Algebra

Solving simultaneous equations using inverse of matrix

Systems of linear equations are succinctly expressed in matrix form as $AX = B$, yielding unique solutions via the matrix inverse formula $X = A^{-1}B$ whenever $\det(A) \neq 0$.

For: UP PGT Mathematics UP TGT Mathematics DSSSB PGT / TGT KVS / NVS PGT LT Grade

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Definition and statement

Matrix Inversion Theorem for Linear Systems

Let a system of $n$ linear equations in $n$ variables be represented in matrix form as $AX = B$, where $A$ is an $n \times n$ coefficient matrix, $X$ is the $n \times 1$ column vector of variables, and $B$ is the $n \times 1$ constant column vector. If $A$ is a non-singular matrix (i.e., $\det(A) \neq 0$), then the system possesses a unique solution given explicitly by $X = A^{-1}B$, where $A^{-1} = \frac{\text{adj}(A)}{\det(A)}$.

📐 Geometric Interpretation

Geometrically, each linear equation in two variables represents a straight line, and in three variables, a plane. A non-singular coefficient matrix implies that these lines or planes intersect at a unique, distinct point in $\mathbb{R}^n$, representing the unique coordinate solution $(x, y, z)$.

⚙️ Algebraic & Structural Meaning

It establishes an isomorphism between solving systems of linear equations and algebraic matrix operations, reducing simultaneous equations to the computation of determinants, adjugates, and matrix-vector multiplications.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition	Domain / Interval	Necessity	What Fails If Condition Dropped
Square Matrix Condition ($n \times n$)	`Matrix dimension domain`	Strictly Mandatory / Necessary	If the number of equations does not equal the number of variables, matrix $A$ is non-square, making the standard inverse $A^{-1}$ undefined.
Non-Singularity Condition ($\det(A) \neq 0$)	`Determinant domain $\mathbb{R} \setminus \{0\}$`	Strictly Mandatory	If $\det(A) = 0$, the matrix is singular, $A^{-1}$ does not exist, and the system either has infinitely many solutions or no solution (inconsistent).
Field of Entries ($\mathbb{R}$ or $\mathbb{C}$)	`Scalar field`	Essential for conclusion	Elements of $A$ and $B$ must belong to a defined field where division by $\det(A)$ is valid.

Useful results and shortcuts

Use these identities and results when their conditions apply:

Inverse of $2 \times 2$ Matrix

$$ \text{If } A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}, \text{ then } A^{-1} = \frac{1}{ad-bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} $$

Rule: Swap the principal diagonal elements, change the signs of the off-diagonal elements, and divide by the determinant $ad - bc$.

Cramer's Rule Equivalence

$$ x_i = \frac{\det(A_i)}{\det(A)} $$

Rule: When only one specific variable (e.g., $y$ or $z$) is required from a $3 \times 3$ system, use Cramer's rule instead of finding the full inverse $A^{-1}$.

Inconsistency / Infinite Solutions Test

$$ \text{If } \det(A) = 0 \text{ and } ((\text{adj } A)B = O \implies \text{ infinitely many solutions}; \text{ else } \text{ no solution}) $$

Rule: Check $\det(A)$ first. If zero, immediately evaluate $(\text{adj } A)B$ to classify consistency without attempting to invert.

Inverse of Diagonal Matrix

$$ \text{If } A = \text{diag}(d_1, d_2, \dots, d_n), \text{ then } A^{-1} = \text{diag}(d_1^{-1}, d_2^{-1}, \dots, d_n^{-1}) $$

Rule: Reciprocate each diagonal element instantly if the system matrix is diagonal.

Application: Used in decoupled system equations in higher linear algebra sections.

Inverse of Orthogonal Matrix

$$ \text{If } A^T A = I, \text{ then } A^{-1} = A^T $$

Rule: If the coefficient matrix is orthogonal, the inverse is simply its transpose.

Application: Appears in advanced matrix transformation items in UP PGT.

Rapid Option Verification

$$ \text{Substitute given options } (x, y, z) \text{ directly into equations} $$

Rule: For 4-option multiple choice questions, back-substitute the options into the equations if determinant calculation is cumbersome.

Common errors and counterexamples

These examples show how a conclusion can fail when a condition is missing:

⚠️ COMMON ERROR #01

Assuming Unique Solution when $\det(A) = 0$

Common Pitfall: Candidates attempt to write $X = A^{-1}B$ even when $\det(A) = 0$, leading to division by zero errors.

Counterexample: $x + y = 1$ and $2x + 2y = 3$, where $\det(A) = 0$

Why it fails: Matrix $A$ is singular; lines are parallel and distinct, resulting in an inconsistent system with zero solutions.

🛡️ Exam Defense: Always evaluate $\det(A)$ first before attempting any inversion.

⚠️ COMMON ERROR #02

Confusing Homogeneous and Non-Homogeneous Systems

Common Pitfall: Applying $X = A^{-1}B$ directly to homogeneous systems ($B = O$) and claiming a unique non-zero solution when $\det(A) = 0$.

Counterexample: $AX = O$ with $\det(A) = 0$

Why it fails: When $B = O$ and $\det(A) = 0$, the system has infinitely many solutions (trivial solution plus non-trivial solutions), not no solution.

🛡️ Exam Defense: For homogeneous systems $AX = O$: $\det(A) \neq 0 \implies$ unique trivial solution ($X=O$); $\det(A) = 0 \implies$ infinitely many solutions.

Comparison

Concept / Theorem	Hypotheses / Assumptions	Key Formula / Outcome	Geometric Meaning	Exam Identification Clue
Unique Solution	$\det(A) \neq 0$	$X = A^{-1}B$	Intersecting planes/lines at a single point	'Find the value of $x, y, z$'
No Solution (Inconsistent)	$\det(A) = 0$ and $(\text{adj } A)B \neq O$	No inverse; parallel planes	Parallel non-intersecting planes	'System is inconsistent for $k = ?$'
Infinite Solutions (Dependent)	$\det(A) = 0$ and $(\text{adj } A)B = O$	Parameter-dependent solution family	Coincident planes	'System has non-trivial solutions'
Homogeneous System	$B = O$, square matrix $A$	$\det(A) = 0$ for non-trivial solutions	Planes passing through the origin	'Find values of $\lambda$ for non-zero solution'

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

If the system of equations $x + y + z = 6$, $x + 2y + 3z = 14$, and $x + 4y + 7z = 30$ is solved using matrix inverse, what is the value of $y$?

(A) 1(B) 2(C) 3(D) 4

📚 Method A: Formal Textbook Derivation

Step 1: Write $AX = B$ where $A = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 4 & 7 \end{pmatrix}$. Step 2: Compute $\det(A) = 1(14-12) - 1(7-3) + 1(4-2) = 2 - 4 + 2 = 0$. Wait, let us check: $\det(A) = 1(14-12) - 1(7-3) + 1(4-2) = 2 - 4 + 2 = 0$. Since $\det(A) = 0$, check $(\text{adj } A)B$. Let us use Cramer's rule or back-substitution: From $x+y+z=6$, $x+2y+3z=14$, subtracting first from second gives $y+2z=8$. Subtracting second from third: $2y+4z=16$, which is identical. Thus $y = 8 - 2z$. For $z=2$, $y=4$; for $z=3$, $y=2$. Let us verify with options: if $y=3$, $2z = 5$ (not integer). If $y=2$, $2z=6 \implies z=3$. Then $x + 2 + 3 = 6 \implies x=1$. Check equation 3: $1 + 4(2) + 7(3) = 1 + 8 + 21 = 30$ (Valid!).

⚡ Method B: 25-Second Exam Shortcut

Use Cramer's Rule for $y$: $y = \frac{\det(A_y)}{\det(A)}$ or direct option back-substitution in 15 seconds.

💡 Tip: Always test options directly in linear systems if $\det(A) = 0$ or if full inversion is tedious.

CANONICAL PROBLEM 02 Dual-Method Breakdown

For what value of $k$ does the system of equations $x + y + z = 1$, $x + 2y + 3z = 2$, $x + 2y + kz = 4$ have no solution?

(A) $k = 3$(B) $k \neq 3$(C) $k = 0$(D) $k = 1$

📚 Method A: Formal Textbook Derivation

Step 1: Form coefficient matrix $A = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 2 & k \end{pmatrix}$. Step 2: Set $\det(A) = 0$ for singularity. $\det(A) = 1(2k - 6) - 1(k - 3) + 1(2 - 2) = 2k - 6 - k + 3 = k - 3$. Setting $k - 3 = 0 \implies k = 3$. Step 3: For $k = 3$, check $(\text{adj } A)B$. The third equation becomes $x + 2y + 3z = 4$, while the second is $x + 2y + 3z = 2$, which is a contradiction ($2 = 4$). Hence, no solution exists at $k = 3$.

⚡ Method B: 25-Second Exam Shortcut

Set $\det(A) = 0$ to find critical parameter $k = 3$, then verify inconsistency.

💡 Tip: A value of $k$ making $\det(A)=0$ almost always corresponds to 'no solution' or 'infinite solutions' in UP PGT questions.

Practice questions

Choose an answer to check it and see the solution:

If $AX = B$ is a system of $n$ linear equations in $n$ variables and $\det(A) \neq 0$, then the system has:

A No solution

B Unique solution

C Infinitely many solutions

D Exactly two solutions

Derivation & Analysis:

By the Matrix Inversion Theorem, if $\det(A) \neq 0$, $A^{-1}$ exists uniquely, yielding the unique solution $X = A^{-1}B$.

The system of equations $kx + y = 1$ and $x + ky = 1$ has infinitely many solutions if:

A $k = 1$

B $k = -1$

C $k = 2$

D $k = 0$

Derivation & Analysis:

For infinite solutions, the determinant of the augmented and coefficient matrices must vanish, and rank condition requires $k = 1$. When $k=1$, both equations reduce to $x+y=1$.

If $A = \begin{pmatrix} 2 & -3 \\ 1 & 2 \end{pmatrix}$, then $A^{-1}$ is:

A $\frac{1}{7} \begin{pmatrix} 2 & 3 \\ -1 & 2 \end{pmatrix}$

B $\frac{1}{5} \begin{pmatrix} 2 & 3 \\ -1 & 2 \end{pmatrix}$

C $\frac{1}{7} \begin{pmatrix} 2 & -3 \\ 1 & 2 \end{pmatrix}$

D $\begin{pmatrix} 2 & 3 \\ -1 & 2 \end{pmatrix}$

Derivation & Analysis:

Determinant $\det(A) = (2)(2) - (-3)(1) = 4 + 3 = 7$. Adjugate swaps diagonal and negates off-diagonal: $\begin{pmatrix} 2 & 3 \\ -1 & 2 \end{pmatrix}$. Thus $A^{-1} = \frac{1}{7} \begin{pmatrix} 2 & 3 \\ -1 & 2 \end{pmatrix}$.

A homogeneous system of linear equations $AX = O$ with $A$ being a $3 \times 3$ matrix has a non-trivial solution if:

A $\det(A) \neq 0$

B $\det(A) = 0$

C $\text{adj}(A) = O$

D $A = I$

Derivation & Analysis:

A homogeneous system has non-trivial (non-zero) solutions if and only if the coefficient matrix is singular, i.e., $\det(A) = 0$.

If the system $x + 2y = 3$ and $2x + 4y = c$ has infinitely many solutions, the value of $c$ is:

A 3

B 6

C 0

D 9

Derivation & Analysis:

For infinite solutions, the equations must be identical multiples. Multiplying $x+2y=3$ by 2 gives $2x+4y=6$, hence $c = 6$.

If $A$ is a non-singular square matrix of order $3$, then $\det(A^{-1})$ is equal to:

A $\det(A)$

B $(\det(A))^{-1}$

C $3 \det(A)$

D $0$

Derivation & Analysis:

Since $AA^{-1} = I$, taking determinants gives $\det(A)\det(A^{-1}) = 1$, so $\det(A^{-1}) = \frac{1}{\det(A)} = (\det(A))^{-1}$.

If $\det(A) = 0$ for a system $AX = B$, what can be deduced about the system?

A Always inconsistent

B Always consistent with infinite solutions

C Either inconsistent or consistent with infinite solutions

D Unique solution exists

Derivation & Analysis:

When $\det(A) = 0$, $A^{-1}$ does not exist. Depending on $(\text{adj } A)B$, the system is either inconsistent (no solution) or has infinitely many solutions.

For the matrix equation $\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 5 \\ 11 \end{pmatrix}$, the value of $x$ is:

A 1

B 2

C 3

D 0

Derivation & Analysis:

Using inverse or elimination: $x + 2y = 5$ and $3x + 4y = 11$. From $x = 5 - 2y$, substitute into second: $3(5-2y) + 4y = 11 \implies 15 - 2y = 11 \implies 2y = 4 \implies y = 2$. Then $x = 5 - 4 = 1$.

If $A$ is a symmetric matrix, then $A^{-1}$ (when it exists) is:

A Skew-symmetric

B Symmetric

C Diagonal

D Identity

Derivation & Analysis:

Since $(A^{-1})^T = (A^T)^{-1}$ and $A^T = A$ for a symmetric matrix, $(A^{-1})^T = A^{-1}$, proving $A^{-1}$ is symmetric.

Q10

In solving $AX = B$ via matrix inverse, the computation of $A^{-1}$ primarily relies on:

A Eigenvalues only

B Adjugate matrix and determinant

C Trace of matrix

D Transpose only

Derivation & Analysis:

The explicit formula for the inverse is $A^{-1} = \frac{1}{\det(A)} \text{adj}(A)$, which relies directly on the determinant and the adjugate matrix.

Common questions

No. The matrix inverse $A^{-1}$ is defined exclusively for square matrices ($n \times n$). For rectangular systems (more equations than variables or vice versa), techniques like generalized inverses or Gaussian elimination are required.

Candidates frequently fail to check whether $\det(A) = 0$ before attempting to compute $A^{-1}$, leading to wasted time or incorrect conclusions on inconsistent/dependent systems.

For 3-variable systems, back-substituting options or using Cramer's rule for a single variable is significantly faster than calculating a full $3 \times 3$ adjugate and inverse matrix.

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