Chapter 3: Matrices (Algebra)

(i) 3 rows and 4 columns: order $3\times 4$.

(ii) $3\times 4=12$ elements.

(iii) $a_{13}=19$, $a_{21}=35$, $a_{33}=-5$, $a_{24}=12$, $a_{23}=\frac{5}{2}$.

(i) 3 × 4. (ii) 12. (iii) 19, 35, −5, 12, 5/2.

The order $m\times n$ must satisfy $mn=24$: $1\times 24$, $24\times 1$, $2\times 12$, $12\times 2$, $3\times 8$, $8\times 3$, $4\times 6$, $6\times 4$.

For 13 elements (13 is prime): $1\times 13$ and $13\times 1$.

24: 1×24, 24×1, 2×12, 12×2, 3×8, 8×3, 4×6, 6×4. 13: 1×13, 13×1.

$mn=18$: $1\times 18$, $18\times 1$, $2\times 9$, $9\times 2$, $3\times 6$, $6\times 3$.

For 5 elements: $1\times 5$ and $5\times 1$.

18: 1×18, 18×1, 2×9, 9×2, 3×6, 6×3. 5: 1×5, 5×1.

(i) $a_{11}=\frac{4}{2}=2$, $a_{12}=\frac{9}{2}$, $a_{21}=\frac{9}{2}$, $a_{22}=\frac{16}{2}=8$:

$$A=\left[\begin{array}{cc}2& \frac{9}{2}\\ \frac{9}{2}& 8\end{array}\right]$$

(ii) $a_{11}=1$, $a_{12}=\frac{1}{2}$, $a_{21}=2$, $a_{22}=1$:

$$A=\left[\begin{array}{cc}1& \frac{1}{2}\\ 2& 1\end{array}\right]$$

(iii) $a_{11}=\frac{9}{2}$, $a_{12}=\frac{25}{2}$, $a_{21}=\frac{16}{2}=8$, $a_{22}=\frac{36}{2}=18$:

$$A=\left[\begin{array}{cc}\frac{9}{2}& \frac{25}{2}\\ 8& 18\end{array}\right]$$

As constructed above.

(i) Row 1 ($i=1$): $\frac{1}{2}|-2|=1$, $\frac{1}{2}|-1|=\frac{1}{2}$, $0$, $\frac{1}{2}$. Row 2: $\frac{5}{2},2,\frac{3}{2},1$. Row 3: $4,\frac{7}{2},3,\frac{5}{2}$.

$$A=\left[\begin{array}{cccc}1& \frac{1}{2}& 0& \frac{1}{2}\\ \frac{5}{2}& 2& \frac{3}{2}& 1\\ 4& \frac{7}{2}& 3& \frac{5}{2}\end{array}\right]$$

(ii)

$$A=\left[\begin{array}{cccc}1& 0& -1& -2\\ 3& 2& 1& 0\\ 5& 4& 3& 2\end{array}\right]$$

As constructed above.

(i) Equating entries: $x=1$, $y=4$, $z=3$.

(ii) $5+z=5\Rightarrow z=0$; $x+y=6$ and $xy=8$, so $x,y$ are roots of $t^2-6t+8=0$: $t=2,4$. So $x=2,y=4$ or $x=4,y=2$.

(iii) $x+y+z=9$, $x+z=5$, $y+z=7$. Then $y=9-5=4$, $x=9-7=2$, $z=5-2=3$.

(i) x = 1, y = 4, z = 3. (ii) z = 0 and (x, y) = (2, 4) or (4, 2). (iii) x = 2, y = 4, z = 3.

$a-b=-1$ and $2a-b=0$. Subtracting: $a=1$, then $b=2$.

$2a+c=5\Rightarrow c=3$; $3c+d=13\Rightarrow d=4$.

a = 1, b = 2, c = 3, d = 4.

A square matrix has as many rows as columns.

(C)

$3x+7=0\Rightarrow x=-\frac{7}{3}$, but $2-3x=4\Rightarrow x=-\frac{2}{3}$. The two values of x contradict each other, so the matrices cannot be made equal.

(B)

There are 9 entries, each with 2 choices: $2^9=512$.

(D)

(i) $A+B=\left[\begin{array}{cc}3& 7\\ 1& 7\end{array}\right]$

(ii) $A-B=\left[\begin{array}{cc}1& 1\\ 5& -3\end{array}\right]$

(iii) $3A-C=\left[\begin{array}{cc}6& 12\\ 9& 6\end{array}\right]-\left[\begin{array}{cc}-2& 5\\ 3& 4\end{array}\right]=\left[\begin{array}{cc}8& 7\\ 6& 2\end{array}\right]$

(iv) $AB=\left[\begin{array}{cc}2-8& 6+20\\ 3-4& 9+10\end{array}\right]=\left[\begin{array}{cc}-6& 26\\ -1& 19\end{array}\right]$

(v) $BA=\left[\begin{array}{cc}2+9& 4+6\\ -4+15& -8+10\end{array}\right]=\left[\begin{array}{cc}11& 10\\ 11& 2\end{array}\right]$

As computed above (note AB ≠ BA).

(i) $\left[\begin{array}{cc}2a& 2b\\ 0& 2a\end{array}\right]$

(ii) $\left[\begin{array}{cc}(a+b{)}^2& (b+c{)}^2\\ (a-c{)}^2& (a-b{)}^2\end{array}\right]$

(iii) $\left[\begin{array}{ccc}11& 11& 0\\ 16& 5& 21\\ 5& 10& 9\end{array}\right]$

(iv) Each entry is ${\sin }^{2}x+{\cos }^{2}x=1$: $\left[\begin{array}{cc}1& 1\\ 1& 1\end{array}\right]$

As computed above.

(i) $\left[\begin{array}{cc}{a}^2+b^2& -ab+ab\\ -ab+ab& b^2+a^2\end{array}\right]=\left[\begin{array}{cc}{a}^2+b^2& 0\\ 0& a^2+b^2\end{array}\right]$

(ii) $\left[\begin{array}{ccc}2& 3& 4\\ 4& 6& 8\\ 6& 9& 12\end{array}\right]$

(iii) $\left[\begin{array}{ccc}1-4& 2-6& 3-2\\ 2+6& 4+9& 6+3\end{array}\right]=\left[\begin{array}{ccc}-3& -4& 1\\ 8& 13& 9\end{array}\right]$

(iv) For example, row 1 × column 1: $2+0+12=14$. In full:

$$\left[\begin{array}{ccc}14& 0& 42\\ 18& -1& 56\\ 22& -2& 70\end{array}\right]$$

(v) $\left[\begin{array}{ccc}2-1& 0+2& 2+1\\ 3-2& 0+4& 3+2\\ -1-1& 0+2& -1+1\end{array}\right]=\left[\begin{array}{ccc}1& 2& 3\\ 1& 4& 5\\ -2& 2& 0\end{array}\right]$

(vi) $\left[\begin{array}{cc}6-1+9& -9+0+3\\ -2+0+6& 3+0+2\end{array}\right]=\left[\begin{array}{cc}14& -6\\ 4& 5\end{array}\right]$

As computed above.

$$A+B=\left[\begin{array}{ccc}4& 1& -1\\ 9& 2& 7\\ 3& -1& 4\end{array}\right],B-C=\left[\begin{array}{ccc}-1& -2& 0\\ 4& -1& 3\\ 1& 2& 0\end{array}\right]$$

$$A+(B-C)=\left[\begin{array}{ccc}0& 0& -3\\ 9& -1& 5\\ 2& 1& 1\end{array}\right]=(A+B)-C$$

Both sides equal $\left[\begin{array}{ccc}0& 0& -3\\ 9& -1& 5\\ 2& 1& 1\end{array}\right]$.

$$3A=\left[\begin{array}{ccc}2& 3& 5\\ 1& 2& 4\\ 7& 6& 2\end{array}\right],5B=\left[\begin{array}{ccc}2& 3& 5\\ 1& 2& 4\\ 7& 6& 2\end{array}\right]$$

$$3A-5B=\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]$$

The zero matrix O.

$$\left[\begin{array}{cc}{\cos }^2\theta +{\sin }^2\theta & \cos \theta \sin \theta -\sin \theta \cos \theta \\ -\sin \theta \cos \theta +\sin \theta \cos \theta & {\cos }^2\theta +{\sin }^2\theta \end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$$

The identity matrix I.

(i) Adding: $2X=\left[\begin{array}{cc}10& 0\\ 2& 8\end{array}\right]$, so $X=\left[\begin{array}{cc}5& 0\\ 1& 4\end{array}\right]$. Then $Y=(X+Y)-X=\left[\begin{array}{cc}2& 0\\ 1& 1\end{array}\right]$.

(ii) Let $P=2X+3Y$, $Q=3X+2Y$. Then $3P-2Q=5Y$ and $3Q-2P=5X$:

$$Y=\frac{1}{5}\left(\left[\begin{array}{cc}6& 9\\ 12& 0\end{array}\right]-\left[\begin{array}{cc}4& -4\\ -2& 10\end{array}\right]\right)=\left[\begin{array}{cc}\frac{2}{5}& \frac{13}{5}\\ \frac{14}{5}& -2\end{array}\right]$$

$$X=\frac{1}{5}\left(\left[\begin{array}{cc}6& -6\\ -3& 15\end{array}\right]-\left[\begin{array}{cc}4& 6\\ 8& 0\end{array}\right]\right)=\left[\begin{array}{cc}\frac{2}{5}& -\frac{12}{5}\\ -\frac{11}{5}& 3\end{array}\right]$$

(i) $X=\left[\begin{array}{cc}5& 0\\ 1& 4\end{array}\right]$, $Y=\left[\begin{array}{cc}2& 0\\ 1& 1\end{array}\right]$. (ii) $X=\left[\begin{array}{cc}\frac{2}{5}& -\frac{12}{5}\\ -\frac{11}{5}& 3\end{array}\right]$, $Y=\left[\begin{array}{cc}\frac{2}{5}& \frac{13}{5}\\ \frac{14}{5}& -2\end{array}\right]$.

$$2X=\left[\begin{array}{cc}1& 0\\ -3& 2\end{array}\right]-\left[\begin{array}{cc}3& 2\\ 1& 4\end{array}\right]=\left[\begin{array}{cc}-2& -2\\ -4& -2\end{array}\right]\Rightarrow X=\left[\begin{array}{cc}-1& -1\\ -2& -1\end{array}\right]$$

$X=\left[\begin{array}{cc}-1& -1\\ -2& -1\end{array}\right]$

$\left[\begin{array}{cc}2+y& 6\\ 1& 2x+2\end{array}\right]=\left[\begin{array}{cc}5& 6\\ 1& 8\end{array}\right]$, so $y=3$ and $2x+2=8\Rightarrow x=3$.

x = 3, y = 3.

$\left[\begin{array}{cc}2x+3& 2z-3\\ 2y& 2t+6\end{array}\right]=\left[\begin{array}{cc}9& 15\\ 12& 18\end{array}\right]$

$x=3$, $z=9$, $y=6$, $t=6$.

x = 3, y = 6, z = 9, t = 6.

$2x-y=10$ and $3x+y=5$. Adding: $5x=15$, so $x=3$ and $y=5-9=-4$.

x = 3, y = −4.

$\left[\begin{array}{cc}3x& 3y\\ 3z& 3w\end{array}\right]=\left[\begin{array}{cc}x+4& 6+x+y\\ -1+z+w& 2w+3\end{array}\right]$

• $3x=x+4\Rightarrow x=2$
• $3w=2w+3\Rightarrow w=3$
• $3y=6+x+y\Rightarrow 2y=8\Rightarrow y=4$
• $3z=-1+z+w\Rightarrow 2z=2\Rightarrow z=1$

x = 2, y = 4, z = 1, w = 3.

$$F(x)F(y)=\left[\begin{array}{ccc}\cos x\cos y-\sin x\sin y& -\cos x\sin y-\sin x\cos y& 0\\ \sin x\cos y+\cos x\sin y& -\sin x\sin y+\cos x\cos y& 0\\ 0& 0& 1\end{array}\right]$$

$$=\left[\begin{array}{ccc}\cos (x+y)& -\sin (x+y)& 0\\ \sin (x+y)& \cos (x+y)& 0\\ 0& 0& 1\end{array}\right]=F(x+y)$$

Multiplying and using the addition formulas gives F(x + y).

(i) LHS $=\left[\begin{array}{cc}10-3& 5-4\\ 12+21& 6+28\end{array}\right]=\left[\begin{array}{cc}7& 1\\ 33& 34\end{array}\right]$; RHS $=\left[\begin{array}{cc}10+6& -2+7\\ 15+24& -3+28\end{array}\right]=\left[\begin{array}{cc}16& 5\\ 39& 25\end{array}\right]$. They differ.

(ii) LHS $=\left[\begin{array}{ccc}5& 8& 14\\ 0& -1& 1\\ -1& 0& 1\end{array}\right]$; RHS $=\left[\begin{array}{ccc}-1& -1& -3\\ 1& 0& 0\\ 6& 11& 6\end{array}\right]$. They differ.

The products differ in each case, so matrix multiplication is not commutative.

$$A^2=\left[\begin{array}{ccc}5& -1& 2\\ 9& -2& 5\\ 0& -1& -2\end{array}\right],5A=\left[\begin{array}{ccc}10& 0& 5\\ 10& 5& 15\\ 5& -5& 0\end{array}\right],6I=\left[\begin{array}{ccc}6& 0& 0\\ 0& 6& 0\\ 0& 0& 6\end{array}\right]$$

$$A^2-5A+6I=\left[\begin{array}{ccc}1& -1& -3\\ -1& -1& -10\\ -5& 4& 4\end{array}\right]$$

$\left[\begin{array}{ccc}1& -1& -3\\ -1& -1& -10\\ -5& 4& 4\end{array}\right]$

$$A^2=\left[\begin{array}{ccc}5& 0& 8\\ 2& 4& 5\\ 8& 0& 13\end{array}\right],A^3=A^{2}A=\left[\begin{array}{ccc}21& 0& 34\\ 12& 8& 23\\ 34& 0& 55\end{array}\right]$$

$$A^3-6A^2+7A+2I=\left[\begin{array}{ccc}21-30+7+2& 0& 34-48+14\\ 12-12+0& 8-24+14+2& 23-30+7\\ 34-48+14& 0& 55-78+21+2\end{array}\right]=O$$

Every entry works out to 0, so $A^3-6A^2+7A+2I=O$.

$$A^2=\left[\begin{array}{cc}9-8& -6+4\\ 12-8& -8+4\end{array}\right]=\left[\begin{array}{cc}1& -2\\ 4& -4\end{array}\right],kA-2I=\left[\begin{array}{cc}3k-2& -2k\\ 4k& -2k-2\end{array}\right]$$

$3k-2=1\Rightarrow k=1$; check: $-2k=-2$, $4k=4$, $-2k-2=-4$ ✔.

k = 1

Let $t=\tan \frac{\alpha }{2}$, so $\cos \alpha =\frac{1-t^2}{1+t^2}$ and $\sin \alpha =\frac{2t}{1+t^2}$. Then

$$I+A=\left[\begin{array}{cc}1& -t\\ t& 1\end{array}\right],I-A=\left[\begin{array}{cc}1& t\\ -t& 1\end{array}\right]$$

$$(I-A)\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right]=\left[\begin{array}{cc}\cos \alpha +t\sin \alpha & -\sin \alpha +t\cos \alpha \\ -t\cos \alpha +\sin \alpha & t\sin \alpha +\cos \alpha \end{array}\right]$$

• $\cos \alpha +t\sin \alpha =\frac{1-t^2+2t^2}{1+t^2}=1$
• $-\sin \alpha +t\cos \alpha =\frac{-2t+t-t^3}{1+t^2}=\frac{-t(1+t^2)}{1+t^2}=-t$
• $\sin \alpha -t\cos \alpha =t$ (similarly), and the last entry is 1.

So the product is $\left[\begin{array}{cc}1& -t\\ t& 1\end{array}\right]=I+A$.

Writing cos α and sin α in terms of $t=\tan \frac{\alpha }{2}$, the product equals I + A.

Let ₹x go into the first bond and ₹(30000 − x) into the second. Interest:

$$\left[\begin{array}{cc}x& 30000-x\end{array}\right]\left[\begin{array}{c}0.05\\ 0.07\end{array}\right]=0.05x+0.07(30000-x)=2100-0.02x$$

(a) $2100-0.02x=1800\Rightarrow x=15000$: ₹15,000 in each bond.

(b) $2100-0.02x=2000\Rightarrow x=5000$: ₹5,000 in the first bond and ₹25,000 in the second.

(a) ₹15,000 and ₹15,000. (b) ₹5,000 (5%) and ₹25,000 (7%).

Numbers of books: 120, 96, 120.

$$\left[\begin{array}{ccc}120& 96& 120\end{array}\right]\left[\begin{array}{c}80\\ 60\\ 40\end{array}\right]=9600+5760+4800=20160$$

₹20,160

PY needs the columns of P (k) to equal the rows of Y (3): $k=3$, giving order $p\times k$. WY ($n\times 3$ times $3\times k$) has order $n\times k$. For the sum, $p\times k=n\times k$, so $p=n$.

(A)

X is $2\times n$ and Z is $2\times p=2\times n$, so $7X-5Z$ is $2\times n$.

(B)

(i) $\left[\begin{array}{ccc}5& \frac{1}{2}& -1\end{array}\right]$ (ii) $\left[\begin{array}{cc}1& 2\\ -1& 3\end{array}\right]$ (iii) $\left[\begin{array}{ccc}-1& \sqrt{3}& 2\\ 5& 5& 3\\ 6& 6& -1\end{array}\right]$

Rows become columns, as shown.

$$A^{\prime }=\left[\begin{array}{ccc}-1& 5& -2\\ 2& 7& 1\\ 3& 9& 1\end{array}\right],B^{\prime }=\left[\begin{array}{ccc}-4& 1& 1\\ 1& 2& 3\\ -5& 0& 1\end{array}\right]$$

(i) $A+B=\left[\begin{array}{ccc}-5& 3& -2\\ 6& 9& 9\\ -1& 4& 2\end{array}\right]$, so $(A+B{)}^{\prime }=\left[\begin{array}{ccc}-5& 6& -1\\ 3& 9& 4\\ -2& 9& 2\end{array}\right]=A^{\prime }+B^{\prime }$.

(ii) $A-B=\left[\begin{array}{ccc}3& 1& 8\\ 4& 5& 9\\ -3& -2& 0\end{array}\right]$, so $(A-B{)}^{\prime }=\left[\begin{array}{ccc}3& 4& -3\\ 1& 5& -2\\ 8& 9& 0\end{array}\right]=A^{\prime }-B^{\prime }$.

Both identities hold (computed above).

$A=\left[\begin{array}{ccc}3& -1& 0\\ 4& 2& 1\end{array}\right]$ and $B^{\prime }=\left[\begin{array}{cc}-1& 1\\ 2& 2\\ 1& 3\end{array}\right]$.

(i) $A+B=\left[\begin{array}{ccc}2& 1& 1\\ 5& 4& 4\end{array}\right]$, $(A+B{)}^{\prime }=\left[\begin{array}{cc}2& 5\\ 1& 4\\ 1& 4\end{array}\right]=A^{\prime }+B^{\prime }$.

(ii) $A-B=\left[\begin{array}{ccc}4& -3& -1\\ 3& 0& -2\end{array}\right]$, $(A-B{)}^{\prime }=\left[\begin{array}{cc}4& 3\\ -3& 0\\ -1& -2\end{array}\right]=A^{\prime }-B^{\prime }$.

Both identities hold (computed above).

$A=\left[\begin{array}{cc}-2& 1\\ 3& 2\end{array}\right]$; $A+2B=\left[\begin{array}{cc}-4& 1\\ 5& 6\end{array}\right]$; so $(A+2B{)}^{\prime }=\left[\begin{array}{cc}-4& 5\\ 1& 6\end{array}\right]$.

$\left[\begin{array}{cc}-4& 5\\ 1& 6\end{array}\right]$

(i) $AB=\left[\begin{array}{ccc}-1& 2& 1\\ 4& -8& -4\\ -3& 6& 3\end{array}\right]$, so $(AB{)}^{\prime }=\left[\begin{array}{ccc}-1& 4& -3\\ 2& -8& 6\\ 1& -4& 3\end{array}\right]$. And $B^{\prime }{A}^{\prime }=\left[\begin{array}{c}-1\\ 2\\ 1\end{array}\right]\left[\begin{array}{ccc}1& -4& 3\end{array}\right]=\left[\begin{array}{ccc}-1& 4& -3\\ 2& -8& 6\\ 1& -4& 3\end{array}\right]$. Equal.

(ii) $AB=\left[\begin{array}{ccc}0& 0& 0\\ 1& 5& 7\\ 2& 10& 14\end{array}\right]$, so $(AB{)}^{\prime }=\left[\begin{array}{ccc}0& 1& 2\\ 0& 5& 10\\ 0& 7& 14\end{array}\right]$. And $B^{\prime }{A}^{\prime }=\left[\begin{array}{c}1\\ 5\\ 7\end{array}\right]\left[\begin{array}{ccc}0& 1& 2\end{array}\right]=\left[\begin{array}{ccc}0& 1& 2\\ 0& 5& 10\\ 0& 7& 14\end{array}\right]$. Equal.

In both cases (AB)′ and B′A′ are the same matrix.

(i) $A^{\prime }A=\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right]\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{array}\right]=\left[\begin{array}{cc}{\cos }^2\alpha +{\sin }^2\alpha & \cos \alpha \sin \alpha -\sin \alpha \cos \alpha \\ \sin \alpha \cos \alpha -\cos \alpha \sin \alpha & {\sin }^2\alpha +{\cos }^2\alpha \end{array}\right]=I$

(ii) $A^{\prime }A=\left[\begin{array}{cc}\sin \alpha & -\cos \alpha \\ \cos \alpha & \sin \alpha \end{array}\right]\left[\begin{array}{cc}\sin \alpha & \cos \alpha \\ -\cos \alpha & \sin \alpha \end{array}\right]=\left[\begin{array}{cc}{\sin }^2\alpha +{\cos }^2\alpha & \sin \alpha \cos \alpha -\cos \alpha \sin \alpha \\ \cos \alpha \sin \alpha -\sin \alpha \cos \alpha & {\cos }^2\alpha +{\sin }^2\alpha \end{array}\right]=I$

In both cases A′A works out to I.

(i) $A^{\prime }=\left[\begin{array}{ccc}1& -1& 5\\ -1& 2& 1\\ 5& 1& 3\end{array}\right]=A$: symmetric.

(ii) $A^{\prime }=\left[\begin{array}{ccc}0& -1& 1\\ 1& 0& -1\\ -1& 1& 0\end{array}\right]=-A$: skew symmetric.

(i) A′ = A. (ii) A′ = −A.

$A^{\prime }=\left[\begin{array}{cc}1& 6\\ 5& 7\end{array}\right]$.

(i) $A+A^{\prime }=\left[\begin{array}{cc}2& 11\\ 11& 14\end{array}\right]$, which equals its transpose: symmetric.

(ii) $A-A^{\prime }=\left[\begin{array}{cc}0& -1\\ 1& 0\end{array}\right]$, whose transpose is $\left[\begin{array}{cc}0& 1\\ -1& 0\end{array}\right]=-(A-A^{\prime })$: skew symmetric.

A + A′ is symmetric and A − A′ is skew symmetric.

$A^{\prime }=\left[\begin{array}{ccc}0& -a& -b\\ a& 0& -c\\ b& c& 0\end{array}\right]=-A$.