MHT-CET Maths · Determinants and Matrices

Systems of Linear Equations and Symmetric, Skew-Symmetric Matrices

AX = B is solved by elimination or by X = A⁻¹B; a homogeneous system has non-trivial solutions exactly when |A| = 0; and any square matrix splits uniquely into a symmetric plus a skew-symmetric part.

All Determinants and Matrices notes MHT-CET Maths strategy

Why this matters

14 PYQs at 29% HARD. The 3 × 3 system AX = B has been set every year — always with small integer solutions, so elimination beats the inverse — and the answer is usually a combination like 2a − 3b + 4c or x² + y² + z², so the solving must be complete. The HARD ones are the classification questions: a homogeneous system with a parameter (non-trivial solutions need a vanishing determinant) and a skew-symmetric coefficient matrix, which is singular whenever its order is odd.

Concept 1 of 3: Solving AX = B: Elimination, or X = A⁻¹B

AX = B

is three linear equations written compactly. With small integer coefficients, subtracting equations is faster than building

A^{-1}

; the matrix form is there to make the question look harder than it is.

Definition

Read the rows of $A$ as equations: $A = \begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & 3 \\ 1 & -2 & 1 \end{pmatrix}$ , $B = \begin{pmatrix} 6 \\ 11 \\ 0 \end{pmatrix}$ is $a + b + c = 6$ , $b + 3c = 11$ , $a - 2b + c = 0$ .
Eliminate: subtract rows to isolate one unknown, back-substitute. Here $R_1 - R_3$ : $3b = 6$ , $b = 2$ ; then $c = 3$ , $a = 1$ .
Matrix method when $|A| \ne 0$ : $X = A^{-1}B$ — correct but slow by hand for $3 \times 3$ ; use it only when $A^{-1}$ is given.
Answer the combination asked ( $2a + b + 2c = 10$ , $x^2 + y^2 + z^2 = 14$ ); the solution triple itself is rarely an option.
A unique solution exists iff $|A| \ne 0$ . If $|A| = 0$ , the system has either no solution or infinitely many.

Linear system in matrix form

AX = B,\quad |A| \ne 0 \ \Rightarrow\ X = A^{-1}B \ \text{(unique)}

Worked example

Solve

\begin{pmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 4 & 9 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 3 \\ 6 \\ 14 \end{pmatrix}

and find

x + 2y + 3z

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 9th May Shift 2 · Q143Moderate

Example 1 · Determinants and Matrices · Systems of Linear Equations and Symmetric, Skew-Symmetric Matrices

Let

A=\begin{bmatrix}1&1&1\\0&1&3\\1&-2&1\end{bmatrix},\ B=\begin{bmatrix}6\\11\\0\end{bmatrix}

and

X=\begin{bmatrix}a\\b\\c\end{bmatrix}

, if

AX=B

, then the value of

2a+b+2c

Trusting a solution without checking every equation

One sitting's stored key gave

(2, 1, 1)

for a system whose first equation that triple fails. Substitute the solution into ALL three equations before choosing — it takes ten seconds and catches both your slips and the paper's.

Drill 4 more on solving ax = b: elimination, or x = a⁻¹b

Concept 2 of 3: Homogeneous Systems: Non-Trivial Solutions Need |A| = 0

AX = O

always has the trivial solution

X = O

. If

|A| \ne 0

that is the only one; if

|A| = 0

there are infinitely many. So 'find

\lambda

so that

x, y, z

are not all zero' means 'set the determinant to zero'.

Definition

$AX = O$ : unique (trivial) solution iff $|A| \ne 0$ ; infinitely many (non-trivial) iff $|A| = 0$ . A homogeneous system is never inconsistent.
A vector equation $\overline{a} + \overline{b} - \overline{c} = \overline{0}$ with $\overline{a}, \overline{b}, \overline{c}$ linear in $x, y, z$ is three homogeneous equations in $x, y, z$ : collect the $\hat{i}, \hat{j}, \hat{k}$ components as rows, then set the $3 \times 3$ determinant to $0$ and solve for the parameter.
For the rows $(\lambda, 1, 1)$ , $(1, 1, 2)$ , $(\lambda + 1, 3, 4)$ : the determinant is $-\lambda$ , so $\lambda = 0$ .
Non-homogeneous $AX = B$ with $|A| = 0$ needs the extra test (consistency): either no solution or infinitely many.

Homogeneous system

AX = O \text{ has a non-trivial solution} \iff |A| = 0

Worked example

Find

k

x + 2y + z = 0

2x + y + kz = 0

x - y + 2z = 0

has a non-trivial solution.

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 26 April Shift I · Q130Hard

Example 2 · Determinants and Matrices · Systems of Linear Equations and Symmetric, Skew-Symmetric Matrices

\overline{a}=\lambda x\hat{i}+y\hat{j}+4z\hat{k}

\overline{b}=y\hat{i}+x\hat{j}+3y\hat{k}

and

\overline{c}=-z\hat{i}-2z\hat{j}-(\lambda+1)x\hat{k}

are the sides of the triangle ABC, where

x,y,z

are not all zero, such that

\overline{a}+\overline{b}-\overline{c}=\overline{0}

, then the value of

\lambda

Reading |A| = 0 as 'no solution'

For a homogeneous system a zero determinant means infinitely many solutions, never none —

X = O

always works. 'No solution' is only possible for

AX = B

with

B \ne O

Concept 3 of 3: Symmetric + Skew-Symmetric: The Unique Split, and Why Odd-Order Skew Is Singular

Any square matrix

M

is half of

M + M^T

(symmetric) plus half of

M - M^T

(skew-symmetric). A skew-symmetric matrix has zeros on its diagonal and, in odd order, determinant

0

— so any system it governs has infinitely many solutions.

Definition

$M = \underbrace{\dfrac{M + M^T}{2}}_{\text{symmetric}} + \underbrace{\dfrac{M - M^T}{2}}_{\text{skew-symmetric}}$ , uniquely. For $M = \begin{pmatrix} 1 & t \\ -t & 1 \end{pmatrix}$ : symmetric part $I$ , skew part $\begin{pmatrix} 0 & t \\ -t & 0 \end{pmatrix}$ .
Skew-symmetric $B$ : $B^T = -B$ , diagonal entries $0$ . For odd $n$ , $|B| = |B^T| = |-B| = (-1)^n|B| = -|B|$ , so $|B| = 0$ .
Products: if $A$ is symmetric and $B$ skew-symmetric, then $A^2$ and $B^2$ are both symmetric, and $A^2B^2 - B^2A^2$ is skew-symmetric — so for $3 \times 3$ it is singular and $(A^2B^2 - B^2A^2)X = O$ has infinitely many solutions.
Inverse of the $2 \times 2$ skew part: $\begin{pmatrix} 0 & t \\ -t & 0 \end{pmatrix}^{-1} = \begin{pmatrix} 0 & -1/t \\ 1/t & 0 \end{pmatrix}$ ; so $B + B^{-1} = \begin{pmatrix} 0 & t - 1/t \\ -(t - 1/t) & 0 \end{pmatrix}$ , and with $t = \tan\frac{\pi}{12} = 2 - \sqrt3$ , $t - \frac1t = -2\sqrt3$ .

Symmetric and skew-symmetric parts

M = \frac{M + M^T}{2} + \frac{M - M^T}{2} \qquad B^T = -B,\ n \text{ odd} \ \Rightarrow\ |B| = 0

Worked example

Write

M = \begin{pmatrix} 2 & 5 \\ 1 & 4 \end{pmatrix}

as a symmetric matrix plus a skew-symmetric matrix.

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 10th May Shift 1 · Q115Hard

Example 3 · Determinants and Matrices · Systems of Linear Equations and Symmetric, Skew-Symmetric Matrices

A+B=\begin{pmatrix}1 & \tan\frac{\theta}{2}\\-\tan\frac{\theta}{2} & 1\end{pmatrix}

, where A is symmetric and B is skew-symmetric matrix, then the matrix

A^{-1}B+AB^{-1}

\theta=\frac{\pi}{6}

is given by

Expecting a 2 × 2 skew-symmetric matrix to be singular

\begin{pmatrix} 0 & t \\ -t & 0 \end{pmatrix}

has determinant

t^2 \ne 0

. The 'skew ⇒ singular' rule is for ODD order only; in even order the skew part is invertible whenever

t \ne 0

Drill 1 more on symmetric + skew-symmetric: the unique split, and why odd-order skew is singular

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Solving AX = B: Elimination, or X = A⁻¹B
Linear system in matrix form
$AX = B,\quad |A| \ne 0 \ \Rightarrow\ X = A^{-1}B \ \text{(unique)}$
Homogeneous Systems: Non-Trivial Solutions Need |A| = 0
Homogeneous system
$AX = O \text{ has a non-trivial solution} \iff |A| = 0$
Symmetric + Skew-Symmetric: The Unique Split, and Why Odd-Order Skew Is Singular
Symmetric and skew-symmetric parts
$M = \frac{M + M^T}{2} + \frac{M - M^T}{2} \qquad B^T = -B,\ n \text{ odd} \ \Rightarrow\ |B| = 0$

Watch out for (3)

Trusting a solution without checking every equation
→ Solving AX = B: Elimination, or X = A⁻¹B
Reading |A| = 0 as 'no solution'
→ Homogeneous Systems: Non-Trivial Solutions Need |A| = 0
Expecting a 2 × 2 skew-symmetric matrix to be singular
→ Symmetric + Skew-Symmetric: The Unique Split, and Why Odd-Order Skew Is Singular

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