Matrices & Determinants — Introduction

Q1 (10)

(a) Entries: diagonal $a_{11}=1,a_{22}=4,a_{33}=9$; off-diagonal $a_{ij}=i+j$. $A=\begin{pmatrix}1&3&4\\3&4&5\\4&5&9\end{pmatrix}$ (3) Why: $a_{12}=1+2=3$, etc.; note $A$ is already symmetric here.

(b) $A^{T}=A$ (since off-diagonal $i+j$ is symmetric and diagonal is symmetric). So $S=\tfrac12(A+A^{T})=A$ and $K=\tfrac12(A-A^{T})=0$ (zero matrix). (5) Why: Recognizing $A$ is symmetric gives $A^T=A$ immediately (2 for computing $A^T$, 3 for $S,K$).

(c) $\det(K)=\det(0)=0$. Any $3\times3$ skew-symmetric matrix has determinant 0 (odd order); here $K$ is the zero matrix so trivially $\det=0$. (2)

Q2 (10)

(a) $AB=\begin{pmatrix}2&-1\\1&3\end{pmatrix}\begin{pmatrix}0&4\\-2&1\end{pmatrix}=\begin{pmatrix}2&7\\-6&7\end{pmatrix}$. $(AB)^{T}=\begin{pmatrix}2&-6\\7&7\end{pmatrix}$. $B^{T}=\begin{pmatrix}0&-2\\4&1\end{pmatrix},\ A^{T}=\begin{pmatrix}2&1\\-1&3\end{pmatrix}$, $B^{T}A^{T}=\begin{pmatrix}0&-6\\7&7\end{pmatrix}$... recompute: $B^TA^T=\begin{pmatrix}0&-2\\4&1\end{pmatrix}\begin{pmatrix}2&1\\-1&3\end{pmatrix}=\begin{pmatrix}2&-6\\7&7\end{pmatrix}$. Matches. (5) Marks: AB (2), transpose (1), $B^TA^T$ (2).

(b) $BA=\begin{pmatrix}0&4\\-2&1\end{pmatrix}\begin{pmatrix}2&-1\\1&3\end{pmatrix}=\begin{pmatrix}4&12\\-3&5\end{pmatrix}$. $AB\ne BA$. $AB-BA=\begin{pmatrix}2-4&7-12\\-6+3&7-5\end{pmatrix}=\begin{pmatrix}-2&-5\\-3&2\end{pmatrix}$. (3)

(c) $\det A=2\cdot3-(-1)(1)=7$; $\det B=0\cdot1-4(-2)=8$; product $=56$. $\det(AB)=2\cdot7-7\cdot(-6)=14+42=56$. Equal. (2)

Q3 (10)

(a) $\det K=3\cdot1-2\cdot1=1$. $K^{-1}=\tfrac1{1}\begin{pmatrix}1&-2\\-1&3\end{pmatrix}=\begin{pmatrix}1&-2\\-1&3\end{pmatrix}$. (3)

(b) $\mathbf{x}=K^{-1}\mathbf{y}=\begin{pmatrix}1&-2\\-1&3\end{pmatrix}\begin{pmatrix}19\\7\end{pmatrix}=\begin{pmatrix}19-14\\-19+21\end{pmatrix}=\begin{pmatrix}5\\2\end{pmatrix}$. (4) Check: $K\mathbf x=\begin{pmatrix}3\cdot5+2\cdot2\\5+2\end{pmatrix}=\begin{pmatrix}19\\7\end{pmatrix}$ ✓.

(c) Need $ad-bc=0$ with $a+d=5$. Example $a=2,d=3,b=6,c=1$: $\det=6-6=0$; singular hence non-invertible. (Any valid singular example accepted.) (3)

Q4 (10)

(a) Coefficient determinant $D=\begin{vmatrix}2&k\\3&-1\end{vmatrix}=-2-3k$. $D_x=\begin{vmatrix}5&k\\4&-1\end{vmatrix}=-5-4k$; $D_y=\begin{vmatrix}2&5\\3&4\end{vmatrix}=8-15=-7$. $x=\frac{-5-4k}{-2-3k}=\frac{5+4k}{2+3k},\quad y=\frac{-7}{-2-3k}=\frac{7}{2+3k}.$ (5)

(b) No unique solution when $D=0$: $-2-3k=0\Rightarrow k=-\tfrac23$. (2)

(c) At $k=-\tfrac23$: equations become $2x-\tfrac23y=5$, i.e. $6x-2y=15$, and $3x-y=4\Rightarrow 6x-2y=8$. Left sides equal but $15\ne8$: inconsistent (parallel lines, no solution). (3)

Q5 (10)

(a) Expand along column 1: $\det M = 1\cdot\begin{vmatrix}1&4\\6&0\end{vmatrix}-0+5\cdot\begin{vmatrix}2&3\\1&4\end{vmatrix}$ $=1(0-24)+5(8-3)=-24+25=1$. (4)

(b) $\det(2M)=2^3\det M=8\cdot1=8$ (order 3, scalar factor cubed). $\det(M^{T})=\det M=1$ (transpose preserves determinant). (3)

(c) Row swap multiplies det by $-1$; scaling a row by $-3$ multiplies by $-3$. $\det N=(-1)(-3)\det M=3\cdot1=3$. (3)

[
{"claim":"Q2 det(AB)=det(A)det(B)=56","code":"A=Matrix([[2,-1],[1,3]]); B=Matrix([[0,4],[-2,1]]); result = ((A*B).det()==A.det()*B.det()) and ((A*B).det()==56)"},
{"claim":"Q3 recovered block x=(5,2)","code":"K=Matrix([[3,2],[1,1]]); x=K.inv()*Matrix([19,7]); result = x==Matrix([5,2])"},
{"claim":"Q4 x,y formulas and D=0 at k=-2/3","code":"k=symbols('k'); D=Matrix([[2,k],[3,-1]]).det(); Dx=Matrix([[5,k],[4,-1]]).det(); Dy=Matrix([[2,5],[3,4]]).det(); result = (simplify(Dx/D - (5+4*k)/(2+3*k))==0) and (simplify(Dy/D - 7/(2+3*k))==0) and (solve(D,k)==[Rational(-2,3)])"},
{"claim":"Q5 det(M)=1, det(2M)=8, det(N)=3","code":"M=Matrix([[1,2,3],[0,1,4],[5,6,0]]); result = (M.det()==1) and ((2*M).det()==8) and (3*M.det()==3)"}
]