Inverse matrices and systems of equations – further maths test answers to download as PDF

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1. Answer to Question 1 (4 marks)

   1. \(\det=8-3=5\) [1]; \(\mathbf A^{-1}=\) \(\dfrac15\begin{pmatrix}4&-3\\-1&2\end{pmatrix}\) [1]
   2. \(\mathbf x=\dfrac15\begin{pmatrix}4&-3\\-1&2\end{pmatrix}\begin{pmatrix}5\\0\end{pmatrix}=\dfrac15\begin{pmatrix}20\\-5\end{pmatrix}\) [1], so \(x=4\), \(y=-1\) [1]

   Method: for \(\begin{pmatrix}a&b\\c&d\end{pmatrix}\), swap \(a\) and \(d\), change the signs of \(b\) and \(c\), and divide by the determinant \(ad-bc\). Write the system as \(\mathbf A\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}5\\0\end{pmatrix}\) and premultiply by \(\mathbf A^{-1}\); the order matters because matrix multiplication is not commutative.

   Check: \(2\times4+3\times(-1)=5\) ✓ and \(4+4\times(-1)=0\) ✓. A quick test of the inverse is \(\mathbf A\mathbf A^{-1}=\mathbf I\).

2. Answer to Question 2 (5 marks)

   1. \(\det=1(1-0)-0+2(6-0)=1+12=\) 13 [1]
   2. Cofactors: \(\begin{pmatrix}1&-2&6\\6&1&-3\\-2&4&1\end{pmatrix}\) [1 for method, 1 for accuracy]. Transpose gives the adjugate; \(\mathbf A^{-1}=\dfrac1{13}\begin{pmatrix}1&6&-2\\-2&1&4\\6&-3&1\end{pmatrix}\) [1]
   3. \(\mathbf x=\dfrac1{13}\begin{pmatrix}7+24-18\\-14+4+36\\42-12+9\end{pmatrix}=\dfrac1{13}\begin{pmatrix}13\\26\\39\end{pmatrix}\), so \(\mathbf x=(1,2,3)\) [1]

   Method: find the matrix of minors, apply the sign pattern to get the cofactors, transpose to obtain the adjugate, then divide by the determinant. In (a), expand along the first row; the zero in position \((1,2)\) saves one calculation.

   Check (c): \(1+0+2\times3=7\), \(2\times1+2=4\) and \(3\times2+3=9\) ✓. Forgetting to transpose the cofactor matrix is the most frequent error and loses the accuracy mark.

3. Answer to Question 3 (4 marks)

   1. The third row of the coefficient matrix equals the sum of the first two, so \(\det=0\) [1]
   2. The right-hand sides must satisfy \(2+5=k\): \(k=7\) [1]
   3. Subtract the first two equations: \(x+2z=3\), so \(x=3-2z\) [1]. Then \(y=2-x-z=z-1\). With \(z=t\): \((x,y,z)=(3-2t,\,t-1,\,t)\) [1]

   Geometrically, the three planes form a sheaf when \(k=7\): they share a common line. For any other value of \(k\) they form a triangular prism with no common point. Check the general solution in the third equation: \(3(3-2t)+2(t-1)+4t=7\) ✓.

4. Answer to Question 4 (4 marks)

   1. The normal of \(\Pi_2\) is twice the normal of \(\Pi_1\) [1]. But \(\Pi_1\) is equivalent to \(2x+4y-2z=2\), not 5, so the planes are distinct and parallel [1].
   2. Any common point would lie on both \(\Pi_1\) and \(\Pi_2\), which do not meet [1]. So there is no solution: two parallel planes are cut by the third plane in two parallel lines [1].

5. Answer to Question 5 (3 marks)

   1. \(\begin{pmatrix}3&2\\2&5\end{pmatrix}\begin{pmatrix}a\\b\end{pmatrix}=\begin{pmatrix}34\\52\end{pmatrix}\), \(\det=11\), inverse \(\dfrac1{11}\begin{pmatrix}5&-2\\-2&3\end{pmatrix}\) [2]
   2. \(\dfrac1{11}\begin{pmatrix}170-104\\-68+156\end{pmatrix}=\dfrac1{11}\begin{pmatrix}66\\88\end{pmatrix}\): A costs £6 and B costs £8 [1]

   Method: let \(a\) and \(b\) be the prices in pounds of one bag of A and one bag of B. The coefficient matrix has determinant \(3\times5-2\times2=11\ne0\), so it is non-singular and the solution is unique; premultiply both sides by the inverse.

   Check: \(3\times6+2\times8=34\) ✓ and \(2\times6+5\times8=52\) ✓. The answer must be given in context, with units, to earn the final mark.

Total : 20 marks