Matrices and linear transformations – further maths worksheet answers to download as PDF

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1. Answer to Question 1

   1. Row by column: \(\mathbf{AB}=\begin{pmatrix}3\cdot1+(-1)\cdot0&3\cdot2+(-1)(-3)\\2\cdot1+4\cdot0&2\cdot2+4(-3)\end{pmatrix}=\) \(\begin{pmatrix}3&9\\2&-8\end{pmatrix}\). Similarly \(\mathbf{BA}=\begin{pmatrix}3+4&-1+8\\0-6&0-12\end{pmatrix}=\) \(\begin{pmatrix}7&7\\-6&-12\end{pmatrix}\). The two products are different, so matrix multiplication is not commutative.
   2. \(\mathbf A\) is \(2\times2\) and \(\mathbf C\) is \(2\times3\), so \(\mathbf{AC}\) is \(2\times3\): first row \((3-3,\;0+1,\;6-1)\), second row \((2+12,\;0-4,\;4+4)\), giving \(\begin{pmatrix}0&1&5\\14&-4&8\end{pmatrix}\). For \(\mathbf{CA}\) the number of columns of \(\mathbf C\) (3) does not equal the number of rows of \(\mathbf A\) (2), so the product is not defined.
   3. \(\mathbf A^2=\begin{pmatrix}9-2&-3-4\\6+8&-2+16\end{pmatrix}=\begin{pmatrix}7&-7\\14&14\end{pmatrix}\). Then \(\mathbf A^2-7\mathbf A=\begin{pmatrix}7&-7\\14&14\end{pmatrix}-\begin{pmatrix}21&-7\\14&28\end{pmatrix}=\begin{pmatrix}-14&0\\0&-14\end{pmatrix}=-14\mathbf I\) ✓
   4. Multiply \(\mathbf A^2=7\mathbf A-14\mathbf I\) by \(\mathbf A\): \(\mathbf A^3=7\mathbf A^2-14\mathbf A=7(7\mathbf A-14\mathbf I)-14\mathbf A=\) \(35\mathbf A-98\mathbf I\). Check: both give \(\begin{pmatrix}7&-35\\70&42\end{pmatrix}\).

2. Answer to Question 2

   1. \(5\times4-(-2)\times3=20+6=\) 26
   2. \(2\begin{vmatrix}3&-1\\4&2\end{vmatrix}-0+1\begin{vmatrix}1&3\\0&4\end{vmatrix}=2(6+4)+1(4-0)=20+4=\) 24
   3. \(1(2k-1)-2(0-3)+k(0-3k)=2k-1+6-3k^2=\) \(-3k^2+2k+5\) ✓
   4. Singular when \(3k^2-2k-5=0\), i.e. \((3k-5)(k+1)=0\), so \(k=-1\) or \(k=\tfrac53\).

3. Answer to Question 3

   1. \(\mathbf i\mapsto\binom01\), \(\mathbf j\mapsto\binom{-1}0\), so the matrix is \(\begin{pmatrix}0&-1\\1&0\end{pmatrix}\). \(P\mapsto\binom{0\cdot2+(-1)(-1)}{1\cdot2+0}=\) \((1,2)\).
   2. \(\mathbf i\mapsto\binom0{-1}\) and \(\mathbf j\mapsto\binom{-1}0\): \(\begin{pmatrix}0&-1\\-1&0\end{pmatrix}\). \(P\mapsto\binom{0+1}{-2+0}=\) \((1,-2)\).
   3. \(\begin{pmatrix}3&0\\0&3\end{pmatrix}\); \(P\mapsto\) \((6,-3)\).
   4. \(\mathbf i\mapsto\binom20\), \(\mathbf j\) unchanged: \(\begin{pmatrix}2&0\\0&1\end{pmatrix}\); \(P\mapsto\) \((4,-1)\) (only the \(x\)-coordinate is doubled).

4. Answer to Question 4

   1. \(\mathbf i\mapsto\mathbf j\) and \(\mathbf j\mapsto\mathbf i\): reflection in the line \(y=x\). Determinant \(0-1=-1\): area is unchanged and orientation is reversed.
   2. The matrix has the form \(\begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix}\) with \(\cos\theta=\tfrac12\), \(\sin\theta=\tfrac{\sqrt3}2\), so \(\theta=60^\circ\): rotation of \(60^\circ\) anticlockwise about \(O\). Determinant \(\tfrac14+\tfrac34=1\).
   3. \(\mathbf i\) is fixed and \(\mathbf j\mapsto\binom31\): points move parallel to the \(x\)-axis by 3 times their \(y\)-coordinate. Shear with the \(x\)-axis invariant, \((0,1)\mapsto(3,1)\). Determinant \(1\): area unchanged.
   4. Every vector is multiplied by \(-2\): enlargement with centre \(O\) and scale factor \(-2\) (equivalently a rotation of \(180^\circ\) followed by an enlargement of scale factor 2). Determinant \(4\): areas are multiplied by 4.

5. Answer to Question 5

   1. \(\mathbf P=\begin{pmatrix}1&0\\0&-1\end{pmatrix}\), \(\mathbf Q=\begin{pmatrix}0&-1\\1&0\end{pmatrix}\)
   2. "\(\mathbf P\) then \(\mathbf Q\)" is the product \(\mathbf{QP}\) (the first transformation is on the right): \(\mathbf{QP}=\begin{pmatrix}0&-1\\1&0\end{pmatrix}\begin{pmatrix}1&0\\0&-1\end{pmatrix}=\) \(\begin{pmatrix}0&1\\1&0\end{pmatrix}\), a reflection in \(y=x\).
   3. \(\mathbf{PQ}=\begin{pmatrix}1&0\\0&-1\end{pmatrix}\begin{pmatrix}0&-1\\1&0\end{pmatrix}=\) \(\begin{pmatrix}0&-1\\-1&0\end{pmatrix}\), a reflection in \(y=-x\). The order matters.
   4. \((3,1)\xrightarrow{\mathbf P}(3,-1)\xrightarrow{\mathbf Q}(1,3)\), and the reflection in \(y=x\) sends \((3,1)\) to \((1,3)\) ✓

6. Answer to Question 6

   1. Invariant points satisfy \(\mathbf M\mathbf x=\mathbf x\), i.e. \(3x+y=0\) and \(2x+2y=0\). The matrix \(\mathbf M-\mathbf I=\begin{pmatrix}3&1\\2&2\end{pmatrix}\) has determinant \(6-2=4\ne0\), so the only solution is \(x=y=0\).
   2. \((x,mx)\mapsto(4x+mx,\;2x+3mx)\). For the image to lie on \(y=mx\): \(2x+3mx=m(4x+mx)\), so \(2+3m=4m+m^2\), i.e. \(m^2+m-2=0\), \((m+2)(m-1)=0\). The invariant lines are \(y=x\) and \(y=-2x\).
   3. No. On \(y=x\), \((1,1)\mapsto(5,5)\), so points move along the line (scale factor 5). On \(y=-2x\), \((1,-2)\mapsto(2,-4)\) (scale factor 2). Each line maps to itself but its points are not fixed.

7. Answer to Question 7

   1. Each point \((x,y)\mapsto(x+\tfrac12y,\;y)\): \((0,0)\), \((4,0)\), \((5.5,3)\), \((1.5,3)\).
   2. \(\det\mathbf S=1\times1-\tfrac12\times0=1\), so the area scale factor is 1: the area stays 12 cm² (a parallelogram with the same base 4 and height 3).
   3. Shear first, so the stretch goes on the left: \(\mathbf{WS}=\begin{pmatrix}2&0\\0&1\end{pmatrix}\begin{pmatrix}1&\frac{1}{2}\\0&1\end{pmatrix}=\) \(\begin{pmatrix}2&1\\0&1\end{pmatrix}\).
   4. \(\det(\mathbf{WS})=2\), so the area is \(12\times2=\) 24 cm². \((4,3)\mapsto(2\times4+3,\;3)=\) \((11,3)\).

8. Answer to Question 8

   1. \(A\mapsto\binom31\), \(B\mapsto\binom{3-1}{1+2}=\binom23\), \(C\mapsto\binom{-1}2\): \(A'(3,1)\), \(B'(2,3)\), \(C'(-1,2)\).
   2. Area factor \(=\det\mathbf M=3\times2-(-1)\times1=\) 7, so the image tile has area 7 square units.
   3. One square unit is \(30\times30=900\) cm², so the area is \(7\times900=\) 6300 cm².
   4. \(9.45\) m² \(=94\,500\) cm². \(94\,500\div6300=\) 15 tiles.

9. Answer to Question 9

   1. \(\mathbf R_{30}=\begin{pmatrix}\frac{\sqrt3}2&-\frac12\\\frac12&\frac{\sqrt3}2\end{pmatrix}\), \(\mathbf R_{45}=\begin{pmatrix}\frac{\sqrt2}2&-\frac{\sqrt2}2\\\frac{\sqrt2}2&\frac{\sqrt2}2\end{pmatrix}\)
   2. Top-left entry: \(\tfrac{\sqrt2}2\cdot\tfrac{\sqrt3}2-\tfrac{\sqrt2}2\cdot\tfrac12=\tfrac{\sqrt6-\sqrt2}4\); bottom-left: \(\tfrac{\sqrt2}2\cdot\tfrac{\sqrt3}2+\tfrac{\sqrt2}2\cdot\tfrac12=\tfrac{\sqrt6+\sqrt2}4\). The product is \(\begin{pmatrix}\frac{\sqrt6-\sqrt2}4&-\frac{\sqrt6+\sqrt2}4\\\frac{\sqrt6+\sqrt2}4&\frac{\sqrt6-\sqrt2}4\end{pmatrix}\), which has the rotation pattern with \(\cos\theta\) on the diagonal: a rotation of \(30^\circ+45^\circ=75^\circ\).
   3. A rotation of \(75^\circ\) has top-left entry \(\cos75^\circ\), and comparing with part (b) gives \(\cos75^\circ=\dfrac{\sqrt6-\sqrt2}4\) ✓
   4. The tool moves to \((10\cos75^\circ,\;10\sin75^\circ)=\) \((2.59,\;9.66)\) cm.
   5. No: rotations about the same centre commute, since both orders give a rotation of \(75^\circ\) (\(\mathbf R_{30}\mathbf R_{45}=\mathbf R_{45}\mathbf R_{30}\)).

10. Answer to Question 10

    1. Enlargement \(\begin{pmatrix}2&0\\0&2\end{pmatrix}\); reflection in the \(y\)-axis \(\begin{pmatrix}-1&0\\0&1\end{pmatrix}\).
    2. Enlargement first, so it goes on the right: \(\begin{pmatrix}-1&0\\0&1\end{pmatrix}\begin{pmatrix}2&0\\0&2\end{pmatrix}=\) \(\begin{pmatrix}-2&0\\0&2\end{pmatrix}\).
    3. \((0,0)\mapsto(0,0)\), \((6,0)\mapsto(-12,0)\), \((0,4)\mapsto(0,8)\): \((0,0)\), \((-12,0)\), \((0,8)\).
    4. Original area \(\tfrac12\times6\times4=12\) m². \(|\det|=|-4|=4\), so the new bed has area 48 m² (check: \(\tfrac12\times12\times8=48\) ✓). The negative determinant shows that the orientation is reversed by the reflection.

11. Answer to Question 11

    1. \(2\theta=60^\circ\): \(\mathbf R=\begin{pmatrix}\frac12&\frac{\sqrt3}2\\\frac{\sqrt3}2&-\frac12\end{pmatrix}\)
    2. \(\mathbf R\binom20=\binom{1}{\sqrt3}\): \((1,\sqrt3)\). (Both points are 2 units from \(O\), at angles \(0^\circ\) and \(60^\circ\), symmetric about the \(30^\circ\) line.)
    3. \(\mathbf R\binom{\sqrt3}1=\binom{\frac{\sqrt3}2+\frac{\sqrt3}2}{\frac32-\frac12}=\binom{\sqrt3}1\) ✓. The point lies on the mirror line, since \(1=\sqrt3\tan30^\circ\), and points on the mirror do not move.
    4. \(\mathbf R^2=\begin{pmatrix}\frac14+\frac34&\frac{\sqrt3}4-\frac{\sqrt3}4\\\frac{\sqrt3}4-\frac{\sqrt3}4&\frac34+\frac14\end{pmatrix}=\mathbf I\) ✓. Reflecting twice brings the character back to its starting position, so the reflection is its own inverse.

12. Answer to Question 12

    1. \(\det\mathbf F=2\times2-1\times4=\) 0, so every area is multiplied by 0: the printed image has zero area (it is squashed onto a line).
    2. \((x,y)\mapsto(2x+y,\;4x+2y)=(2x+y,\;2(2x+y))\), so the image \((X,Y)\) always satisfies \(Y=2X\).
    3. \(2x+y=0\): every point on the line \(y=-2x\) is printed at the origin, for example \((1,-2)\mapsto(0,0)\).
    4. The matrix is singular (no inverse): many points share the same image, for example every point on \(y=-2x\) goes to \(O\), so the image does not tell us which point it came from.

13. Answer to Question 13

    1. \(\mathbf R_z=\begin{pmatrix}0&-1&0\\1&0&0\\0&0&1\end{pmatrix}\), \(\mathbf R_x=\begin{pmatrix}1&0&0\\0&0&-1\\0&1&0\end{pmatrix}\) (the columns are the images of \(\mathbf i\), \(\mathbf j\), \(\mathbf k\)).
    2. \(\mathbf R_z\begin{pmatrix}1\\2\\3\end{pmatrix}=\) \((-2,1,3)\); the height \(z=3\) does not change.
    3. \(\mathbf R_x\begin{pmatrix}-2\\1\\3\end{pmatrix}=\begin{pmatrix}-2\\-3\\1\end{pmatrix}\): \((-2,-3,1)\).
    4. \(\mathbf R_x(1,2,3)=(1,-3,2)\), then \(\mathbf R_z(1,-3,2)=\) \((3,1,2)\). The answers are different: rotations about different axes do not commute.

14. Answer to Question 14

    1. \(\mathbf D\begin{pmatrix}1\\1\\1\end{pmatrix}=\begin{pmatrix}2+1+0\\0+3+1\\1+0+1\end{pmatrix}=\) \((3,4,2)\)
    2. \(2(3\cdot1-1\cdot0)-1(0\cdot1-1\cdot1)+0=6+1=\) 7
    3. The volume scale factor is \(|\det\mathbf D|=7\). Original volume \(2\times3\times4=24\) cm³, so the image has volume 168 cm³.
    4. A transformation that flattens the box onto a plane or a line gives zero volume, so its determinant would be 0 (the matrix would be singular).

15. Answer to Question 15

    1. The columns are the images of \(\mathbf i\) and \(\mathbf j\), which are \(A'\) and \(C'\): \(\mathbf M=\begin{pmatrix}3&1\\1&2\end{pmatrix}\). Check: \(\mathbf M\binom11=\binom43=B'\) ✓
    2. \(\det\mathbf M=6-1=5\), and the unit square has area 1, so the area is 5 square units.
    3. \(\mathbf M^{-1}=\dfrac15\begin{pmatrix}2&-1\\-1&3\end{pmatrix}\), so \(\mathbf x=\dfrac15\binom{14-4}{-7+12}=\dfrac15\binom{10}5\): \((2,1)\). Check: \(\mathbf M\binom21=\binom{7}{4}\) ✓
    4. \((x,mx)\mapsto(3x+mx,\;x+2mx)\). Need \(1+2m=m(3+m)\), so \(m^2+m-1=0\) and \(m=\dfrac{-1\pm\sqrt5}2\).
    5. A rotation followed by an enlargement keeps angles, so \(\mathbf i\) and \(\mathbf j\) would map to perpendicular vectors of equal length. Here \(\binom31\cdot\binom12=5\ne0\) and the lengths \(\sqrt{10}\) and \(\sqrt5\) differ, so \(T\) is not a rotation with an enlargement.

16. Answer to Question 16

    1. \(\mathbf R\binom31=\dfrac15\binom{-9+4}{12+3}=\) \((-1,3)\); \(\mathbf R\binom12=\dfrac15\binom{-3+8}{4+6}=\) \((1,2)\).
    2. \(\mathbf R\binom t{2t}=\dfrac15\binom{-3t+8t}{4t+6t}=\binom t{2t}\) ✓ for every \(t\), so \(y=2x\) is a line of invariant points.
    3. \(\mathbf R\binom{2t}{-t}=\dfrac15\binom{-6t-4t}{8t-3t}=\binom{-2t}{t}\), which lies on \(y=-\tfrac12x\), so the line is invariant. But \((2,-1)\mapsto(-2,1)\), so points on it move (they are reversed through \(O\)).
    4. \(\det\mathbf R=\tfrac1{25}(-9-16)=-1\) and \(\mathbf R^2=\tfrac1{25}\begin{pmatrix}9+16&-12+12\\-12+12&16+9\end{pmatrix}=\mathbf I\). With a line of invariant points and determinant \(-1\): reflection in the line \(y=2x\).
    5. \(\cos2\theta=\dfrac{1-\tan^2\theta}{1+\tan^2\theta}=\dfrac{1-4}{5}=-\dfrac35\) and \(\sin2\theta=\dfrac{2\tan\theta}{1+\tan^2\theta}=\dfrac45\). The reflection matrix \(\begin{pmatrix}\cos2\theta&\sin2\theta\\\sin2\theta&-\cos2\theta\end{pmatrix}\) is exactly \(\mathbf R\) ✓

17. Answer to Question 17

    1. \(\mathbf E=\begin{pmatrix}2&0\\0&2\end{pmatrix}\), \(\mathbf R=\begin{pmatrix}0&-1\\1&0\end{pmatrix}\), \(\mathbf F=\begin{pmatrix}0&1\\1&0\end{pmatrix}\)
    2. The first transformation acts first on the vector, so it is written next to the vector, on the right: \(\mathbf F(\mathbf R(\mathbf E\mathbf x))=(\mathbf{FRE})\mathbf x\). \(\mathbf{FR}=\begin{pmatrix}1&0\\0&-1\end{pmatrix}\), so \(\mathbf{FRE}=2\begin{pmatrix}1&0\\0&-1\end{pmatrix}=\) \(\begin{pmatrix}2&0\\0&-2\end{pmatrix}\).
    3. \((x,y)\mapsto(2x,-2y)\): a reflection in the \(x\)-axis combined with an enlargement, centre \(O\), scale factor 2 (in either order).
    4. \(|\det|=|{-4}|=4\), so the area is \(6\times4=\) 24 cm².
    5. Solve \((2x,-2y)=(6,-2)\): \((3,1)\).

18. Answer to Question 18

    1. Reflection in \(x=y\) swaps the \(x\)- and \(y\)-coordinates: \(\mathbf i\leftrightarrow\mathbf j\), \(\mathbf k\) fixed, so \(\mathbf A=\begin{pmatrix}0&1&0\\1&0&0\\0&0&1\end{pmatrix}\). Reflection in \(z=0\) changes the sign of \(z\): \(\mathbf B=\begin{pmatrix}1&0&0\\0&1&0\\0&0&-1\end{pmatrix}\).
    2. \(\mathbf C=\mathbf{BA}=\) \(\begin{pmatrix}0&1&0\\1&0&0\\0&0&-1\end{pmatrix}\)
    3. \(\det\mathbf C=\det\mathbf B\det\mathbf A=(-1)(-1)=\) 1: volumes are unchanged and orientation is preserved (two reflections).
    4. \(\mathbf C\mathbf x=\mathbf x\) gives \(y=x\), \(x=y\), \(-z=z\), so \(z=0\) and \(x=y\): the invariant points are \((t,t,0)\), the line through \(O\) with direction \(\mathbf i+\mathbf j\).
    5. \(\mathbf C^2=\mathbf I\). A transformation with determinant 1, a line of invariant points and \(\mathbf C^2=\mathbf I\) is a rotation of \(180^\circ\) about the line \(x=y,\ z=0\).

19. Answer to Question 19

    1. \(\mathbf S^2=\begin{pmatrix}1&2\\0&1\end{pmatrix}\), \(\mathbf S^3=\begin{pmatrix}1&3\\0&1\end{pmatrix}\), \(\mathbf S^4=\begin{pmatrix}1&4\\0&1\end{pmatrix}\) (each time the top-right entry increases by 1).
    2. \(\mathbf S^n=\begin{pmatrix}1&n\\0&1\end{pmatrix}\): each shear moves a point \(y\) units to the right, so \(n\) shears move it \(ny\) units. (This can be proved by induction.)
    3. \(\mathbf S^{10}\binom23=\binom{2+30}{3}\): \((32,3)\).
    4. \(\mathbf S^{-10}=\begin{pmatrix}1&-10\\0&1\end{pmatrix}\) and \(\begin{pmatrix}1&-10\\0&1\end{pmatrix}\binom{32}3=\binom{32-30}3=\binom23\) ✓
    5. \(\det\mathbf S^n=(\det\mathbf S)^n=1^n=1\), so every area is multiplied by 1.

20. Answer to Question 20

    1. \(\mathbf M_\beta\mathbf M_\alpha\) has top-left entry \(\cos2\beta\cos2\alpha+\sin2\beta\sin2\alpha=\cos(2\beta-2\alpha)\) and bottom-left entry \(\sin2\beta\cos2\alpha-\cos2\beta\sin2\alpha=\sin(2\beta-2\alpha)\). Similarly the second column is \(\binom{-\sin(2\beta-2\alpha)}{\cos(2\beta-2\alpha)}\). So the product is a rotation through \(2(\beta-\alpha)\) about \(O\).
    2. With \(\beta-\alpha=30^\circ\) the angle is \(2\times30^\circ=\) \(60^\circ\). A kaleidoscope with mirrors at \(30^\circ\) therefore shows copies rotated by multiples of \(60^\circ\), giving 6-fold symmetry.
    3. \(\mathbf{XS}=\begin{pmatrix}p&p+q\\r&r+s\end{pmatrix}\) and \(\mathbf{SX}=\begin{pmatrix}p+r&q+s\\r&s\end{pmatrix}\). Equating: \(p=p+r\) gives \(r=0\); \(p+q=q+s\) gives \(s=p\). So \(\mathbf X=\begin{pmatrix}p&q\\0&p\end{pmatrix}\) for any \(p\), \(q\).
    4. Every matrix that commutes with \(\mathbf S\) has determinant \(p^2\ge0\), but a reflection has determinant \(-1\). So no reflection commutes with \(\mathbf S\).