Questions (FAQ)

Q1: What is the condition for a matrix to be invertible?

A square matrix $A$ is invertible if and only if its determinant is non-zero ($\det(A) \neq 0$). A matrix with $\det(A) = 0$ is called a singular matrix.

Q2: What is the Cayley-Hamilton Theorem?

The Cayley-Hamilton Theorem states that every square matrix satisfies its own characteristic equation (i.e., if $p(\lambda) = \det(A – \lambda I) = 0$, then $p(A) = 0$).