Matrices and determinants – university Year 1 maths test answers to download as PDF

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1. Answer to Question 1 (4 marks)

   1. Multiply rows by columns: \(PQ=\) \(\begin{pmatrix}2+2&8+0\\0+6&0+0\end{pmatrix}\)\(=\) \(\begin{pmatrix}4&8\\6&0\end{pmatrix}\) (1) and \(QP=\) \(\begin{pmatrix}2+0&1+12\\4+0&2+0\end{pmatrix}\)\(=\) \(\begin{pmatrix}2&13\\4&2\end{pmatrix}\) (1). The products are different.
   2. The transpose swaps rows and columns: \((PQ)^T=\) \(\begin{pmatrix}4&6\\8&0\end{pmatrix}\) (1), which also equals \(Q^TP^T\).
   3. \(\det(PQ)=4\times0-8\times6=-48\). Also \(\det P=6\) and \(\det Q=-8\), so \(\det P\det Q=-48\), as expected (1).

   In (a), \(PQ\ne QP\) shows that matrix multiplication is not commutative. In (b), a common error is to write \((PQ)^T=P^TQ^T\); the order reverses. In (c), \(\det(QP)=4-52=-48\) as well.

2. Answer to Question 2 (4 marks)

   Expand along the first row, using the signs \(+,-,+\) and the \(2\times2\) minors: \(1(1\times2-4\times5)-2(3\times2-4\times0)+0\times(\dots)\) (2). This gives \(1(-18)-2(6)=-18-12=-30\) (1). Since \(\det=-30\ne0\), the matrix is invertible (1): its rows are linearly independent, and any system with this coefficient matrix has a unique solution. The zero in the first row saved one minor.

   Check with a cofactor expansion down the first column: \(1(2-20)-3(4-0)+0=-30\). A common error is to forget the minus sign on the middle term.

3. Answer to Question 3 (4 marks)

   1. \(\det A=4\times2-3\times3=8-9=-1\) (1). Swapping the diagonal entries, negating the others and dividing by the determinant, \(A^{-1}=-1\times\)\(\begin{pmatrix}2&-3\\-3&4\end{pmatrix}\)\(=\) \(\begin{pmatrix}-2&3\\3&-4\end{pmatrix}\) (1).
   2. The system is \(A\binom xy=\binom11\), so \(\binom xy=A^{-1}\binom11=\binom{-2+3}{3-4}=\binom1{-1}\) (1). Hence \(x=1\), \(y=-1\) (1). Check by substituting into both equations: \(4(1)+3(-1)=1\) and \(3(1)+2(-1)=1\). Because \(\det A\ne0\), this is the only solution.

   In (a), check: \(AA^{-1}=\begin{pmatrix}-8+9&12-12\\-6+6&9-8\end{pmatrix}=I\). Because the determinant is \(-1\), the inverse has integer entries.

4. Answer to Question 4 (4 marks)

   1. Let £\(a\) be the adult price and £\(c\) the child price. Then \(\begin{pmatrix}2&3\\3&2\end{pmatrix}\)\(\begin{pmatrix}a\\c\end{pmatrix}\)\(=\) \(\begin{pmatrix}62\\68\end{pmatrix}\) (1).
   2. The determinant is \(2\times2-3\times3=-5\) (1). Replacing the first column by the right-hand side, \(a=\dfrac{62\times2-3\times68}{-5}=\dfrac{124-204}{-5}=16\) (1). Replacing the second column, \(c=\dfrac{2\times68-3\times62}{-5}=\dfrac{136-186}{-5}=10\) (1). An adult ticket costs £16 and a child ticket £10. Check: \(32+30=62\).

   In (b), Cramer's rule applies only because the determinant is non-zero. Check the second equation: \(3\times16+2\times10=48+20=68\).

5. Answer to Question 5 (4 marks)

   1. A matrix is singular when its determinant is zero. \(\det M=1\times6-2k=6-2k\), and \(6-2k=0\) gives \(k=3\) (2).
   2. With \(k=3\) the equations are \(x+3y=3\) and \(2x+6y=6\). The second is exactly twice the first, so the two equations describe the same line and the system has infinitely many solutions (2): \(x=3-3y\) for any \(y\). If the right-hand sides were not in the same ratio, there would be no solution.

Total : 20 marks