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

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1. Answer to Question 1

   1. Determinant \(=6-5=1\). Inverse \(=\) \(\begin{pmatrix}2&-1\\-5&3\end{pmatrix}\). Check: \(\begin{pmatrix}3&1\\5&2\end{pmatrix}\begin{pmatrix}2&-1\\-5&3\end{pmatrix}=\begin{pmatrix}6-5&-3+3\\10-10&-5+6\end{pmatrix}=\mathbf I\) ✓
   2. Determinant \(=4+6=10\). Inverse \(=\) \(\dfrac1{10}\begin{pmatrix}1&2\\-3&4\end{pmatrix}\).
   3. Determinant \(=6-6=0\), so the matrix is singular and has no inverse (its columns are parallel).
   4. Singular when \(k(k+1)-12=0\), i.e. \(k^2+k-12=(k+4)(k-3)=0\), so \(k=3\) or \(k=-4\).

2. Answer to Question 2

   1. \(\begin{pmatrix}3&1\\5&2\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}7\\12\end{pmatrix}\). \(\mathbf x=\begin{pmatrix}2&-1\\-5&3\end{pmatrix}\begin{pmatrix}7\\12\end{pmatrix}=\begin{pmatrix}14-12\\-35+36\end{pmatrix}\). \(x=2\), \(y=1\). Check: \(6+1=7\) ✓, \(10+2=12\) ✓.
   2. \(\mathbf x=\dfrac1{10}\begin{pmatrix}1&2\\-3&4\end{pmatrix}\begin{pmatrix}10\\5\end{pmatrix}=\dfrac1{10}\begin{pmatrix}20\\-10\end{pmatrix}\). \(x=2\), \(y=-1\). Check: \(8+2=10\) ✓, \(6-1=5\) ✓.
   3. The matrix \(\begin{pmatrix}2&6\\1&3\end{pmatrix}\) has determinant 0, so there is no inverse. The equations are inconsistent: the first says \(x+3y=2\) and the second says \(x+3y=5\), so there is no solution.

3. Answer to Question 3

   1. \(\det\mathbf A=5-6=-1\), so \(\mathbf A^{-1}=\begin{pmatrix}-5&2\\3&-1\end{pmatrix}\). \(\det\mathbf B=6\), so \(\mathbf B^{-1}=\dfrac16\begin{pmatrix}3&0\\-1&2\end{pmatrix}\).
   2. Multiply on the left: \(\mathbf X=\mathbf A^{-1}\mathbf C=\begin{pmatrix}-5&2\\3&-1\end{pmatrix}\begin{pmatrix}1&4\\2&-1\end{pmatrix}=\) \(\begin{pmatrix}-1&-22\\1&13\end{pmatrix}\).
   3. \(\mathbf{AB}=\begin{pmatrix}4&6\\11&15\end{pmatrix}\), determinant \(60-66=-6\). Directly: \(\dfrac1{-6}\begin{pmatrix}15&-6\\-11&4\end{pmatrix}=\begin{pmatrix}-\frac52&1\\\frac{11}6&-\frac23\end{pmatrix}\). And \(\mathbf B^{-1}\mathbf A^{-1}=\dfrac16\begin{pmatrix}3&0\\-1&2\end{pmatrix}\begin{pmatrix}-5&2\\3&-1\end{pmatrix}=\dfrac16\begin{pmatrix}-15&6\\11&-4\end{pmatrix}\), the same. Reverse order of the inverses.

4. Answer to Question 4

   1. \(\det\mathbf M=1(1\cdot1-3\cdot0)-2(0\cdot1-3\cdot2)+0=1+12=\) 13
   2. Cofactors: \(C_{11}=1\), \(C_{12}=-(0-6)=6\), \(C_{13}=0-2=-2\); \(C_{21}=-(2-0)=-2\), \(C_{22}=1-0=1\), \(C_{23}=-(0-4)=4\); \(C_{31}=6-0=6\), \(C_{32}=-(3-0)=-3\), \(C_{33}=1\). Cofactor matrix \(\begin{pmatrix}1&6&-2\\-2&1&4\\6&-3&1\end{pmatrix}\), so \(\operatorname{adj}\mathbf M=\begin{pmatrix}1&-2&6\\6&1&-3\\-2&4&1\end{pmatrix}\).
   3. \(\mathbf M^{-1}=\dfrac1{13}\begin{pmatrix}1&-2&6\\6&1&-3\\-2&4&1\end{pmatrix}\)
   4. First row of \(\mathbf M\) is \((1,2,0)\). Times column 1: \(\tfrac1{13}(1+12+0)=1\); column 2: \(\tfrac1{13}(-2+2+0)=0\); column 3: \(\tfrac1{13}(6-6+0)=0\). The first row of \(\mathbf I\). ✓

5. Answer to Question 5

   1. \(\mathbf A=\begin{pmatrix}1&1&1\\2&-1&1\\1&2&-1\end{pmatrix}\), \(\mathbf b=\begin{pmatrix}6\\3\\2\end{pmatrix}\). \(\det\mathbf A=1(1-2)-1(-2-1)+1(4+1)=-1+3+5=\) 7
   2. Cofactors: \(C_{11}=-1\), \(C_{12}=3\), \(C_{13}=5\), \(C_{21}=3\), \(C_{22}=-2\), \(C_{23}=-1\), \(C_{31}=2\), \(C_{32}=1\), \(C_{33}=-3\). The adjugate is the transpose: \(\begin{pmatrix}-1&3&2\\3&-2&1\\5&-1&-3\end{pmatrix}\).
   3. \(\mathbf A^{-1}=\dfrac17\begin{pmatrix}-1&3&2\\3&-2&1\\5&-1&-3\end{pmatrix}\)
   4. \(\mathbf x=\dfrac17\begin{pmatrix}-6+9+4\\18-6+2\\30-3-6\end{pmatrix}=\dfrac17\begin{pmatrix}7\\14\\21\end{pmatrix}\). \(x=1\), \(y=2\), \(z=3\). Check: \(1+2+3=6\) ✓; \(2-2+3=3\) ✓; \(1+4-3=2\) ✓.

6. Answer to Question 6

   1. \(\det=1(8-9)-1(4-6)+1(3-4)=-1+2-1=\) 0
   2. Adding the first two equations gives \(2x+3y+4z=3\). So the third equation says the same thing as the sum of the first two when \(k=3\); for any other \(k\) the equations contradict each other, and there is no solution.
   3. Subtract the first two equations: \(y+2z=1\), so \(y=1-2z\). Then \(x=1-y-z=z\). The solution is \((x,y,z)=(t,1-2t,t)\) for any \(t\): the three planes meet along a line (a sheaf).
   4. For \(k=5\) the third equation conflicts with the sum of the first two (\(3\ne5\)): no solution. The three planes form a triangular prism.

7. Answer to Question 7

   1. \(\begin{pmatrix}10&8&6\\12&5&8\\7&10&4\end{pmatrix}\begin{pmatrix}c\\t\\k\end{pmatrix}=\begin{pmatrix}70\\78\\57\end{pmatrix}\)
   2. \(\det=10(20-80)-8(48-56)+6(120-35)=-600+64+510=\) \(-26\), which is not zero.
   3. Using the inverse (calculator or cofactors) \(\mathbf x=\mathbf A^{-1}\mathbf b=\begin{pmatrix}3\\2\\4\end{pmatrix}\): coffee £3, tea £2, cake £4.
   4. Monday: \(10\times3+8\times2+6\times4=30+16+24=70\) ✓. Tuesday: \(36+10+32=78\) ✓. Wednesday: \(21+20+16=57\) ✓.

8. Answer to Question 8

   1. Protein: \(20x+10y+5z=100\). Carbohydrate: \(5x+30y+10z=115\). Fat: \(2x+4y+8z=46\).
   2. \(\det=20(240-40)-10(40-20)+5(20-60)=4000-200-200=\) 3600
   3. \(\mathbf x=\begin{pmatrix}20&10&5\\5&30&10\\2&4&8\end{pmatrix}^{-1}\begin{pmatrix}100\\115\\46\end{pmatrix}=\begin{pmatrix}3\\2\\4\end{pmatrix}\): 3 portions of X, 2 of Y, 4 of Z. Check protein: \(60+20+20=100\) ✓.
   4. \(3+2+4=9\) portions of 100 g: 900 g.

9. Answer to Question 9

   1. \(\begin{pmatrix}1&-1&-1\\4&6&0\\0&6&3\end{pmatrix}\begin{pmatrix}I_1\\I_2\\I_3\end{pmatrix}=\begin{pmatrix}0\\18\\12\end{pmatrix}\)
   2. \(\det=1(18-0)-(-1)(12-0)+(-1)(24-0)=18+12-24=\) 6
   3. Inverse gives \(\mathbf I=\begin{pmatrix}3\\1\\2\end{pmatrix}\): \(I_1=3\) A, \(I_2=1\) A, \(I_3=2\) A. Check: \(12+6=18\) ✓; \(6+6=12\) ✓.
   4. \(I_1-I_2-I_3=3-1-2=0\) ✓, so the current entering the junction equals the current leaving.

10. Answer to Question 10

    1. \(\begin{pmatrix}1&1&0\\0&1&1\\1&0&-1\end{pmatrix}\) has determinant \(1(-1-0)-1(0-1)=-1+1=\) 0.
    2. Subtracting the first equation from the second gives \(z-x=20\), i.e. \(x-z=-20\), which is the third equation. So the system is consistent with infinitely many solutions: \(x=50-y\), \(z=70-y\), with \(y\) free.
    3. \(x\ge0\) gives \(y\le50\); \(z\ge0\) gives \(y\le70\); \(y\ge0\). So \(0\le y\le50\).
    4. \(x=50-30=\) 20 and \(z=70-30=\) 40 cars per minute.

11. Answer to Question 11

    1. \(\begin{pmatrix}1&2&1\\2&-1&1\\1&1&-1\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}7\\2\\1\end{pmatrix}\). \(\det=1(1-1)-2(-2-1)+1(2+1)=0+6+3=\) 9
    2. Since \(\det\ne0\) the system has a unique solution: the three planes meet at exactly one point (like the corner of a room).
    3. \(\mathbf x=\mathbf A^{-1}\mathbf b=(1,2,2)\). The planes meet at \((1,2,2)\).
    4. \(1+4+2=7\) ✓; \(2-2+2=2\) ✓; \(1+2-2=1\) ✓.

12. Answer to Question 12

    1. \(\det\mathbf M=2-1=1\), so \(\mathbf M^{-1}=\begin{pmatrix}1&-1\\-1&2\end{pmatrix}\). \(\mathbf M^{-1}\begin{pmatrix}5\\3\end{pmatrix}=\begin{pmatrix}2\\1\end{pmatrix}\): the original pixel was at \((2,1)\). Check: \(\mathbf M\begin{pmatrix}2\\1\end{pmatrix}=\begin{pmatrix}5\\3\end{pmatrix}\) ✓.
    2. The area scale factor is \(|\det\mathbf M|=\) 1: areas are preserved (the picture is sheared and stretched but keeps its area).
    3. \(|\det\mathbf N|=6\), so the image has area \(6\times3=\) 18 cm\(^2\).
    4. Vertices map to \((0,0)\), \((6,0)\) and \((3,6)\). Area \(=\tfrac12|6\times6-0\times3|=\) 18 ✓.

13. Answer to Question 13

    1. H = 7 and I = 8. \(\mathbf K\begin{pmatrix}7\\8\end{pmatrix}=\begin{pmatrix}21+16\\7+8\end{pmatrix}=\begin{pmatrix}37\\15\end{pmatrix}\equiv\begin{pmatrix}11\\15\end{pmatrix}\), which is the pair LP.
    2. \(\det\mathbf K=3-2=1\), so \(\mathbf K^{-1}=\begin{pmatrix}1&-2\\-1&3\end{pmatrix}\). Adding 26 to negative entries: \(-2\equiv24\) and \(-1\equiv25\).
    3. P = 15 and B = 1. \(\begin{pmatrix}1&24\\25&3\end{pmatrix}\begin{pmatrix}15\\1\end{pmatrix}=\begin{pmatrix}39\\378\end{pmatrix}\equiv\begin{pmatrix}13\\14\end{pmatrix}\), which is the pair NO.
    4. To decode, the matrix must have an inverse that uses only whole numbers modulo 26. If the determinant shared a factor with 26, there would be no such inverse, and different messages would give the same code.

14. Answer to Question 14

    1. \(\mathbf I-\mathbf A=\begin{pmatrix}0.8&-0.3\\-0.4&0.9\end{pmatrix}\), \(\det=0.72-0.12=\) 0.6
    2. \((\mathbf I-\mathbf A)^{-1}=\dfrac1{0.6}\begin{pmatrix}0.9&0.3\\0.4&0.8\end{pmatrix}\)
    3. \(\mathbf x=\dfrac1{0.6}\begin{pmatrix}90+60\\40+160\end{pmatrix}=\dfrac1{0.6}\begin{pmatrix}150\\200\end{pmatrix}\), so \(x_1=250\) and \(x_2\approx333.3\) units.
    4. Part of each industry’s output is used up as input by the industries themselves, so more than the final demand has to be produced.

15. Answer to Question 15

    1. \(\det=1(2k-9)-1(k-6)+1(3-4)=2k-9-k+6-1=\) \(k-4\)
    2. When \(k\ne4\) the determinant is non-zero, so there is a unique solution: the planes meet at a single point.
    3. For \(k=4\), the third equation \(2x+3y+4z=3\) is the sum of the first two equations (\(1+2=3\)), so it adds no new information: consistent. Subtracting the first two gives \(y+2z=1\). Putting \(z=t\): \(y=1-2t\) and \(x=1-y-z=t\). \((x,y,z)=(t,1-2t,t)\).
    4. The three planes are all different and none are parallel, and they meet along a common line with direction vector \((1,-2,1)\) (a sheaf of planes).

16. Answer to Question 16

    1. The normals are \((1,1,1)\), \((2,1,-1)\) and \((3,2,0)\). No normal is a multiple of another, so no two planes are parallel.
    2. \(\det=1(0+2)-1(0+3)+1(4-3)=2-3+1=\) 0
    3. Adding \(\Pi_1\) and \(\Pi_2\) gives \(3x+2y=3\), but \(\Pi_3\) says \(3x+2y=5\): a contradiction, so no solution.
    4. Each pair of planes meets in a line, and the three lines are parallel (all with direction \((2,-3,1)\) here) but never coincide: the planes form a triangular prism.

17. Answer to Question 17

    1. \(\det=1(1-0)-2(0-2)+k(0-2)=1+4-2k=\) \(5-2k\). Singular when \(k=\tfrac52\).
    2. For \(k=1\), \(\det=3\). The adjugate is \(\begin{pmatrix}1&-2&1\\2&-1&-1\\-2&4&1\end{pmatrix}\), so \(\mathbf A^{-1}=\dfrac13\begin{pmatrix}1&-2&1\\2&-1&-1\\-2&4&1\end{pmatrix}\).
    3. \(\mathbf x=\dfrac13\begin{pmatrix}1-4+3\\2-2-3\\-2+8+3\end{pmatrix}=\dfrac13\begin{pmatrix}0\\-3\\9\end{pmatrix}\), so \((x,y,z)=(0,-1,3)\). Check: \(0-2+3=1\) ✓; \(-1+3=2\) ✓; \(0+0+3=3\) ✓.
    4. For \(k=\tfrac52\) the equations are \(x+2y+\tfrac52z=1\), \(y+z=2\), \(2x+z=3\). From the second \(y=2-z\); the first becomes \(x+4-2z+\tfrac52z=1\), i.e. \(x+\tfrac12z=-3\), so \(2x+z=-6\). This contradicts the third equation \(2x+z=3\). No solution.

18. Answer to Question 18

    1. \(\mathbf A^2=\begin{pmatrix}1+2&1+1\\2+2&2+1\end{pmatrix}=\begin{pmatrix}3&2\\4&3\end{pmatrix}\) and \(2\mathbf A+\mathbf I=\begin{pmatrix}2+1&2\\4&2+1\end{pmatrix}=\begin{pmatrix}3&2\\4&3\end{pmatrix}\) ✓
    2. \(\mathbf A^2-2\mathbf A=\mathbf I\), i.e. \(\mathbf A(\mathbf A-2\mathbf I)=\mathbf I\). Hence \(\mathbf A^{-1}=\mathbf A-2\mathbf I=\begin{pmatrix}-1&1\\2&-1\end{pmatrix}\).
    3. \(\det\mathbf A=1-2=-1\), so \(\mathbf A^{-1}=\dfrac1{-1}\begin{pmatrix}1&-1\\-2&1\end{pmatrix}=\begin{pmatrix}-1&1\\2&-1\end{pmatrix}\) ✓
    4. \(\mathbf A^3=\mathbf A\cdot\mathbf A^2=\mathbf A(2\mathbf A+\mathbf I)=2\mathbf A^2+\mathbf A=2(2\mathbf A+\mathbf I)+\mathbf A=5\mathbf A+2\mathbf I=\begin{pmatrix}7&5\\10&7\end{pmatrix}\).

19. Answer to Question 19

    1. The columns are the images of the unit vectors: \(\mathbf M=\begin{pmatrix}2&-1\\1&3\end{pmatrix}\).
    2. \(\det\mathbf M=6+1=\) 7. The image has area \(7\times4=\) 28.
    3. \(\mathbf M^{-1}=\dfrac17\begin{pmatrix}3&1\\-1&2\end{pmatrix}\)
    4. \(\mathbf M^{-1}\begin{pmatrix}3\\5\end{pmatrix}=\dfrac17\begin{pmatrix}14\\7\end{pmatrix}=\begin{pmatrix}2\\1\end{pmatrix}\), the point \((2,1)\). Check: \(\mathbf M\begin{pmatrix}2\\1\end{pmatrix}=\begin{pmatrix}4-1\\2+3\end{pmatrix}=\begin{pmatrix}3\\5\end{pmatrix}\) ✓.

20. Answer to Question 20

    1. Take determinants of both sides of \(\mathbf A\operatorname{adj}\mathbf A=(\det\mathbf A)\mathbf I\): \(\det\mathbf A\cdot\det(\operatorname{adj}\mathbf A)=(\det\mathbf A)^3\det\mathbf I=(\det\mathbf A)^3\). Divide by \(\det\mathbf A\): \(\det(\operatorname{adj}\mathbf A)=(\det\mathbf A)^2\).
    2. \(\det(\operatorname{adj}\mathbf M)=1(1+12)+2(6-6)+6(24+2)=13+0+156=169=13^2\) ✓.
    3. Apply the identity to the matrix \(\operatorname{adj}\mathbf A\): \(\operatorname{adj}\mathbf A\cdot\operatorname{adj}(\operatorname{adj}\mathbf A)=\det(\operatorname{adj}\mathbf A)\mathbf I=(\det\mathbf A)^2\mathbf I\). Multiply on the left by \(\mathbf A\): \((\det\mathbf A)\mathbf I\cdot\operatorname{adj}(\operatorname{adj}\mathbf A)=(\det\mathbf A)^2\mathbf A\), so \(\operatorname{adj}(\operatorname{adj}\mathbf A)=(\det\mathbf A)\mathbf A\).
    4. For \(2\times2\) the identity gives \(\det\mathbf A\det(\operatorname{adj}\mathbf A)=(\det\mathbf A)^2\), so \(\det(\operatorname{adj}\mathbf A)=\det\mathbf A\) to the power \(1\), not \(2\). In general the power is \(n-1\) for an \(n\times n\) matrix.