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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.

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Why this matters

8 PYQs at 38% 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

Intuition

AX=BAX = B is three linear equations written compactly. With small integer coefficients, subtracting equations is faster than building A−1A^{-1}; the matrix form is there to make the question look harder than it is.

Definition

  • Read the rows of AA as equations: A=(1110131−21)A = \begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & 3 \\ 1 & -2 & 1 \end{pmatrix}, B=(6110)B = \begin{pmatrix} 6 \\ 11 \\ 0 \end{pmatrix} is a+b+c=6a + b + c = 6, b+3c=11b + 3c = 11, a−2b+c=0a - 2b + c = 0.
  • Eliminate: subtract rows to isolate one unknown, back-substitute. Here R1−R3R_1 - R_3: 3b=63b = 6, b=2b = 2; then c=3c = 3, a=1a = 1.
  • Matrix method when ∣A∣≠0|A| \ne 0: X=A−1BX = A^{-1}B — correct but slow by hand for 3×33 \times 3; use it only when A−1A^{-1} is given.
  • Answer the combination asked (2a+b+2c=102a + b + 2c = 10, x2+y2+z2=14x^2 + y^2 + z^2 = 14); the solution triple itself is rarely an option.
  • A unique solution exists iff ∣A∣≠0|A| \ne 0. If ∣A∣=0|A| = 0, the system has either no solution or infinitely many.

Linear system in matrix form

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

Worked example

Solve (111123149)(xyz)=(3614)\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+3zx + 2y + 3z.
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 1Determinants and MatricesMODERATE
Let A=[1110131−21], B=[6110]A=\begin{bmatrix}1&1&1\\0&1&3\\1&-2&1\end{bmatrix},\ B=\begin{bmatrix}6\\11\\0\end{bmatrix} and X=[abc]X=\begin{bmatrix}a\\b\\c\end{bmatrix}, if AX=BAX=B, then the value of 2a+b+2c2a+b+2c is

[Q143 · 9th May Shift 2 · 2024]

Trusting a solution without checking every equation

One sitting's stored key gave (2,1,1)(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

Intuition

AX=OAX = O always has the trivial solution X=OX = O. If ∣A∣≠0|A| \ne 0 that is the only one; if ∣A∣=0|A| = 0 there are infinitely many. So 'find λ\lambda so that x,y,zx, y, z are not all zero' means 'set the determinant to zero'.

Definition

  • AX=OAX = O: unique (trivial) solution iff ∣A∣≠0|A| \ne 0; infinitely many (non-trivial) iff ∣A∣=0|A| = 0. A homogeneous system is never inconsistent.
  • A vector equation a‾+b‾−c‾=0‾\overline{a} + \overline{b} - \overline{c} = \overline{0} with a‾,b‾,c‾\overline{a}, \overline{b}, \overline{c} linear in x,y,zx, y, z is three homogeneous equations in x,y,zx, y, z: collect the i^,j^,k^\hat{i}, \hat{j}, \hat{k} components as rows, then set the 3×33 \times 3 determinant to 00 and solve for the parameter.
  • For the rows (λ,1,1)(\lambda, 1, 1), (1,1,2)(1, 1, 2), (λ+1,3,4)(\lambda + 1, 3, 4): the determinant is −λ-\lambda, so λ=0\lambda = 0.
  • Non-homogeneous AX=BAX = B with ∣A∣=0|A| = 0 needs the extra test (consistency): either no solution or infinitely many.

Homogeneous system

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

Worked example

Find kk if x+2y+z=0x + 2y + z = 0, 2x+y+kz=02x + y + kz = 0, x−y+2z=0x - y + 2z = 0 has a non-trivial solution.
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 2Determinants and MatricesHARD
If a‾=λxi^+yj^+4zk^\overline{a}=\lambda x\hat{i}+y\hat{j}+4z\hat{k}, b‾=yi^+xj^+3yk^\overline{b}=y\hat{i}+x\hat{j}+3y\hat{k} and c‾=−zi^−2zj^−(λ+1)xk^\overline{c}=-z\hat{i}-2z\hat{j}-(\lambda+1)x\hat{k} are the sides of the triangle ABC, where x,y,zx,y,z are not all zero, such that a‾+b‾−c‾=0‾\overline{a}+\overline{b}-\overline{c}=\overline{0}, then the value of λ\lambda is

[Q130 · 26 April Shift I · 2025]

Reading |A| = 0 as 'no solution'

For a homogeneous system a zero determinant means infinitely many solutions, never none — X=OX = O always works. 'No solution' is only possible for AX=BAX = B with B≠OB \ne O.

Concept 3 of 3

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

Intuition

Any square matrix MM is half of M+MTM + M^T (symmetric) plus half of M−MTM - M^T (skew-symmetric). A skew-symmetric matrix has zeros on its diagonal and, in odd order, determinant 00 — so any system it governs has infinitely many solutions.

Definition

  • M=M+MT2⏟symmetric+M−MT2⏟skew-symmetricM = \underbrace{\dfrac{M + M^T}{2}}_{\text{symmetric}} + \underbrace{\dfrac{M - M^T}{2}}_{\text{skew-symmetric}}, uniquely. For M=(1t−t1)M = \begin{pmatrix} 1 & t \\ -t & 1 \end{pmatrix}: symmetric part II, skew part (0t−t0)\begin{pmatrix} 0 & t \\ -t & 0 \end{pmatrix}.
  • Skew-symmetric BB: BT=−BB^T = -B, diagonal entries 00. For odd nn, ∣B∣=∣BT∣=∣−B∣=(−1)n∣B∣=−∣B∣|B| = |B^T| = |-B| = (-1)^n|B| = -|B|, so ∣B∣=0|B| = 0.
  • Products: if AA is symmetric and BB skew-symmetric, then A2A^2 and B2B^2 are both symmetric, and A2B2−B2A2A^2B^2 - B^2A^2 is skew-symmetric — so for 3×33 \times 3 it is singular and (A2B2−B2A2)X=O(A^2B^2 - B^2A^2)X = O has infinitely many solutions.
  • Inverse of the 2×22 \times 2 skew part: (0t−t0)−1=(0−1/t1/t0)\begin{pmatrix} 0 & t \\ -t & 0 \end{pmatrix}^{-1} = \begin{pmatrix} 0 & -1/t \\ 1/t & 0 \end{pmatrix}; so B+B−1=(0t−1/t−(t−1/t)0)B + B^{-1} = \begin{pmatrix} 0 & t - 1/t \\ -(t - 1/t) & 0 \end{pmatrix}, and with t=tan⁡π12=2−3t = \tan\frac{\pi}{12} = 2 - \sqrt3, t−1t=−23t - \frac1t = -2\sqrt3.

Symmetric and skew-symmetric parts

M=M+MT2+M−MT2BT=−B, n odd ⇒ ∣B∣=0M = \frac{M + M^T}{2} + \frac{M - M^T}{2} \qquad B^T = -B,\ n \text{ odd} \ \Rightarrow\ |B| = 0

Worked example

Write M=(2514)M = \begin{pmatrix} 2 & 5 \\ 1 & 4 \end{pmatrix} as a symmetric matrix plus a skew-symmetric matrix.
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 3Determinants and MatricesHARD
If A+B=(1tan⁡θ2−tan⁡θ21)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−1B+AB−1A^{-1}B+AB^{-1} at θ=π6\theta=\frac{\pi}{6} is given by

[Q115 · 10th May Shift 1 · 2023]

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

(0t−t0)\begin{pmatrix} 0 & t \\ -t & 0 \end{pmatrix} has determinant t2≠0t^2 \ne 0. The 'skew ⇒ singular' rule is for ODD order only; in even order the skew part is invertible whenever t≠0t \ne 0.
Drill 1 more on symmetric + skew-symmetric: the unique split, and why odd-order skew is singular

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