Level 5 — MasteryMatrices & Determinants — Introduction

Matrices & Determinants — Introduction

⏱ 75 minutes50 marksprintable — key stays hidden on paper

LEVEL 5 — Mastery Paper (Cross-domain: Math + Physics + Coding)

Time limit: 75 minutes
Total marks: 50
Instructions: Answer all questions. Show full working. Proofs must be rigorous. Pseudocode/code may be in any clear syntax.

Question 1 — Structure, Proof & Computation (18 marks)

Let $A$ be any square matrix of order $n$ .

(a) Prove that any square matrix $A$ can be written uniquely as the sum of a symmetric matrix $S$ and a skew-symmetric matrix $K$ . State $S$ and $K$ explicitly in terms of $A$ and $A^T$ . (5)

(b) For $A = \begin{pmatrix} 2 & 5 & -1 \\ 1 & 3 & 4 \\ 0 & 2 & 6 \end{pmatrix},$ find $S$ and $K$ explicitly and verify $A = S + K$ . (4)

(c) Prove that for any skew-symmetric matrix $K$ of odd order, $\det(K) = 0$ . Hence state $\det(K)$ for the matrix $K$ found in part (b) without direct expansion. (5)

(d) Write pseudocode for a function decompose(A) that returns $(S, K)$ for an $n\times n$ input matrix, using only element-wise operations and transpose. Comment on its time complexity in terms of $n$ . (4)

Question 2 — Physics Application: Coupled System via Cramer & Inversion (18 marks)

Two masses on frictionless rails are connected by springs. At static equilibrium the displacements $x_1, x_2$ (in metres) satisfy the stiffness equations from applied forces $F_1 = 12\ \text{N}$ , $F_2 = 6\ \text{N}$ :

$\begin{pmatrix} 5 & -2 \\ -2 & 4 \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \begin{pmatrix} 12 \\ 6 \end{pmatrix}$

(units: stiffness in N/m).

(a) Compute the determinant of the stiffness matrix $K$ and explain physically why $\det K \neq 0$ guarantees a unique equilibrium. (3)

(b) Solve for $x_1, x_2$ using Cramer's rule. (4)

(c) Find $K^{-1}$ (the compliance matrix) and re-solve the system by matrix inversion, confirming the answer in (b). (5)

(d) The elastic potential energy is $U = \tfrac{1}{2}\mathbf{x}^T K \mathbf{x}$ where $\mathbf{x}=(x_1,x_2)^T$ . Using your solution, compute $U$ in joules. Also verify the identity $U = \tfrac12 \mathbf{x}^T\mathbf{F}$ and explain why these agree. (6)

Question 3 — Non-commutativity & a Structural Claim (14 marks)

(a) Give an explicit pair of $2\times2$ matrices $A, B$ with $AB \neq BA$ , and show both products. (3)

(b) Prove the general property $(AB)^T = B^T A^T$ for conformable matrices $A$ ( $m\times p$ ) and $B$ ( $p\times n$ ), working from the definition of the $(i,j)$ entry. (6)

(c) A student claims: "For all $2\times2$ matrices, $\det(A+B) = \det A + \det B$ ." Disprove this with a counterexample, and separately prove the correct multiplicative property $\det(AB)=\det A\,\det B$ holds for the specific matrices $A=\begin{pmatrix}1&2\\0&3\end{pmatrix}$ , $B=\begin{pmatrix}4&1\\1&2\end{pmatrix}$ by computing both sides. (5)

Answer keyMark scheme & solutions

Question 1

(a) [5 marks] Define $S = \tfrac12(A+A^T)$ and $K = \tfrac12(A-A^T)$ . (1) Then $S + K = \tfrac12(A+A^T) + \tfrac12(A-A^T) = A$ . (1) $S^T = \tfrac12(A^T + A) = S$ ⇒ symmetric. $K^T = \tfrac12(A^T - A) = -K$ ⇒ skew-symmetric. (1, why: transpose is linear and $(A^T)^T=A$ ) Uniqueness: suppose $A = S'+K'$ with $S'$ symmetric, $K'$ skew. Take transpose: $A^T = S'-K'$ . Adding/subtracting gives $S'=\tfrac12(A+A^T)=S$ , $K'=\tfrac12(A-A^T)=K$ . (2)

(b) [4 marks] $A^T = \begin{pmatrix}2&1&0\\5&3&2\\-1&4&6\end{pmatrix}$ . (1) $S=\tfrac12(A+A^T)=\begin{pmatrix}2&3&-\tfrac12\\3&3&3\\-\tfrac12&3&6\end{pmatrix}$ . (1.5) $K=\tfrac12(A-A^T)=\begin{pmatrix}0&2&-\tfrac12\\-2&0&1\\ \tfrac12&-1&0\end{pmatrix}$ . (1.5) (Sum check: $S+K=A$ — award full if verified.)

(c) [5 marks] For skew-symmetric $K$ : $K^T=-K$ . Then $\det(K^T)=\det(-K)$ . (1) Left: $\det(K^T)=\det K$ . (1) Right: $\det(-K)=(-1)^n\det K$ for order $n$ . (1) So $\det K = (-1)^n \det K$ . For odd $n$ , $(-1)^n=-1$ , giving $\det K = -\det K \Rightarrow 2\det K = 0 \Rightarrow \det K = 0$ . (1) Here $n=3$ (odd), so $\det K = 0$ . (1)

(d) [4 marks]

function decompose(A):        # A is n x n
    n = rows(A)
    AT = transpose(A)
    S = zeros(n,n); K = zeros(n,n)
    for i in 1..n:
        for j in 1..n:
            S[i][j] = 0.5*(A[i][j] + AT[i][j])
            K[i][j] = 0.5*(A[i][j] - AT[i][j])
    return (S, K)

Transpose is $O(n^2)$ , the double loop is $O(n^2)$ ; total $O(n^2)$ . (loop correctness 2, complexity 2)

Question 2

(a) [3 marks] $\det K = (5)(4)-(-2)(-2)=20-4=16$ . (1) $\det K\neq0$ ⇒ $K$ invertible ⇒ system has a unique solution $\mathbf{x}=K^{-1}\mathbf F$ ; physically a unique equilibrium configuration exists (no rigid-body zero modes / positive-definite stable equilibrium). (2)

(b) [4 marks] $D=16$ . $D_1=\det\begin{pmatrix}12&-2\\6&4\end{pmatrix}=48-(-12)=60$ ; $x_1=60/16=15/4=3.75$ m. (2) $D_2=\det\begin{pmatrix}5&12\\-2&6\end{pmatrix}=30-(-24)=54$ ; $x_2=54/16=27/8=3.375$ m. (2)

(c) [5 marks] $K^{-1}=\tfrac1{16}\begin{pmatrix}4&2\\2&5\end{pmatrix}$ . (2) $\mathbf x=K^{-1}\mathbf F=\tfrac1{16}\begin{pmatrix}4&2\\2&5\end{pmatrix}\begin{pmatrix}12\\6\end{pmatrix}=\tfrac1{16}\begin{pmatrix}48+12\\24+30\end{pmatrix}=\tfrac1{16}\begin{pmatrix}60\\54\end{pmatrix}=\begin{pmatrix}3.75\\3.375\end{pmatrix}$ . (2) Matches (b). (1)

(d) [6 marks] $K\mathbf x=\mathbf F=(12,6)^T$ by construction, so $U=\tfrac12\mathbf x^T K\mathbf x = \tfrac12 \mathbf x^T\mathbf F$ . (identity, 2) $U=\tfrac12(x_1\cdot12+x_2\cdot6)=\tfrac12(3.75\cdot12+3.375\cdot6)=\tfrac12(45+20.25)=\tfrac12(65.25)=32.625$ J. (3) They agree because $K\mathbf x=\mathbf F$ substituted into $\tfrac12\mathbf x^TK\mathbf x$ directly yields $\tfrac12\mathbf x^T\mathbf F$ . (1)

Question 3

(a) [3 marks] Take $A=\begin{pmatrix}1&1\\0&1\end{pmatrix}$ , $B=\begin{pmatrix}1&0\\1&1\end{pmatrix}$ . $AB=\begin{pmatrix}2&1\\1&1\end{pmatrix}$ , $BA=\begin{pmatrix}1&1\\1&2\end{pmatrix}$ . $AB\neq BA$ . (3)

(b) [6 marks] $(AB)_{ij}=\sum_{k=1}^p A_{ik}B_{kj}$ . (1) $((AB)^T)_{ij}=(AB)_{ji}=\sum_k A_{jk}B_{ki}$ . (2) $(B^TA^T)_{ij}=\sum_k (B^T)_{ik}(A^T)_{kj}=\sum_k B_{ki}A_{jk}=\sum_k A_{jk}B_{ki}$ . (2) Both equal the same scalar sum for every $(i,j)$ ; orders match ( $n\times m$ ), so $(AB)^T=B^TA^T$ . (1)

(c) [5 marks] Counterexample: $A=I$ , $B=I$ : $\det(A+B)=\det(2I)=4$ but $\det A+\det B=1+1=2\neq4$ . Claim false. (2) Multiplicative check: $\det A=3$ , $\det B=8-1=7$ , product $=21$ . (1) $AB=\begin{pmatrix}1&2\\0&3\end{pmatrix}\begin{pmatrix}4&1\\1&2\end{pmatrix}=\begin{pmatrix}6&5\\3&6\end{pmatrix}$ , $\det(AB)=36-15=21$ . (2) Equal ✓.

[
{"claim":"Q1(c): odd-order skew matrix has zero determinant (n=3 example)","code":"K=Matrix([[0,2,Rational(-1,2)],[-2,0,1],[Rational(1,2),-1,0]]); result = (K.det()==0)"},
{"claim":"Q2(b): Cramer solution x1=15/4, x2=27/8","code":"K=Matrix([[5,-2],[-2,4]]); F=Matrix([12,6]); x=K.solve(F); result = (x[0]==Rational(15,4) and x[1]==Rational(27,8))"},
{"claim":"Q2(d): U=32.625 J via both formulas","code":"K=Matrix([[5,-2],[-2,4]]); F=Matrix([12,6]); x=K.solve(F); U1=(x.T*K*x)[0]/2; U2=(x.T*F)[0]/2; result = (U1==U2 and U1==Rational(261,8))"},
{"claim":"Q3(c): det(AB)=detA*detB=21 and det(A+B)!=detA+detB for A=B=I","code":"A=Matrix([[1,2],[0,3]]); B=Matrix([[4,1],[1,2]]); prod=(A.det()*B.det()); I=eye(2); result = ((A*B).det()==21 and prod==21 and (I+I).det()!=I.det()+I.det())"}
]