System Of Linear Equations

MathsMatricesFor JEE aspirants

A system of $n$ linear equations in $n$ unknowns can be written in matrix form as $A X = B$ , where $A$ is the coefficient matrix, $X$ is the column of unknowns, and $B$ is the column of constants. In JEE Mathematics, when $∣ A ∣ \neq = 0$ the system has a unique solution $X = A^{- 1} B$ ; when $∣ A ∣ = 0$ , the system is either inconsistent (no solution) or consistent with infinitely many solutions, depending on the value of $(adj A) B$ . This concept covers the matrix method, all three consistency cases, and an important geometric application - the matrix of rotation of axes.

Key Formulas - Quick Reference

Matrix form of $n$ equations in $n$ unknowns: $A X = B$ , where $A$ is $n \times n$ , $X$ is $n \times 1$ , $B$ is $n \times 1$ .
Unique solution (when $∣ A ∣ \neq = 0$ ): $X = A^{- 1} B = \frac{1}{∣ A ∣} adj (A) B$ .
Consistency criteria (for square systems): unique solution if $∣ A ∣ \neq = 0$ ; inconsistent if $∣ A ∣ = 0$ and $(adj A) B \neq = O$ ; infinite solutions if $∣ A ∣ = 0$ and $(adj A) B = O$ .
Rotation of axes through angle $θ$ : $[x y] = [cos θ sin θ - sin θ cos θ] [X Y]$ .
Rotation matrix $R (θ) = [cos θ sin θ - sin θ cos θ]$ is orthogonal with $∣ R ∣ = 1$ ; its inverse is $R (- θ) = R^{T}$ .

1. Matrix Form of a Linear System

Consider the following system of $n$ linear equations in $n$ unknowns:

$a_{11} x_{1} + a_{12} x_{2} + \dots + a_{1 n} x_{n} = b_{1}$

$a_{21} x_{1} + a_{22} x_{2} + \dots + a_{2 n} x_{n} = b_{2}$

$⋮$

$a_{n 1} x_{1} + a_{n 2} x_{2} + \dots + a_{nn} x_{n} = b_{n}$

This system can be written in matrix form as:

$a_{11} a_{21} ⋮ a_{n 1} a_{12} a_{22} ⋮ a_{n 2} \dots \dots ⋱ \dots a_{1 n} a_{2 n} ⋮ a_{nn} x_{1} x_{2} ⋮ x_{n} = b_{1} b_{2} ⋮ b_{n}$

or compactly $A X = B$ . Here $A$ is the coefficient matrix, $X$ is the column of unknowns, and $B$ is the column of constants.

2. Solving by Matrix Inverse

If $∣ A ∣ \neq = 0$ , then $A^{- 1}$ exists, and premultiplying both sides of $A X = B$ by $A^{- 1}$ gives:

$X = A^{- 1} B$

Using the formula $A^{- 1} = \frac{1}{∣ A ∣} adj (A)$ derived in the Algebra Of Matrices concept:

$X = \frac{1}{∣ A ∣} adj (A) B$

This gives the unique solution vector, provided $∣ A ∣ \neq = 0$ .

Step-by-Step Method

Write the system in matrix form $A X = B$ .
Compute $∣ A ∣$ . If $∣ A ∣ = 0$ , go to Section 3 (consistency analysis); otherwise proceed.
Compute the cofactor matrix, then transpose it to get $adj (A)$ .
Form $A^{- 1} = adj (A) /∣ A ∣$ .
Multiply: $X = A^{- 1} B$ . Read off $x_{1}, x_{2}, \dots, x_{n}$ from the column vector $X$ .

3. Consistency of the System

The classification depends on $∣ A ∣$ and the product $(adj A) B$ :

Condition	Type	Number of solutions
$∣ A ∣ \neq = 0$	Consistent, independent	Unique solution $X = A^{- 1} B$
$∣ A ∣ = 0$ and $(adj A) B \neq = O$	Inconsistent	No solution
$∣ A ∣ = 0$ and $(adj A) B = O$	Consistent, dependent	Infinitely many solutions

Why this test works: multiplying

A X = B

adj (A)

gives

∣ A ∣ X = (adj A) B

. When

∣ A ∣ = 0

, the left side is zero regardless of

X

, so consistency requires

(adj A) B = O

. When that also holds, the system reduces to fewer independent equations than unknowns, yielding infinitely many solutions.

Solved Example 1

Problem: With the help of matrices, solve the equations

3 x + y + 2 z = 3

2 x - 3 y - z = - 3

x + 2 y + z = 4

Solution:

Step 1: Matrix form. Write $A X = B$ where

$A = 321 1 - 3 2 2 - 1 1, X = x y z, B = 3 - 3 4$

Step 2: Compute $∣ A ∣$ . Expand along $R_{1}$ :

$∣ A ∣ = 3 [(- 3) (1) - (- 1) (2)] - 1 [(2) (1) - (- 1) (1)] + 2 [(2) (2) - (- 3) (1)]$

$= 3 (- 3 + 2) - 1 (2 + 1) + 2 (4 + 3) = - 3 - 3 + 14 = 8 \neq = 0$ .

Since $∣ A ∣ \neq = 0$ , the system has a unique solution.

Step 3: Compute the cofactors.

$C_{11} = - 3 + 2 = - 1$ ; $C_{12} = - (2 + 1) = - 3$ ; $C_{13} = 4 + 3 = 7$ .

$C_{21} = - (1 - 4) = 3$ ; $C_{22} = 3 - 2 = 1$ ; $C_{23} = - (6 - 1) = - 5$ .

$C_{31} = - 1 + 6 = 5$ ; $C_{32} = - (- 3 - 4) = 7$ ; $C_{33} = - 9 - 2 = - 11$ .

Cofactor matrix $C = - 1 35 - 3 17 7 - 5 - 11$ .

Step 4: Adjoint = $C^{T}$ :

$adj (A) = - 1 - 3 7 31 - 5 57 - 11$

Step 5: Inverse.

$A^{- 1} = \frac{1}{8} - 1 - 3 7 31 - 5 57 - 11$

Step 6: Solution $X = A^{- 1} B$ .

$A^{- 1} B = \frac{1}{8} - 3 - 9 + 20 - 9 - 3 + 28 21 + 15 - 44 = \frac{1}{8} 816 - 8 = 12 - 1$

Hence $x = 1$ , $y = 2$ , $z = - 1$ .

Solved Example 2

Problem: Show that the system

x + 2 y + 3 z = 4

2 x + 3 y + 4 z = 5

3 x + 4 y + 5 z = 6

has infinitely many solutions.

Solution:

Coefficient matrix $A = 123234345$ ; constants $B = 456$ .

Observe: each row of $A$ is in A.P. with common difference $1$ , and each row (including $B$ ) forms an A.P. across the augmented matrix as well. By Property 9 of determinants, $∣ A ∣ = 0$ .

Also, expanding $∣ A ∣$ : $1 (15 - 16) - 2 (10 - 12) + 3 (8 - 9) = - 1 + 4 - 3 = 0$ . ✓

Now check $(adj A) B$ . Since each equation is a linear combination of the others (equation 2 is the average of 1 and 3 on both sides), the constants are consistent with the coefficient structure, giving $(adj A) B = O$ .

Concretely, subtracting equation 1 from equation 2: $x + y + z = 1$ . Subtracting equation 2 from equation 3: $x + y + z = 1$ . Both reduce to the same relation, so effectively there are only two independent equations in three unknowns: infinitely many solutions.

Parametrising: let $z = t$ . Then $x + y + t = 1$ and $x + 2 y + 3 t = 4$ give $y = 3 - 2 t$ (from subtraction), and $x = 1 - y - t = 1 - (3 - 2 t) - t = - 2 + t$ .

General solution: $(x, y, z) = (- 2 + t, 3 - 2 t, t)$ for any $t \in R$ .

4. Application: Rotation of Axes

An important geometric application of matrices is coordinate transformation. If the $x$ - and $y$ -axes are rotated through an angle $θ$ about the origin, the new coordinates $(X, Y)$ and old coordinates $(x, y)$ are related by:

$x = X cos θ - Y sin θ$

$y = X sin θ + Y cos θ$

In matrix form:

$[x y] = [cos θ sin θ - sin θ cos θ] [X Y]$

The matrix $R (θ) = [cos θ sin θ - sin θ cos θ]$ is called the rotation matrix through angle $θ$ .

Properties of the Rotation Matrix

$∣ R (θ) ∣ = cos^{2} θ + sin^{2} θ = 1$ , so $R (θ)$ is non-singular for every $θ$ .
$R (θ)$ is orthogonal: $R (θ)^{T} R (θ) = I$ . Verify by direct computation.
Inverse of a rotation is a rotation in the opposite direction: $R (θ)^{- 1} = R (- θ) = R (θ)^{T}$ .
Composition of rotations: $R (α) \cdot R (β) = R (α + β)$ - a matrix version of the angle-addition identities for $sin$ and $cos$ .

Solved Example 3

Problem: Show that the rotation matrix

R (θ) = [cos θ sin θ - sin θ cos θ]

is orthogonal, and hence find

R (θ)^{- 1}

Solution:

Compute $R^{T}$ :

$R^{T} = [cos θ - sin θ sin θ cos θ]$

Compute $R^{T} R$ :

$R^{T} R = [cos^{2} θ + sin^{2} θ - sin θ cos θ + cos θ sin θ - cos θ sin θ + sin θ cos θ sin^{2} θ + cos^{2} θ] = [1001] = I$

Similarly $R R^{T} = I$ . So $R$ is orthogonal, and the inverse of an orthogonal matrix is its transpose:

$R^{- 1} = R^{T} = [cos θ - sin θ sin θ cos θ] = R (- θ)$

This confirms that undoing a rotation by $θ$ is the same as rotating by $- θ$ .

Common Mistakes to Avoid

Watch out

Trying to solve when $∣ A ∣ = 0$ using $A^{- 1}$ . The inverse does not exist when $∣ A ∣ = 0$ . Switch to consistency analysis via $(adj A) B$ or Cramer's rule (from the Determinants concept).
Confusing $X = A^{- 1} B$ with $X = B A^{- 1}$ . Matrix multiplication is not commutative, and $A^{- 1} B$ is generally different from $B A^{- 1}$ . The correct formula from $A X = B$ is $X = A^{- 1} B$ (premultiplication).
Forgetting to check the consistency conditions before concluding "no solution". When $∣ A ∣ = 0$ , the system might have no solution OR infinitely many solutions - do not conclude one without checking $(adj A) B$ .
Applying the matrix method to non-square systems. The inverse method works only for $n$ equations in $n$ unknowns. For $m$ equations in $n$ unknowns with $m \neq = n$ , use rank-based analysis or Gaussian elimination.
Confusing the rotation matrix with its transpose. $R (θ)$ has $- sin θ$ in the top-right; $R (θ)^{T} = R (- θ)$ has $+ sin θ$ there. Swapping these accidentally produces a rotation in the wrong direction.
Forgetting Cramer's rule as an alternative. For $3 \times 3$ systems in JEE MCQs, Cramer's rule is often faster than computing the full inverse. Use the matrix-inverse method when the problem specifically asks for $A^{- 1}$ or when the same coefficient matrix is used to solve for multiple right-hand sides.

Frequently Asked Questions

When should I use the matrix inverse method versus Cramer's rule?

Use Cramer's rule for a one-off $3 \times 3$ system in an MCQ - it is often faster by hand. Use the matrix inverse method when the problem asks for $A^{- 1}$ explicitly, when you need to solve $A X = B$ for multiple different $B$ (the inverse is computed once and reused), or when the problem uses matrix notation throughout.

How do I check if a system has infinitely many solutions?

For a square system $A X = B$ : if $∣ A ∣ = 0$ AND $(adj A) B = O$ (zero column vector), the system has infinitely many solutions. In practice, this often manifests as one equation being a linear combination of the others.

What is the geometric picture of the three cases?

For three equations in three unknowns, each equation represents a plane. Unique solution: three planes intersect in a single point. No solution: the three planes have no common point (e.g. two parallel planes cut by a third). Infinitely many solutions: three planes share a common line or coincide.

Why is the rotation matrix orthogonal?

Because rotation preserves lengths and angles - it is a rigid motion. Algebraically, the columns of $R (θ)$ are unit vectors that are orthogonal to each other (they are the images of the standard basis vectors under rotation), which is exactly the definition of an orthogonal matrix.

Can the matrix method solve non-square systems?

Not directly with the inverse formula, which requires $A$ to be square and non-singular. For non-square systems ( $m$ equations, $n$ unknowns, $m \neq = n$ ), use rank-based methods: compare $rank (A)$ with $rank ([A ∣ B])$ to classify the system as consistent or not.

What does $det (A^{- 1}) = 1/ det (A)$ mean geometrically?

Determinants measure volume scaling. If $A$ scales volumes by a factor $∣ A ∣$ , then $A^{- 1}$ must scale them back by $1/∣ A ∣$ to recover the original. The product $A \cdot A^{- 1} = I$ scales volumes by $∣ A ∣ \cdot 1/∣ A ∣ = 1$ , i.e. leaves them unchanged.

How do rotations compose in matrix form?

$R (α) \cdot R (β) = R (α + β)$ . Rotating by $β$ and then by $α$ is the same as rotating by $α + β$ . Expanding the matrix product recovers the angle-addition identities for $sin$ and $cos$ - a nice example of how algebra of matrices encodes geometric structure.

What is the difference between rotation of axes and rotation of points?

Rotation of axes keeps the geometric point fixed while rotating the coordinate frame - the point gets new coordinates. Rotation of points moves the point while keeping the coordinate frame fixed. The two operations are inverses of each other: rotating axes by $θ$ has the same coordinate effect as rotating the point by $- θ$ .

Previous year questions on System Of Linear Equations

4 questions from past papers, each with a step-by-step solution.

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1. Matrix Form of a Linear System

2. Solving by Matrix Inverse

Step-by-Step Method

3. Consistency of the System

4. Application: Rotation of Axes

Properties of the Rotation Matrix

Common Mistakes to Avoid

Frequently Asked Questions

When should I use the matrix inverse method versus Cramer's rule?

How do I check if a system has infinitely many solutions?

What is the geometric picture of the three cases?

Why is the rotation matrix orthogonal?

Can the matrix method solve non-square systems?

What does det(A−1)=1/det(A) mean geometrically?

How do rotations compose in matrix form?

What is the difference between rotation of axes and rotation of points?

Previous year questions on System Of Linear Equations

Ready to master Matrices?

What does $det (A^{- 1}) = 1/ det (A)$ mean geometrically?