Linear Equations, Matrices, Determinants & Cramer's Rule

1. Introduction to Systems of Linear Equations

In Class 11, you learned about linear equations with two variables, like $x+y=5$. In Class 12, we deal with systems of equations. This means we have multiple equations with multiple unknowns (variables) that must be true at the same time.

For example, consider these two equations:

1. $2x+3y=8$
2. $x-y=1$

Here, $x$ and $y$ are unknowns. We need to find the specific values of $x$ and $y$ that satisfy both equations simultaneously.

Matrix Form

To make calculations easier, we use matrices. A system of linear equations can be written as:

$$AX=B$$

Where:

• $A$ is the Coefficient Matrix (contains the numbers multiplying the variables).
• $X$ is the Variable Matrix (contains the unknowns $x,y,z,\dots$).
• $B$ is the Constant Matrix (contains the numbers on the right side of the equals sign).

Example: For the system: $$2x+3y=8$$ $$x-y=1$$

The matrices are: $$A=\left[\begin{array}{cc}2& 3\\ 1& -1\end{array}\right],X=\left[\begin{array}{c}x\\ y\end{array}\right],B=\left[\begin{array}{c}8\\ 1\end{array}\right]$$

So, $\left[\begin{array}{cc}2& 3\\ 1& -1\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}8\\ 1\end{array}\right]$.

2. Determinants

A determinant is a special number calculated from a square matrix. It is denoted by $|A|$ or $\det (A)$.

2x2 Determinant

For a matrix $A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$: $$|A|=ad-bc$$

3x3 Determinant

For a matrix $A=\left[\begin{array}{ccc}a& b& c\\ d& e& f\\ g& h& i\end{array}\right]$, we expand along the first row: $$|A|=a(ei-fh)-b(di-fg)+c(dh-eg)$$

Worked Example 1: Find the determinant of a 3x3 matrix. Let $A=\left[\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\\ 7& 8& 9\end{array}\right]$.

$$|A|=1(5\times 9-6\times 8)-2(4\times 9-6\times 7)+3(4\times 8-5\times 7)$$ $$|A|=1(45-48)-2(36-42)+3(32-35)$$ $$|A|=1(-3)-2(-6)+3(-3)$$ $$|A|=-3+12-9$$ $$|A|=0$$

Since the determinant is 0, this matrix is singular (it does not have an inverse).

3. Conditions for Solutions

The value of the determinant $|A|$ tells us the nature of the solution for the system $AX=B$.

If $|A|=0$, we must check further to see if it is "no solution" or "infinite solutions". This is usually done by checking the consistency of the equations.

4. Cramer's Rule

Cramer's Rule is a method to solve a system of linear equations using determinants. It works only if $|A|\ne 0$.

For a system: $$a_{1}x+b_{1}y+c_{1}z=d_1$$ $$a_{2}x+b_{2}y+c_{2}z=d_2$$ $$a_{3}x+b_{3}y+c_{3}z=d_3$$

The solutions are: $$x=\frac{|A_x|}{|A|},y=\frac{|A_y|}{|A|},z=\frac{|A_z|}{|A|}$$

Where:

• $|A|$ is the determinant of the coefficient matrix.
• $|A_x|$ is the determinant of matrix $A$ with the x-column replaced by the constant matrix $B$.
• $|A_y|$ is the determinant of matrix $A$ with the y-column replaced by $B$.
• $|A_z|$ is the determinant of matrix $A$ with the z-column replaced by $B$.

Worked Example 2: Solve using Cramer's Rule. $$2x+y-z=8$$ $$-3x-y+2z=-11$$ $$-2x+y+2z=-3$$

Step 1: Find $|A|$ $$A=\left[\begin{array}{ccc}2& 1& -1\\ -3& -1& 2\\ -2& 1& 2\end{array}\right]$$ $$|A|=2(-2-2)-1(-6+4)+(-1)(-3-2)$$ $$|A|=2(-4)-1(-2)-1(-5)$$ $$|A|=-8+2+5=-1$$ Since $|A|\ne 0$, a unique solution exists.

Step 2: Find $|A_x|$ Replace column 1 with $B=\left[\begin{array}{c}8\\ -11\\ -3\end{array}\right]$. $$A_x=\left[\begin{array}{ccc}8& 1& -1\\ -11& -1& 2\\ -3& 1& 2\end{array}\right]$$ $$|A_x|=8(-2-2)-1(-22+6)+(-1)(-11-3)$$ $$|A_x|=8(-4)-1(-16)-1(-14)$$ $$|A_x|=-32+16+14=-2$$

Step 3: Find $|A_y|$ Replace column 2 with $B$. $$A_y=\left[\begin{array}{ccc}2& 8& -1\\ -3& -11& 2\\ -2& -3& 2\end{array}\right]$$ $$|A_y|=2(-22+6)-8(-6+4)+(-1)(9-22)$$ $$|A_y|=2(-16)-8(-2)-1(-13)$$ $$|A_y|=-32+16+13=-3$$

Step 4: Find $|A_z|$ Replace column 3 with $B$. $$A_z=\left[\begin{array}{ccc}2& 1& 8\\ -3& -1& -11\\ -2& 1& -3\end{array}\right]$$ $$|A_z|=2(3+11)-1(9-22)+8(-3-2)$$ $$|A_z|=2(14)-1(-13)+8(-5)$$ $$|A_z|=28+13-40=1$$

Step 5: Calculate $x,y,z$ $$x=\frac{|A_x|}{|A|}=\frac{-2}{-1}=2$$ $$y=\frac{|A_y|}{|A|}=\frac{-3}{-1}=3$$ $$z=\frac{|A_z|}{|A|}=\frac{1}{-1}=-1$$

Solution: $x=2,y=3,z=-1$.

5. Gaussian Elimination Method

Cramer's Rule can be time-consuming for large systems. Gaussian Elimination is often faster. It converts the augmented matrix $[A|B]$ into Row Echelon Form (upper triangular form) using row operations.

Row Operations:

1. Swap two rows ($R_i\leftrightarrow R_j$).
2. Multiply a row by a non-zero constant ($k{R}_i$).
3. Add a multiple of one row to another ($R_i+k{R}_j$).

Worked Example 3: Solve using Gaussian Elimination. $$x+y+z=6$$ $$2x+3y+z=14$$ $$x+y-2z=-1$$

Step 1: Write Augmented Matrix $$\left[\begin{array}{cccc}1& 1& 1& 6\\ 2& 3& 1& 14\\ 1& 1& -2& -1\end{array}\right]$$

Step 2: Eliminate $x$ from Row 2 and Row 3 $R_2\to R_2-2R_1$ $R_3\to R_3-R_1$

$$\left[\begin{array}{cccc}1& 1& 1& 6\\ 0& 1& -1& 2\\ 0& 0& -3& -7\end{array}\right]$$

Step 3: Back Substitution From Row 3: $$-3z=-7⟹z=\frac{7}{3}$$

From Row 2: $$y-z=2⟹y-\frac{7}{3}=2⟹y=2+\frac{7}{3}=\frac{13}{3}$$

From Row 1: $$x+y+z=6⟹x+\frac{13}{3}+\frac{7}{3}=6$$ $$x+\frac{20}{3}=6⟹x=6-\frac{20}{3}=\frac{18-20}{3}=-\frac{2}{3}$$

Solution: $x=-2/3,y=13/3,z=7/3$.

6. Matrix Inverse Method

If $A$ is a non-singular matrix, its inverse $A^{-1}$ exists. The solution to $AX=B$ is: $$X=A^{-1}B$$

To find $A^{-1}$ for a 2x2 matrix $A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$: $$A^{-1}=\frac{1}{ad-bc}\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]$$

Worked Example 4: Solve using Inverse Matrix. $$2x+3y=5$$ $$4x+7y=11$$

Step 1: Find $A^{-1}$ $$A=\left[\begin{array}{cc}2& 3\\ 4& 7\end{array}\right]$$ $$|A|=(2)(7)-(3)(4)=14-12=2$$ $$A^{-1}=\frac{1}{2}\left[\begin{array}{cc}7& -3\\ -4& 2\end{array}\right]=\left[\begin{array}{cc}3.5& -1.5\\ -2& 1\end{array}\right]$$

Step 2: Multiply $A^{-1}B$ $$B=\left[\begin{array}{c}5\\ 11\end{array}\right]$$ $$X=\left[\begin{array}{cc}3.5& -1.5\\ -2& 1\end{array}\right]\left[\begin{array}{c}5\\ 11\end{array}\right]$$ $$x=(3.5\times 5)+(-1.5\times 11)=17.5-16.5=1$$ $$y=(-2\times 5)+(1\times 11)=-10+11=1$$

Solution: $x=1,y=1$.

Comparison of Methods

Applications

1. Engineering: Solving electrical circuits (Kirchhoff's laws).
2. Economics: Input-Output models (Leontief model) to determine production levels.
3. Computer Graphics: Transforming images (rotation, scaling) using matrices.
4. Physics: Calculating forces in trusses and bridges.

Exam Tip

1. Check the Determinant First: Always calculate $|A|$ before choosing a method. If $|A|=0$, do not use Cramer's Rule or Inverse Method. Use Gaussian Elimination to check for consistency.
2. Sign Errors: In Cramer's Rule and Determinants, the most common mistake is sign errors. Double-check your expansion signs $(+-+)$.
3. Verification: After finding the solution, substitute the values back into the original equations to verify. This is a quick way to catch calculation mistakes.
4. NEB Pattern: NEB often asks to "Solve the system using Cramer's Rule" or "Show that the system has no solution." Be prepared to prove consistency/inconsistency by comparing ranks of matrices or using row reduction.
5. Augmented Matrix: Always write the augmented matrix $[A|B]$ clearly. It helps organize your row operations.