Applications of Determinants and Matrices

Consistent and Inconsistent Systems

In this section, we apply the concepts of matrices and determinants to solve systems of linear equations in two or three variables.

A system of linear equations may behave in one of the following ways:

Consistent System

A system is said to be consistent if it has at least one solution. A consistent system may have:

• a unique solution, or
• infinitely many solutions.

Inconsistent System

A system is said to be inconsistent if it has no solution.

For two variables, the geometric meaning is very clear:

• If two lines intersect at one point, the system has a unique solution.
• If two lines coincide, the system has infinitely many solutions.
• If two lines are parallel and distinct, the system has no solution.

For three variables, the equations represent planes. Depending on how the planes intersect, the system may have one solution, infinitely many solutions, or no solution.

---

Matrix Representation of a System of Linear Equations

Consider the system of linear equations: $$a_1x + b_1y + c_1z = d_1$$ $$a_2x + b_2y + c_2z = d_2$$ $$a_3x + b_3y + c_3z = d_3$$ This system can be written in the compact matrix form $$AX = B,$$ where $$A = \begin{bmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{bmatrix}, \qquad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \qquad B = \begin{bmatrix} d_1 \\ d_2 \\ d_3 \end{bmatrix}.$$

Here,

• $A$ is called the coefficient matrix because it contains the coefficients of the variables.
• $X$ is called the variable matrix or unknown matrix.
• $B$ is called the constant matrix.

This form is useful because once the system is written as $AX = B$, matrix operations can be used to study and solve the system.

---

Solution of a System of Linear Equations using Inverse of a Matrix

To solve the system $$AX = B,$$ we examine the determinant of the coefficient matrix $A$.

Case I:

If $A$ is non-singular, i.e. $|A| \neq 0$

Then the inverse of $A$ exists. Pre-multiplying both sides of the equation by $A^{-1}$, we get $$A^{-1}(AX) = A^{-1}B.$$ Using associativity of matrix multiplication, $$(A^{-1}A)X = A^{-1}B.$$ Since $A^{-1}A = I$, $$IX = A^{-1}B,$$ so $$X = A^{-1}B.$$ Thus the system has a unique solution.

Case II:

If $A$ is singular, i.e. $|A| = 0$

Then $A^{-1}$ does not exist. In this case, we examine $(\operatorname{adj} A)B$.

• Subcase (i): If $$(\operatorname{adj} A)B \neq O,$$ then the system is inconsistent, so no solution exists.

• Subcase (ii): If $$(\operatorname{adj} A)B = O,$$ then the system may be consistent. In standard school-level treatment, this generally indicates that the system has infinitely many solutions.

The reason is that if we premultiply $AX=B$ by $\operatorname{adj} A$, then $$(\operatorname{adj} A)AX = (\operatorname{adj} A)B.$$ Using the property $$(\operatorname{adj} A)A = |A|I,$$ and since $|A|=0$, the left-hand side becomes the zero matrix. So if the right-hand side is non-zero, we get a contradiction. If the right-hand side is also zero, the system is dependent and may have infinitely many solutions.

Example 1: Solving a 2x2 System

Solve the system of equations using matrix method: $$2x + 5y = 1$$ $$3x + 2y = 7$$

Solution:

Step 1: Write the system in matrix form $AX=B$: $$A = \begin{bmatrix} 2 & 5 \\ 3 & 2 \end{bmatrix}, \qquad X = \begin{bmatrix} x \\ y \end{bmatrix}, \qquad B = \begin{bmatrix} 1 \\ 7 \end{bmatrix}.$$

Step 2: Find the determinant of $A$: $$|A| = \begin{vmatrix} 2 & 5 \\ 3 & 2 \end{vmatrix} = (2)(2) - (3)(5) = 4 - 15 = -11.$$ Since $|A| \neq 0$, the matrix is non-singular and the system has a unique solution.

Step 3: Find $A^{-1}$: $$\operatorname{adj} A = \begin{bmatrix} 2 & -5 \\ -3 & 2 \end{bmatrix},$$ so $$A^{-1} = \frac{1}{-11}\begin{bmatrix} 2 & -5 \\ -3 & 2 \end{bmatrix}.$$

Step 4: Compute $X = A^{-1}B$: $$X = \frac{-1}{11}\begin{bmatrix} 2 & -5 \\ -3 & 2 \end{bmatrix}\begin{bmatrix} 1 \\ 7 \end{bmatrix} = \frac{-1}{11}\begin{bmatrix} 2-35 \\ -3+14 \end{bmatrix} = \frac{-1}{11}\begin{bmatrix} -33 \\ 11 \end{bmatrix} = \begin{bmatrix} 3 \\ -1 \end{bmatrix}.$$

Answer: $$x=3, \qquad y=-1.$$

---

Example 2: Checking Consistency (Inconsistent System)

Examine the consistency of the system of equations:$$x + 3y = 5$$ $$2x + 6y = 8$$

Solution:

Step 1: Write the system in matrix form: $$A = \begin{bmatrix} 1 & 3 \\ 2 & 6 \end{bmatrix}, \qquad X = \begin{bmatrix} x \\ y \end{bmatrix}, \qquad B = \begin{bmatrix} 5 \\ 8 \end{bmatrix}.$$

Step 2: Find $|A|$: $$|A| = \begin{vmatrix} 1 & 3 \\ 2 & 6 \end{vmatrix} = 1\cdot 6 - 2\cdot 3 = 6-6 = 0.$$ Since $|A|=0$, the system is not guaranteed to have a unique solution. We now examine $(\operatorname{adj} A)B$.

Step 3: Find $\operatorname{adj} A$: $$\operatorname{adj} A = \begin{bmatrix} 6 & -3 \\ -2 & 1 \end{bmatrix}.$$ Now multiply by $B$: $$(\operatorname{adj} A)B = \begin{bmatrix} 6 & -3 \\ -2 & 1 \end{bmatrix}\begin{bmatrix} 5 \\ 8 \end{bmatrix} = \begin{bmatrix} 30-24 \\ -10+8 \end{bmatrix} = \begin{bmatrix} 6 \\ -2 \end{bmatrix}.$$ This is not the zero matrix.

Step 4: Since $|A|=0$ and $(\operatorname{adj} A)B \neq O$, the system is inconsistent.

Answer: The system has no solution and is inconsistent.

---

Example 3: Checking Consistency (Infinitely Many Solutions)

Examine the consistency of the system of equations: $$3x - y = 5$$ $$6x - 2y = 10$$

Solution:

Step 1: Matrix form: $$A = \begin{bmatrix} 3 & -1 \\ 6 & -2 \end{bmatrix}, \qquad B = \begin{bmatrix} 5 \\ 10 \end{bmatrix}.$$

Step 2: Find the determinant: $$|A| = 3(-2) - 6(-1) = -6 + 6 = 0.$$ So the matrix is singular.

Step 3: Find the adjoint: $$\operatorname{adj} A = \begin{bmatrix} -2 & 1 \\ -6 & 3 \end{bmatrix}.$$ Now compute $$(\operatorname{adj} A)B = \begin{bmatrix} -2 & 1 \\ -6 & 3 \end{bmatrix}\begin{bmatrix} 5 \\ 10 \end{bmatrix} = \begin{bmatrix} -10+10 \\ -30+30 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} = O.$$

Step 4: Since $|A|=0$ and $(\operatorname{adj} A)B=O$, the system is consistent and dependent. The second equation is exactly twice the first equation, so both equations represent the same line.

Answer: The system is consistent and has infinitely many solutions.

---

Example 4: Solving a 3x3 System

Solve the following system of equations by matrix method: $$3x - 2y + 3z = 8$$ $$2x + y - z = 1$$ $$4x - 3y + 2z = 4$$

Solution: Step 1: Matrix representation: $$A = \begin{bmatrix} 3 & -2 & 3 \\ 2 & 1 & -1 \\ 4 & -3 & 2 \end{bmatrix}, \qquad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \qquad B = \begin{bmatrix} 8 \\ 1 \\ 4 \end{bmatrix}.$$

Step 2: Find $|A|$ by expanding along the first row: $$|A| = 3\begin{vmatrix} 1 & -1 \\ -3 & 2 \end{vmatrix} - (-2)\begin{vmatrix} 2 & -1 \\ 4 & 2 \end{vmatrix} + 3\begin{vmatrix} 2 & 1 \\ 4 & -3 \end{vmatrix}.$$ So, $$|A| = 3(2-3) + 2(4+4) + 3(-6-4)$$ $$= 3(-1) + 2(8) + 3(-10)$$ $$= -3 + 16 - 30 = -17.$$ Since $|A| \neq 0$, a unique solution exists.

Step 3: Find cofactors: $$A_{11}=-1, \quad A_{12}=-8, \quad A_{13}=-10,$$ $$A_{21}=-5, \quad A_{22}=-6, \quad A_{23}=1,$$ $$A_{31}=-1, \quad A_{32}=9, \quad A_{33}=7.$$ Thus, $$\operatorname{adj} A = \begin{bmatrix} -1 & -5 & -1 \\ -8 & -6 & 9 \\ -10 & 1 & 7 \end{bmatrix}.$$

Step 4: Compute$$X = A^{-1}B = \frac{1}{|A|}(\operatorname{adj} A)B$$ $$X = -\frac{1}{17}\begin{bmatrix} -1 & -5 & -1 \\ -8 & -6 & 9 \\ -10 & 1 & 7 \end{bmatrix}\begin{bmatrix} 8 \\ 1 \\ 4 \end{bmatrix}.$$ Now multiply: $$(\operatorname{adj} A)B = \begin{bmatrix} -8-5-4 \\ -64-6+36 \\ -80+1+28 \end{bmatrix} = \begin{bmatrix} -17 \\ -34 \\ -51 \end{bmatrix}.$$ Therefore, $$X = -\frac{1}{17}\begin{bmatrix} -17 \\ -34 \\ -51 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}.$$

Answer: $$x=1, \qquad y=2, \qquad z=3.$$

---

Example 5: Solving a System with Reciprocals

Solve the system of equations: $$\frac{2}{x} + \frac{3}{y} + \frac{10}{z} = 4$$ $$\frac{4}{x} - \frac{6}{y} + \frac{5}{z} = 1$$ $$\frac{6}{x} + \frac{9}{y} - \frac{20}{z} = 2$$

Solution: Step 1: Substitute $$u = \frac{1}{x}, \qquad v = \frac{1}{y}, \qquad w = \frac{1}{z}.$$ Then the system becomes $$2u + 3v + 10w = 4,$$ $$4u - 6v + 5w = 1,$$ $$6u + 9v - 20w = 2.$$

Step 2: Matrix form: $$A = \begin{bmatrix} 2 & 3 & 10 \\ 4 & -6 & 5 \\ 6 & 9 & -20 \end{bmatrix}, \qquad X = \begin{bmatrix} u \\ v \\ w \end{bmatrix}, \qquad B = \begin{bmatrix} 4 \\ 1 \\ 2 \end{bmatrix}.$$

Step 3: Compute the determinant: $$|A| = 2\begin{vmatrix} -6 & 5 \\ 9 & -20 \end{vmatrix} - 3\begin{vmatrix} 4 & 5 \\ 6 & -20 \end{vmatrix} + 10\begin{vmatrix} 4 & -6 \\ 6 & 9 \end{vmatrix}.$$ So, $$|A| = 2(120-45) - 3(-80-30) + 10(36+36)$$ $$\Rightarrow |A| = 2(75) - 3(-110) + 10(72)$$ $$\Rightarrow |A| = 150 + 330 + 720 = 1200.$$ Since $|A| \neq 0$, a unique solution exists.

Step 4: Cofactors: $$A_{11}=75, \quad A_{12}=110, \quad A_{13}=72,$$ $$A_{21}=150, \quad A_{22}=-100, \quad A_{23}=0,$$ $$A_{31}=75, \quad A_{32}=30, \quad A_{33}=-24.$$ Thus, $$\operatorname{adj} A = \begin{bmatrix} 75 & 150 & 75 \\ 110 & -100 & 30 \\ 72 & 0 & -24 \end{bmatrix}.$$

Step 5: Solve for $X$: $$X = \frac{1}{1200}\begin{bmatrix} 75 & 150 & 75 \\ 110 & -100 & 30 \\ 72 & 0 & -24 \end{bmatrix}\begin{bmatrix} 4 \\ 1 \\ 2 \end{bmatrix}$$ $$X = \frac{1}{1200}\begin{bmatrix} 600 \\ 400 \\ 240 \end{bmatrix} = \begin{bmatrix} 1/2 \\ 1/3 \\ 1/5 \end{bmatrix}.$$ So, $$u = \frac{1}{2}, \qquad v = \frac{1}{3}, \qquad w = \frac{1}{5}.$$ Hence, $$x=2, \qquad y=3, \qquad z=5.$$

Answer: $$x=2, \qquad y=3, \qquad z=5.$$

---

Example 6: Using Product of Matrices to Solve a System

Find the product $\begin{bmatrix} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{bmatrix} \begin{bmatrix} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{bmatrix}$. Hence, solve the system of equations: $$x - y + 2z = 1$$ $$2y - 3z = 1$$ $$3x - 2y + 4z = 2$$

Solution: Let $$A = \begin{bmatrix} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{bmatrix}, \qquad C = \begin{bmatrix} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{bmatrix}.$$ Compute the product: $$AC = \begin{bmatrix} 1(-2)+(-1)9+2(6) & 1(0)+(-1)2+2(1) & 1(1)+(-1)(-3)+2(-2) \\ 0(-2)+2(9)+(-3)6 & 0(0)+2(2)+(-3)1 & 0(1)+2(-3)+(-3)(-2) \\ 3(-2)+(-2)9+4(6) & 3(0)+(-2)2+4(1) & 3(1)+(-2)(-3)+4(-2) \end{bmatrix}.$$ This gives $$AC = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = I.$$ So $C = A^{-1}$.

Now the system is $$AX = B, \qquad B = \begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix}.$$ Hence, $$X = A^{-1}B = CB.$$ So, $$X = \begin{bmatrix} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{bmatrix}\begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix} = \begin{bmatrix} -2+0+2 \\ 9+2-6 \\ 6+1-4 \end{bmatrix} = \begin{bmatrix} 0 \\ 5 \\ 3 \end{bmatrix}.$$

Answer: $$x=0, \qquad y=5, \qquad z=3.$$

---

Example 7: Using $A^{-1}$ to Solve a System

If $A = \begin{bmatrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{bmatrix}$, find $A^{-1}$. Using $A^{-1}$, solve the system of equations: $$2x - 3y + 5z = 11$$ $$3x + 2y - 4z = -5$$ $$x + y - 2z = -3$$

Solution: Step 1: Find the determinant: $$|A| = 2\begin{vmatrix} 2 & -4 \\ 1 & -2 \end{vmatrix} - (-3)\begin{vmatrix} 3 & -4 \\ 1 & -2 \end{vmatrix} + 5\begin{vmatrix} 3 & 2 \\ 1 & 1 \end{vmatrix}.$$ So, $$|A| = 2(-4+4) + 3(-6+4) + 5(3-2) = 0 + 3(-2) + 5 = -1.$$ Thus the inverse exists.

Step 2: Cofactors: $$A_{11}=0, \quad A_{12}=2, \quad A_{13}=1,$$ $$A_{21}=-1, \quad A_{22}=-9, \quad A_{23}=-5,$$ $$A_{31}=2, \quad A_{32}=23, \quad A_{33}=13.$$ Therefore, $$\operatorname{adj} A = \begin{bmatrix} 0 & -1 & 2 \\ 2 & -9 & 23 \\ 1 & -5 & 13 \end{bmatrix}.$$ Hence, $$A^{-1} = -1 \cdot \begin{bmatrix} 0 & -1 & 2 \\ 2 & -9 & 23 \\ 1 & -5 & 13 \end{bmatrix} = \begin{bmatrix} 0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13 \end{bmatrix}.$$

Step 3: Let $$B = \begin{bmatrix} 11 \\ -5 \\ -3 \end{bmatrix}.$$ Then $$X = A^{-1}B = \begin{bmatrix} 0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13 \end{bmatrix}\begin{bmatrix} 11 \\ -5 \\ -3 \end{bmatrix} = \begin{bmatrix} -5+6 \\ -22-45+69 \\ -11-25+39 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}.$$

Answer: $$A^{-1} = \begin{bmatrix} 0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13 \end{bmatrix}, \qquad x=1, \; y=2, \; z=3.$$

---

Example 8: Finding $k$ for Unique Solution

Find the value of $k$ for which the system of equations has a unique solution: $$kx + 2y = 5$$ $$3x + y = 1$$

Solution: The coefficient matrix is $$A = \begin{bmatrix} k & 2 \\ 3 & 1 \end{bmatrix}.$$ For a unique solution, we need $$|A| \neq 0.$$ Now, $$|A| = \begin{vmatrix} k & 2 \\ 3 & 1 \end{vmatrix} = k(1) - 3(2) = k-6.$$ So the condition is $$k-6 \neq 0.$$ Therefore, $$k \neq 6.$$

Answer: The system has a unique solution for all real values of $k$ except $k=6$.

---

Example 9: Word Problem into Matrix Form

The sum of three numbers is 6. If we multiply the third number by 3 and add the second number to it, we get 11. By adding first and third numbers, we get double of the second number. Represent it algebraically and find the numbers using matrix method.

Solution: Let the three numbers be $x, y, z$. Then the equations are: $$x+y+z = 6,$$ $$y+3z = 11,$$ $$x+z = 2y \Rightarrow x-2y+z = 0.$$

So, $$A = \begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 3 \\ 1 & -2 & 1 \end{bmatrix}, \qquad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \qquad B = \begin{bmatrix} 6 \\ 11 \\ 0 \end{bmatrix}.$$

Now, $$|A| = 1\begin{vmatrix} 1 & 3 \\ -2 & 1 \end{vmatrix} - 1\begin{vmatrix} 0 & 3 \\ 1 & 1 \end{vmatrix} + 1\begin{vmatrix} 0 & 1 \\ 1 & -2 \end{vmatrix}.$$ Thus, $$|A| = 1(1+6) - 1(0-3) + 1(0-1) = 7+3-1 = 9 \neq 0.$$ So a unique solution exists.

Cofactors are: $$A_{11}=7, \quad A_{12}=3, \quad A_{13}=-1,$$ $$A_{21}=-3, \quad A_{22}=0, \quad A_{23}=3,$$ $$A_{31}=2, \quad A_{32}=-3, \quad A_{33}=1.$$ Hence, $$\operatorname{adj} A = \begin{bmatrix} 7 & -3 & 2 \\ 3 & 0 & -3 \\ -1 & 3 & 1 \end{bmatrix}.$$ Then, $$X = \frac{1}{9}\begin{bmatrix} 7 & -3 & 2 \\ 3 & 0 & -3 \\ -1 & 3 & 1 \end{bmatrix}\begin{bmatrix} 6 \\ 11 \\ 0 \end{bmatrix}$$ $$X = \frac{1}{9}\begin{bmatrix} 42-33+0 \\ 18+0+0 \\ -6+33+0 \end{bmatrix} = \frac{1}{9}\begin{bmatrix} 9 \\ 18 \\ 27 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}.$$

Answer: The numbers are $$1, \quad 2, \quad 3.$$

---

Example 10: Homogeneous System

Solve the system: $$2x + 3y = 0$$ $$4x + 6y = 0$$

Solution: This is a homogeneous system of the form $$AX = O.$$ Here, $$A = \begin{bmatrix} 2 & 3 \\ 4 & 6 \end{bmatrix}, \qquad B = \begin{bmatrix} 0 \\ 0 \end{bmatrix}.$$ Now, $$|A| = 2\cdot 6 - 4\cdot 3 = 12-12 = 0.$$ Since the determinant is zero, we examine $(\operatorname{adj} A)B$.

$$\operatorname{adj} A = \begin{bmatrix} 6 & -3 \\ -4 & 2 \end{bmatrix}.$$ Then, $$(\operatorname{adj} A)B = \begin{bmatrix} 6 & -3 \\ -4 & 2 \end{bmatrix}\begin{bmatrix} 0 \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}.$$ So the system is consistent and has infinitely many solutions.

From the first equation, $$2x + 3y = 0.$$ Let $$y = k.$$ Then, $$2x = -3k \Rightarrow x = -\frac{3k}{2}.$$

Answer: $$x = -\frac{3k}{2}, \qquad y = k,$$ where $k$ is any real number. Also, $x=0, y=0$ is the trivial solution.