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

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1. Answer to Question 1

   1. Each entry of a product is a row of the first matrix times a column of the second. \(AB=\) \(\begin{pmatrix}1\cdot0+2\cdot1&1\cdot1+2\cdot0\\3\cdot0+4\cdot1&3\cdot1+4\cdot0\end{pmatrix}\)\(=\) \(\begin{pmatrix}2&1\\4&3\end{pmatrix}\) and \(BA=\) \(\begin{pmatrix}3&4\\1&2\end{pmatrix}\). They are different: matrix multiplication is not commutative.
   2. Multiplying by \(B\) on the right (\(AB\)) swaps the columns of \(A\); multiplying on the left (\(BA\)) swaps the rows of \(A\).
   3. \(A^2=AA=\) \(\begin{pmatrix}1+6&2+8\\3+12&6+16\end{pmatrix}\)\(=\) \(\begin{pmatrix}7&10\\15&22\end{pmatrix}\).
   4. No. Expanding carefully, \((A+B)^2=A^2+AB+BA+B^2\). The middle terms combine into \(2AB\) only if \(AB=BA\), and part (a) shows that \(AB\ne BA\).

2. Answer to Question 2

   1. \(\det C=3\times2-1\times5=1\). For a \(2\times2\) matrix the inverse is \(\dfrac1{\det}\times\) (swap the diagonal, negate the other entries), so \(C^{-1}=\) \(\begin{pmatrix}2&-1\\-5&3\end{pmatrix}\).
   2. \(\det D=4-4=0\), so \(D\) is singular: it has no inverse, and its rows are proportional.
   3. \(\det=4-k^2=0\) gives \(k=\) \(2\) or \(-2\).

3. Answer to Question 3

   1. \(\det=2(3\times2-2\times1)-0+1(1\times1-3\times1)=2(4)+(-2)=\) 6.
   2. The matrix is upper triangular, so the determinant is the product of the diagonal entries: \(2\times3\times(-4)=\) \(-24\).
   3. \(R_2\to R_2-4R_1\) gives \((0,-3,-6)\) and \(R_3\to R_3-7R_1\) gives \((0,-6,-12)\). Now \(R_3=2R_2\), so the determinant is 0 (row operations of this type do not change it).

4. Answer to Question 4

   1. The coefficient matrix is \(\begin{pmatrix}2&1\\5&3\end{pmatrix}\), with determinant \(6-5=1\ne0\), so it is invertible with inverse \(\begin{pmatrix}3&-1\\-5&2\end{pmatrix}\). Multiplying both sides by the inverse, \(\begin{pmatrix}x\\y\end{pmatrix}\)\(=\) \(\begin{pmatrix}3&-1\\-5&2\end{pmatrix}\)\(\begin{pmatrix}5\\13\end{pmatrix}\)\(=\) \(\begin{pmatrix}15-13\\-25+26\end{pmatrix}\)\(=\) \(\begin{pmatrix}2\\1\end{pmatrix}\). So \(x=2\), \(y=1\). Check: \(4+1=5\) and \(10+3=13\).
   2. The determinant of the coefficients is \(2\times3-1\times1=5\). Cramer's rule replaces one column by the right-hand side: \(x=\dfrac{\begin{vmatrix}5&1\\10&3\end{vmatrix}}5=\dfrac{15-10}5=1\) and \(y=\dfrac{\begin{vmatrix}2&5\\1&10\end{vmatrix}}5=\dfrac{20-5}5=3\). So \(x=1\), \(y=3\). Check: \(2+3=5\) and \(1+9=10\).

5. Answer to Question 5

   1. \(\det(AB)=\det A\det B=5\times(-2)=\) \(-10\) (product rule).
   2. Multiplying every row of a \(3\times3\) matrix by 2 multiplies the determinant by \(2^3\): \(\det(2A)=8\times5=\) 40.
   3. The determinant is unchanged by transposing: \(\det A^T=\) 5. And \(\det A^{-1}=\dfrac1{\det A}=\) \(\dfrac15\).
   4. \(\det(A^2B^{-1})=(\det A)^2\cdot\dfrac1{\det B}=25\times\dfrac1{-2}=\) \(-12.5\).

6. Answer to Question 6

   1. \(AB=\) \(\begin{pmatrix}2&1\\4&3\end{pmatrix}\) so \((AB)^T=\) \(\begin{pmatrix}2&4\\1&3\end{pmatrix}\). Also \(B^TA^T=\) \(\begin{pmatrix}0&1\\1&0\end{pmatrix}\)\(\begin{pmatrix}1&3\\2&4\end{pmatrix}\)\(=\) \(\begin{pmatrix}2&4\\1&3\end{pmatrix}\). They agree.
   2. \(B^{-1}=B\) (since \(B^2=I\)) and \(A^{-1}=\dfrac1{-2}\)\(\begin{pmatrix}4&-2\\-3&1\end{pmatrix}\). Then \(B^{-1}A^{-1}=\dfrac1{-2}\)\(\begin{pmatrix}-3&1\\4&-2\end{pmatrix}\). Also \(\det(AB)=2\times3-1\times4=2\) and \((AB)^{-1}=\dfrac12\)\(\begin{pmatrix}3&-1\\-4&2\end{pmatrix}\). Dividing by \(-2\) in the first expression gives the same matrix.
   3. \(\det R=0\times0-(-1)(1)=1\), and \(R^2=\) \(\begin{pmatrix}-1&0\\0&-1\end{pmatrix}\)\(=-I\), a rotation through \(180^\circ\).
   4. Area \(=\tfrac12\left|\begin{vmatrix}4&1\\2&5\end{vmatrix}\right|=\tfrac12|20-2|=\) 9.

7. Answer to Question 7

   1. Each equation gives one row: \(3c+2k=11.70\) and \(2c+3k=11.05\). In matrix form, \(\begin{pmatrix}3&2\\2&3\end{pmatrix}\)\(\begin{pmatrix}c\\k\end{pmatrix}\)\(=\) \(\begin{pmatrix}11.70\\11.05\end{pmatrix}\).
   2. \(\det M=3\times3-2\times2=5\ne0\), so \(M\) is invertible. Swapping the diagonal entries and negating the others, \(M^{-1}=\dfrac15\)\(\begin{pmatrix}3&-2\\-2&3\end{pmatrix}\).
   3. \(\mathbf x=M^{-1}\mathbf b\) gives \(c=\dfrac15(3\times11.70-2\times11.05)=\dfrac{35.1-22.1}5=\) £2.60 and \(k=\dfrac15(-2\times11.70+3\times11.05)=\dfrac{-23.4+33.15}5=\) £1.95. Check: \(3\times2.60+2\times1.95=7.80+3.90=11.70\).
   4. \(5\times2.60+4\times1.95=13+7.80=\) £20.80 for 5 coffees and 4 cakes.

8. Answer to Question 8

   1. Put one day per row and one product per column: \(S=\) \(\begin{pmatrix}40&25&30\\35&30&20\\50&20&25\end{pmatrix}\), and list the prices in the same product order: \(\mathbf p=\) \(\begin{pmatrix}1.10\\1.40\\2.20\end{pmatrix}\).
   2. Each row of \(S\) times \(\mathbf p\) gives (quantity \(\times\) price) summed over products: \(S\mathbf p=\) \(\begin{pmatrix}44+35+66\\38.5+42+44\\55+28+55\end{pmatrix}\)\(=\) \(\begin{pmatrix}145.00\\124.50\\138.00\end{pmatrix}\). The entries are the takings on Monday, Tuesday and Wednesday in pounds.
   3. Adding the entries, \(145+124.50+138=\) £407.50 over the three days.
   4. The new price vector is \(1.1\,\mathbf p\), so the new takings are \(S(1.1\,\mathbf p)=1.1\,S\mathbf p\). The total becomes \(1.1\times407.50=\) £448.25.

9. Answer to Question 9

   1. \(0.95+0.05=1\) and \(0.03+0.97=1\): everyone in each place either stays or moves, so the total population is conserved.
   2. \(\mathbf v_1=P\mathbf v_0=\) \(\begin{pmatrix}0.95(600)+0.03(400)\\0.05(600)+0.97(400)\end{pmatrix}\)\(=\) \(\begin{pmatrix}582\\418\end{pmatrix}\).
   3. \(\mathbf v_{10}=P^{10}\mathbf v_0\approx\) \(\begin{pmatrix}472.7\\527.3\end{pmatrix}\) thousand.
   4. From \(-0.05c+0.03s=0\) we get \(c=0.6s\). With \(c+s=1000\): \(1.6s=1000\), \(s=625\) and \(c=375\). In the long run 375 thousand live in the city and 625 thousand in the suburbs.

10. Answer to Question 10

    1. \(T\binom00=\binom00\); \(T\binom20=\binom04\); \(T\binom03=\binom{-6}0\). The vertices are \((0,0)\), \((0,4)\) and \((-6,0)\).
    2. \(\det T=0-(-2)(2)=4\). The original area is \(\tfrac12\times2\times3=3\), so the new area is \(4\times3=\) 12. Check: the new triangle has base 4 and height 6, area \(12\).
    3. \(T=2\times\) \(\begin{pmatrix}0&-1\\1&0\end{pmatrix}\): an anticlockwise rotation through \(90^\circ\) followed by an enlargement with scale factor 2. The area scale factor is \(2^2=4\).
    4. \(T^{-1}=\dfrac14\)\(\begin{pmatrix}0&2\\-2&0\end{pmatrix}\)\(=\) \(\begin{pmatrix}0&1/2\\-1/2&0\end{pmatrix}\).

11. Answer to Question 11

    1. \(\det K=3\times5-3\times2=15-6=9\). Also \(9\times3=27=26+1\equiv1\pmod{26}\), so 3 is the inverse of 9 modulo 26. Because \(\det K\) has an inverse mod 26, \(K\) can be inverted and messages can be decoded.
    2. "HI" is the pair \((7,8)\). \(K\binom78=\binom{21+24}{14+40}=\binom{45}{54}\equiv\binom{19}2\pmod{26}\), which is the pair "TC".
    3. The adjugate of \(K\) is \(\begin{pmatrix}5&-3\\-2&3\end{pmatrix}\). Instead of dividing by 9, multiply by \(9^{-1}\equiv3\): \(\begin{pmatrix}15&-9\\-6&9\end{pmatrix}\)\(\equiv\) \(\begin{pmatrix}15&17\\20&9\end{pmatrix}\) (mod 26), adding 26 to the negative entries.
    4. \(\binom{15\cdot19+17\cdot2}{20\cdot19+9\cdot2}=\binom{319}{398}\equiv\binom78\), since \(319=12\times26+7\) and \(398=15\times26+8\). This is "HI" again, so the decoding matrix works.

12. Answer to Question 12

    1. The three conditions are \(x+y+z=100\), \(2x+5y+z=270\) and \(x+3z=110\). In matrix form, \(\begin{pmatrix}1&1&1\\2&5&1\\1&0&3\end{pmatrix}\)\(\begin{pmatrix}x\\y\\z\end{pmatrix}\)\(=\) \(\begin{pmatrix}100\\270\\110\end{pmatrix}\).
    2. Expanding along the first row, \(\det M=1(15-0)-1(6-1)+1(0-5)=15-5-5=5\). Since \(\det M\ne0\), \(M\) is invertible and the system has exactly one solution.
    3. \(R_2\to R_2-2R_1\) gives \((0,3,-1\,|\,70)\), and \(R_3\to R_3-R_1\) gives \((0,-1,2\,|\,10)\). From the new \(R_3\): \(y=2z-10\). Substituting into \(R_2\): \(3(2z-10)-z=70\), so \(5z=100\) and \(z=20\). Then \(y=30\), and from the first row \(x=50\). So 50 g oats, 30 g nuts, 20 g raisins.
    4. \(50+30+20=100\); \(2(50)+5(30)+20=270\); \(50+3(20)=110\). All three targets are met.

13. Answer to Question 13

    1. \(\begin{pmatrix}1&-1&-1\\2&4&0\\0&4&-6\end{pmatrix}\)\(\begin{pmatrix}I_1\\I_2\\I_3\end{pmatrix}\)\(=\) \(\begin{pmatrix}0\\14\\2\end{pmatrix}\).
    2. \(\det=1(-24-0)-(-1)(-12-0)+(-1)(8-0)=-24-12-8=-44\).
    3. From the third equation \(I_2=\dfrac{2+6I_3}4\). From the first, \(I_1=I_2+I_3\). Substitute into the second: \(2(I_2+I_3)+4I_2=14\), that is \(6I_2+2I_3=14\), so \(6\left(\dfrac{2+6I_3}4\right)+2I_3=14\), giving \(3+9I_3+2I_3=14\) and \(I_3=1\). Then \(I_2=2\) and \(I_1=3\). So \(I_1=3\) A, \(I_2=2\) A, \(I_3=1\) A.
    4. \(3-2-1=0\). The current entering the junction equals the currents leaving it.

14. Answer to Question 14

    1. Rearrange: \(\mathbf x-A\mathbf x=\mathbf d\), so \((I-A)\mathbf x=\mathbf d\).
    2. \(I-A=\) \(\begin{pmatrix}0.8&-0.3\\-0.4&0.9\end{pmatrix}\), with determinant \(0.72-0.12=0.6\).
    3. \((I-A)^{-1}=\dfrac1{0.6}\)\(\begin{pmatrix}0.9&0.3\\0.4&0.8\end{pmatrix}\). Then \(\mathbf x=\dfrac1{0.6}\)\(\begin{pmatrix}90+15\\40+40\end{pmatrix}\)\(=\) \(\begin{pmatrix}175\\133.3\end{pmatrix}\). So steel production is 175 units and energy 133.3 units.
    4. Each industry also uses the other's output, so it must produce extra to supply the inputs of the other industry (and its own). The extra is \(175-100=75\) units of steel and \(133.3-50=83.3\) of energy.

15. Answer to Question 15

    1. \(Q^2=\) \(\begin{pmatrix}2&1\\1&1\end{pmatrix}\) and \(Q^3=Q^2Q=\) \(\begin{pmatrix}3&2\\2&1\end{pmatrix}\). The entries are consecutive Fibonacci numbers: \(Q^3\) contains \(F_4=3\), \(F_3=2\) and \(F_2=1\).
    2. Conjecture: \(Q^n=\) \(\begin{pmatrix}F_{n+1}&F_n\\F_n&F_{n-1}\end{pmatrix}\). Base case \(n=1\): \(Q=\) \(\begin{pmatrix}1&1\\1&0\end{pmatrix}\), which matches \(F_2=1\), \(F_1=1\), \(F_0=0\). Inductive step: assume it holds for \(n\). Then \(Q^{n+1}=Q^nQ=\) \(\begin{pmatrix}F_{n+1}+F_n&F_{n+1}\\F_n+F_{n-1}&F_n\end{pmatrix}\)\(=\) \(\begin{pmatrix}F_{n+2}&F_{n+1}\\F_{n+1}&F_n\end{pmatrix}\) by the recurrence, so it holds for \(n+1\). By induction it holds for all \(n\ge1\).
    3. Using the formula, \(Q^{10}=\) \(\begin{pmatrix}89&55\\55&34\end{pmatrix}\), so \(F_{10}=\) 55.
    4. \(\det Q=-1\) and determinants multiply, so \(\det Q^n=(-1)^n\). Therefore \(F_{n+1}F_{n-1}-F_n^2=(-1)^n\). Check with \(n=10\): \(89\times34-55^2=3026-3025=1\).

16. Answer to Question 16

    1. All entries are non-negative and each column sums to 1 (\(0.7+0.2+0.1\), \(0.3+0.5+0.2\), \(0.1+0.4+0.5\)).
    2. \(\mathbf v_1=\) first column \(=(0.7,0.2,0.1)^T\). \(\mathbf v_2=W\mathbf v_1=(0.49+0.06+0.01,\;0.14+0.10+0.04,\;0.07+0.04+0.05)^T=(0.56,0.28,0.16)^T\).
    3. \((W-I)\mathbf v=0\) gives \(-0.3a+0.3b+0.1c=0\) and \(0.2a-0.5b+0.4c=0\) with \(a+b+c=1\). From the first, \(c=3a-3b\). Substitute into the second: \(0.2a-0.5b+1.2a-1.2b=0\), so \(1.4a=1.7b\) and \(b=\dfrac{14}{17}a\). Then \(c=3a-\dfrac{42}{17}a=\dfrac9{17}a\). The sum \(a\left(1+\dfrac{14}{17}+\dfrac9{17}\right)=\dfrac{40}{17}a=1\) gives \(a=\dfrac{17}{40}=0.425\), \(b=0.35\), \(c=0.225\).
    4. Rain occurs on 22.5 days in 100 in the long run.

17. Answer to Question 17

    1. \(RS=\) \(\begin{pmatrix}0&-1\\1&0\end{pmatrix}\)\(\begin{pmatrix}1&0\\0&-1\end{pmatrix}\)\(=\) \(\begin{pmatrix}0&1\\1&0\end{pmatrix}\): reflection in the line \(y=x\).
    2. \(SR=\) \(\begin{pmatrix}1&0\\0&-1\end{pmatrix}\)\(\begin{pmatrix}0&-1\\1&0\end{pmatrix}\)\(=\) \(\begin{pmatrix}0&-1\\-1&0\end{pmatrix}\): reflection in the line \(y=-x\). This differs from \(RS\).
    3. \(\det S=-1\), \(\det R=1\), \(\det RS=-1\). A negative determinant means that orientation is reversed (a reflection) and the area is multiplied by \(|\det|=1\).
    4. Under \(RS\): \(\binom{0\cdot3+1\cdot1}{1\cdot3+0\cdot1}=\binom13\). Under \(SR\): \(\binom{-1}{-3}\).

18. Answer to Question 18

    1. The matrix is \(\begin{pmatrix}1&k\\k&4\end{pmatrix}\), with determinant \(1\times4-k\times k=4-k^2\).
    2. A unique solution exists when \(k\ne\pm2\). By Cramer's rule: \(x=\dfrac{2\times4-k\times4}{4-k^2}=\dfrac{4(2-k)}{(2-k)(2+k)}=\dfrac4{2+k}\) and \(y=\dfrac{1\times4-k\times2}{4-k^2}=\dfrac{2(2-k)}{(2-k)(2+k)}=\dfrac2{2+k}\).
    3. For \(k=2\) the equations are \(x+2y=2\) and \(2x+4y=4\). The second is twice the first, so there are infinitely many solutions (all points on the line \(x+2y=2\)). This matches \(\dfrac4{2+k}\to1\) only formally; the formulas break down.
    4. For \(k=-2\) the equations are \(x-2y=2\) and \(-2x+4y=4\). The second says \(x-2y=-2\), which contradicts the first, so there is no solution. The determinant is zero in both cases, but the right-hand sides decide whether the system is consistent.

19. Answer to Question 19

    1. \(\begin{pmatrix}2&0&1\\1&3&2\\1&1&2\end{pmatrix}\).
    2. The determinant is \(6\) (computed earlier), so the volume of the parallelepiped is 6 Å\(^3\).
    3. A tetrahedron has \(\dfrac16\) of the parallelepiped's volume: 1 Å\(^3\).
    4. Volumes scale by the cube of the linear factor: \(6\times1.5^3=6\times3.375=\) 20.25 Å\(^3\). Equivalently, the matrix \(1.5\,I_3\) has determinant \(1.5^3\).

20. Answer to Question 20

    1. \(\operatorname{tr}A=1+4=5\) and \(\det A=4-6=-2\), so the theorem says \(A^2-5A-2I=0\). Indeed \(A^2=\) \(\begin{pmatrix}7&10\\15&22\end{pmatrix}\) and \(5A=\) \(\begin{pmatrix}5&10\\15&20\end{pmatrix}\), so \(A^2-5A=\) \(\begin{pmatrix}2&0\\0&2\end{pmatrix}\)\(=2I\), and the theorem holds.
    2. Rearrange: \(A^2-5A=2I\), so \(A(A-5I)=2I\). Dividing by 2, \(A\cdot\dfrac12(A-5I)=I\), hence \(A^{-1}=\dfrac12(A-5I)=\) \(\begin{pmatrix}-2&1\\3/2&-1/2\end{pmatrix}\).
    3. From \(A^2=5A+2I\), multiply by \(A\): \(A^3=5A^2+2A=5(5A+2I)+2A=27A+10I\). So \(p=27\) and \(q=10\).
    4. \(27A+10I=\) \(\begin{pmatrix}37&54\\81&118\end{pmatrix}\). Directly, \(A^3=A\cdot A^2=\) \(\begin{pmatrix}1&2\\3&4\end{pmatrix}\)\(\begin{pmatrix}7&10\\15&22\end{pmatrix}\)\(=\) \(\begin{pmatrix}37&54\\81&118\end{pmatrix}\). They agree.