Matrices & Determinants — Introduction

Q1 (10)

(a) $AX=I$ gives two systems (one per column of $I$): Column 1: $ax_1+bx_3=1$, $cx_1+dx_3=0$. Column 2: $ax_2+bx_4=0$, $cx_2+dx_4=1$. (1 for setting up systems)

Solve column 1: from second eqn $x_3=-\frac{c}{d}x_1$ (or eliminate). Using elimination: $x_1 = \dfrac{d}{ad-bc}$, $x_3=\dfrac{-c}{ad-bc}$. (2) Column 2 similarly: $x_2=\dfrac{-b}{ad-bc}$, $x_4=\dfrac{a}{ad-bc}$. (2)

Hence $A^{-1}=\frac{1}{ad-bc}\begin{pmatrix} d & -b \\ -c & a\end{pmatrix}.$ (1) Condition: exists iff $ad-bc=\det A \neq 0$. (1)

(b) The formula divides by $\det A$; if $\det A=0$ the division is undefined, so no matrix $X$ satisfies $AX=I$. Geometrically the rows/columns are linearly dependent, so $A$ collapses dimensions and cannot be undone. (3: any valid reasoning referencing derivation)

Q2 (12)

(a) Suppose $M=S+K$ with $S^T=S$, $K^T=-K$. Take transpose: $M^T=S^T+K^T=S-K$. (2) Add: $M+M^T=2S \Rightarrow S=\tfrac12(M+M^T)$. (1) Subtract: $M-M^T=2K \Rightarrow K=\tfrac12(M-M^T)$. (1) Check $S^T=\tfrac12(M^T+M)=S$ ✓, $K^T=\tfrac12(M^T-M)=-K$ ✓. (1) Uniqueness: the two equations determine $S,K$ from $M$ uniquely. (1)

(b) $M^T=\begin{pmatrix}4&7\\3&2\end{pmatrix}$. $S=\tfrac12\begin{pmatrix}8&10\\10&4\end{pmatrix}=\begin{pmatrix}4&5\\5&2\end{pmatrix}$. (2) $K=\tfrac12\begin{pmatrix}0&-4\\4&0\end{pmatrix}=\begin{pmatrix}0&-2\\2&0\end{pmatrix}$. (2)

(c) Diagonal entries are all $0$. From $K^T=-K$, entry $k_{ii}=-k_{ii}\Rightarrow 2k_{ii}=0\Rightarrow k_{ii}=0$. (2)

Q3 (12)

(a) Expand along column 1 (entries $2,0,1$): $\det A = 2\cdot C_{11} + 0\cdot C_{21} + 1\cdot C_{31}$. $C_{11}=+\det\begin{pmatrix}4&5\\0&2\end{pmatrix}=8$. (2) $C_{31}=+\det\begin{pmatrix}-1&3\\4&5\end{pmatrix}=(-1)(5)-(3)(4)=-5-12=-17$. (2) (middle term zero.) $\det A = 2(8)+1(-17)=16-17=-1.$ (2)

(b) Expand along row 3 (entries $1,0,2$): $= 1\cdot(+)\det\begin{pmatrix}-1&3\\4&5\end{pmatrix} + 0 + 2\cdot(+)\det\begin{pmatrix}2&-1\\0&4\end{pmatrix}$ $= 1(-17)+2(8)= -17+16 = -1$. Matches. (4)

(c) The cofactor expansion along any line is a way of grouping the same signed permutation terms of the determinant; the determinant is a single well-defined value independent of the chosen line. (2)

Q4 (10)

(a) $AB=\begin{pmatrix}1&2\\0&1\end{pmatrix}\begin{pmatrix}1&0\\3&1\end{pmatrix}=\begin{pmatrix}7&2\\3&1\end{pmatrix}$. (2) $BA=\begin{pmatrix}1&0\\3&1\end{pmatrix}\begin{pmatrix}1&2\\0&1\end{pmatrix}=\begin{pmatrix}1&2\\3&7\end{pmatrix}$. (1) $AB\neq BA$. (1)

(b) $(AB)^T=\begin{pmatrix}7&3\\2&1\end{pmatrix}$. (1) $B^T=\begin{pmatrix}1&3\\0&1\end{pmatrix}$, $A^T=\begin{pmatrix}1&0\\2&1\end{pmatrix}$. $B^TA^T=\begin{pmatrix}1&3\\0&1\end{pmatrix}\begin{pmatrix}1&0\\2&1\end{pmatrix}=\begin{pmatrix}7&3\\2&1\end{pmatrix}$. (2) Equal ✓. (1)

(c) Multiplication combines rows of the first with columns of the second; swapping order changes which rows meet which columns (and may not even be defined dimensionally), so results differ in general. (2)

Q5 (16)

(a) $A=\begin{pmatrix}3&2\\1&-4\end{pmatrix}$, $b=\begin{pmatrix}7\\-9\end{pmatrix}$. $D=3(-4)-2(1)=-14$. (2) $D_x=\det\begin{pmatrix}7&2\\-9&-4\end{pmatrix}=7(-4)-2(-9)=-28+18=-10$. (1) $D_y=\det\begin{pmatrix}3&7\\1&-9\end{pmatrix}=3(-9)-7(1)=-27-7=-34$. (1) $x=D_x/D=\frac{-10}{-14}=\frac{5}{7}$, $y=D_y/D=\frac{-34}{-14}=\frac{17}{7}$. (2)

(b) $\det A=-14$, $A^{-1}=\frac{1}{-14}\begin{pmatrix}-4&-2\\-1&3\end{pmatrix}$. (3) $X=A^{-1}b=\frac{1}{-14}\begin{pmatrix}-4&-2\\-1&3\end{pmatrix}\begin{pmatrix}7\\-9\end{pmatrix}$ $=\frac{1}{-14}\begin{pmatrix}-28+18\\-7-27\end{pmatrix}=\frac{1}{-14}\begin{pmatrix}-10\\-34\end{pmatrix}=\begin{pmatrix}5/7\\17/7\end{pmatrix}$. (4) (matches (a))

(c) If $D=\det A=0$, neither Cramer's rule nor inversion works (division by zero / no inverse). Geometrically the two lines are parallel — either coincident (infinitely many solutions) or distinct (no solution). (3)

[
  {"claim":"det of Q3 matrix A = -1","code":"A=Matrix([[2,-1,3],[0,4,5],[1,0,2]]); result=(A.det()==-1)"},
  {"claim":"Q4 AB != BA","code":"A=Matrix([[1,2],[0,1]]); B=Matrix([[1,0],[3,1]]); result=(A*B!=B*A)"},
  {"claim":"Q4 (AB)^T = B^T A^T","code":"A=Matrix([[1,2],[0,1]]); B=Matrix([[1,0],[3,1]]); result=((A*B).T==B.T*A.T)"},
  {"claim":"Q5 solution x=5/7 y=17/7","code":"A=Matrix([[3,2],[1,-4]]); b=Matrix([7,-9]); X=A.inv()*b; result=(X==Matrix([Rational(5,7),Rational(17,7)]))"},
  {"claim":"Q2 S+K reconstructs M","code":"M=Matrix([[4,3],[7,2]]); S=(M+M.T)/2; K=(M-M.T)/2; result=(S+K==M and S.T==S and K.T==-K)"}
]