[4 marks sum]

Question 14 Marks

If $A =\left[\begin{array}{cc}1 & 2 \\ -2 & 3\end{array}\right], B =\left[\begin{array}{ll}2 & 1 \\ 2 & 3\end{array}\right] C =\left[\begin{array}{cc}-3 & 1 \\ 2 & 0\end{array}\right]$ verify that
$A(B + C) = AB + AC$.

Answer

$B+C=\left[\begin{array}{ll}2 & 1 \\ 2 & 3\end{array}\right]+\left[\begin{array}{cc}-3 & 1 \\ 2 & 0\end{array}\right] $
$ =\left[\begin{array}{ll}2-3 & 1+1 \\ 2+2 & 3+0\end{array}\right]=\left[\begin{array}{cc}-1 & 2 \\ 4 & 3\end{array}\right]$
$A(B+C)=\left[\begin{array}{cc}1 & 2 \\ -2 & 3\end{array}\right]\left[\begin{array}{cc}-1 & 2 \\ 4 & 3\end{array}\right] $
$ =\left[\begin{array}{cc}-1+8 & 2+6 \\ 2+12 & -4+9\end{array}\right]=\left[\begin{array}{cc}7 & 8 \\ 14 & 5\end{array}\right]$
Now $ A B=\left[\begin{array}{ll}6 & 7 \\ 2 & 7\end{array}\right] \\ A C=\left[\begin{array}{cc}1 & 2 \\ -2 & 3\end{array}\right]\left[\begin{array}{cc}-3 & 1 \\ 2 & 0\end{array}\right] $
$ =\left[\begin{array}{cc}-3+4 & 1+0 \\ 6+6 & -2+0\end{array}\right]=\left[\begin{array}{cc}1 & 1 \\ 12 & -2\end{array}\right]$
$ AB + AC =\left[\begin{array}{ll}6 & 7 \\ 2 & 7\end{array}\right]+\left[\begin{array}{cc}1 & 1 \\ 12 & -2\end{array}\right] $
$ =\left[\begin{array}{cc}6+1 & 7+1 \\ 2+12 & 7-2\end{array}\right] $
$ AB + AC =\left[\begin{array}{cc}7 & 8 \\ 14 & 5\end{array}\right] $
Hence $A ( B + C )= AB + AC .$

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Question 24 Marks

Answer

$A B=\left[\begin{array}{cc}1 & 2 \\ -2 & 3\end{array}\right]\left[\begin{array}{ll}2 & 1 \\ 2 & 3\end{array}\right] $
$ =\left[\begin{array}{cc}2+4 & 1+6 \\ -4+6 & -2+9\end{array}\right]=\left[\begin{array}{ll}6 & 7 \\ 2 & 7\end{array}\right]$
$( AB ) C =\left[\begin{array}{ll}6 & 7 \\ 2 & 7\end{array}\right]\left[\begin{array}{cc}-3 & 1 \\ 2 & 0\end{array}\right] $
$=\left[\begin{array}{cc}-18+14 & 6+0 \\ -6+14 & 2+0\end{array}\right]=\left[\begin{array}{cc}-4 & 6 \\ 8 & 2\end{array}\right] $
Now $, BC =\left[\begin{array}{ll}2 & 1 \\ 2 & 3\end{array}\right]\left[\begin{array}{cc}-3 & 1 \\ 2 & 0\end{array}\right] $
$ =\left[\begin{array}{cc}-6+2 & 2+0 \\ -6+6 & 2+0\end{array}\right]=\left[\begin{array}{cc}-4 & 2 \\ 0 & 2\end{array}\right]$
$ A(B C)=\left[\begin{array}{cc} 1 & 2 \\ -2 & 3 \end{array}\right]\left[\begin{array}{cc} -4 & 2 \\
0 & 2 \end{array}\right] $
$=\left[\begin{array}{cc} -4+0 & 2+4 \\ 8+0 & -4+6 \end{array}\right]=\left[\begin{array}{cc} -4 & 6 \\ 8 & 2
\end{array}\right]$
Hence, $(A B) C=A(B C)$.

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Question 34 Marks

Let $A =\left[\begin{array}{ll}1 & 0 \\ 2 & 1\end{array}\right], B =\left[\begin{array}{cc}2 & 3 \\ -1 & 0\end{array}\right]$. Find $A^2 + AB + B^2$.

Answer

$A=\left[\begin{array}{ll}1 & 0 \\ 2 & 1\end{array}\right], $
$B=\left[\begin{array}{cc}2 & 3 \\ -1 & 0\end{array}\right] $
$ A^2=A \times A=\left[\begin{array}{ll}1 & 0 \\ 2 & 1\end{array}\right] \times\left[\begin{array}{ll}1 & 0 \\ 2 & 1\end{array}\right] $
$ =\left[\begin{array}{ll}1 \times 1+0 \times 2 & 1 \times 0+0 \times 1 \\ 2 \times 1+1 \times 2 & 2 \times 0+1 \times 1\end{array}\right] \\ =\left[\begin{array}{ll}1 & 0 \\ 4 & 1\end{array}\right]$
$B^2=B \times B=\left[\begin{array}{cc}2 & 3 \\ -1 & 0\end{array}\right] \times\left[\begin{array}{cc}2 & 3 \\ -1 & 0\end{array}\right] $
$ =\left[\begin{array}{cc}2 \times 2+3 \times-1 & 2 \times 3+3 \times 0 \\ -1 \times 2+0 \times-1 & -1 \times 3+0 \times 0\end{array}\right] $
$ =\left[\begin{array}{cc}1 & 6 \\ -2 & -3\end{array}\right] \\ \therefore A^2+A B+B^2$
$=\left[\begin{array}{ll}1 & 0 \\ 4 & 1\end{array}\right]+\left[\begin{array}{ll}2 & 3 \\ 3 & 6\end{array}\right]+\left[\begin{array}{cc}1 & 6 \\ -2 & -3\end{array}\right] $
$ =\left[\begin{array}{ll}4 & 9 \\ 5 & 4\end{array}\right] .$

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Question 44 Marks

Let $A =\left[\begin{array}{ll}4 & -2 \\ 6 & -3\end{array}\right], B =\left[\begin{array}{cc}0 & 2 \\ 1 & -1\end{array}\right]$ and $C =\left[\begin{array}{cc}-2 & 3 \\ 1 & -1\end{array}\right]$. Find $A^2 - A + BC$.

Answer

$A=\left[\begin{array}{ll}4 & -2 \\ 6 & -3\end{array}\right], B=\left[\begin{array}{cc}0 & 2 \\ 1 & -1\end{array}\right] \text { and } C=\left[\begin{array}{cc}-2 & 3 \\ 1 & -1\end{array}\right] . $
$ \therefore A^2=\left[\begin{array}{ll}4 & -2 \\ 6 & -3\end{array}\right]\left[\begin{array}{ll}4 & -2 \\ 6 & -3\end{array}\right] $
$ =\left[\begin{array}{cc}16-12 & -8+6 \\ 24-18 & -12+9\end{array}\right]$
$BC =\left[\begin{array}{cc}0 & 2 \\ 1 & -1\end{array}\right]\left[\begin{array}{cc}-2 & 3 \\ 1 & -1\end{array}\right]=\left[\begin{array}{cc}0+2 & 0-2 \\ -2-1 & 3+1\end{array}\right] $
$ =\left[\begin{array}{cc}2 & -2 \\ -3 & 4\end{array}\right]$
Now $A ^2- A + BC$
$ =\left[\begin{array}{ll} 4 & -2 \\ 6 & -3 \end{array}\right]-\left[\begin{array}{cc} 4 & -2 \\ 6 & -3
\end{array}\right]+\left[\begin{array}{cc} 2 & -2 \\ -3 & 4 \end{array}\right] $
$=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right]+\left[\begin{array}{cc} 2 & -2 \\ -3 & 4 \end{array}\right]=\left[\begin{array}{cc} 2 & -2 \\-3 & 4 \end{array}\right] $

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Question 54 Marks

If $A=\left[\begin{array}{ll}9 & 1 \\ 5 & 3\end{array}\right]$ and $B=\left[\begin{array}{cc}1 & 5 \\ 7 & -11\end{array}\right]$, find matrix $X$ such that $3A + 5B - 2X = 0.$

Answer

Let $X =\left[\begin{array}{ll}x & y \\ z & u\end{array}\right]$
We have $A=\left[\begin{array}{ll}9 & 1 \\ 5 & 3\end{array}\right]$ and $B=\left[\begin{array}{cc}1 & 5 \\ 7 & -11\end{array}\right]$
$ 3 A =3\left[\begin{array}{ll} 9 & 1 \\ 5 & 3 \end{array}\right]=\left[\begin{array}{ll} 27 & 3 \\
15 & 9 \end{array}\right] $
$5 B =5\left[\begin{array}{cc} 1 & 5 \\ 7 & -11 \end{array}\right]=\left[\begin{array}{cc} 5 & 25 \\ 35 & -55
\end{array}\right] $
Now $3 A+5 B-2 X=0$
$\Rightarrow\left[\begin{array}{ll}27 & 3 \\ 15 & 9\end{array}\right]+\left[\begin{array}{cc}5 & 25 \\ 35 & -55\end{array}\right]+\left[\begin{array}{cc}-2 x & -2 y \\ -2 z & -2 u\end{array}\right]=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right] $
$ \Rightarrow\left[\begin{array}{cc}27+5-2 x & 3+25-2 y \\ 15+35-2 z & 9-55-2 u\end{array}\right]=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right] $
$ \Rightarrow\left[\begin{array}{cc}32-2 x & 28-2 y \\ 50-2 z & -46-2 u\end{array}\right]=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]$
$\Rightarrow 32-2 x=0 $
$\Rightarrow 2 x-32 $
$\Rightarrow x=16= 28-2 y=0 $
$\Rightarrow 2 y=28 $
$\Rightarrow y=14 = 50-2 z=0 $
$\Rightarrow 2 z=50$
$ \Rightarrow z=25-46-2 u=0 $
$\Rightarrow 2 u=-46$
$ \Rightarrow u=-23$

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Question 64 Marks

Find the $2 \times 2$ matrix $X$ which satisfies the equation. $\left[\begin{array}{ll}3 & 7 \\ 2 & 4\end{array}\right]\left[\begin{array}{ll}0 & 2 \\ 5 & 3\end{array}\right]+2 X =\left[\begin{array}{cc}1 & -5 \\ -4 & 6\end{array}\right]$

Answer

${\left[\begin{array}{ll}3 & 7 \\ 2 & 4\end{array}\right]\left[\begin{array}{ll}0 & 2 \\ 5 & 3\end{array}\right]+2 X =\left[\begin{array}{cc}1 & -5 \\ -4 & 6\end{array}\right]} $
$\Rightarrow\left[\begin{array}{l}0+356+24 \\ 0+204+12\end{array}\right]+2 X=\left[\begin{array}{cc}1 & -5 \\ -4 & 6\end{array}\right] $
$\Rightarrow\left[\begin{array}{ll}35 & 27 \\ 20 & 16\end{array}\right]+2 X=\left[\begin{array}{cc}1 & -5 \\ -4 & 6\end{array}\right] $
$ \Rightarrow 2 X =\left[\begin{array}{cc}1 & -5 \\ -4 & 6\end{array}\right] -\left[\begin{array}{ll}35 & 27 \\ 20 & 16\end{array}\right]$
$\Rightarrow 2 X=\left[\begin{array}{ll}-34 & -32 \\ -24 & -10\end{array}\right] $
$ \Rightarrow X=\left[\begin{array}{cc}\frac{-34}{2} \frac{-32}{2} \\ \frac{-24}{2} \frac{-10}{2}\end{array}\right] $
$ \Rightarrow X=\left[\begin{array}{ll}-17 & -16 \\ -12 & -5\end{array}\right]$

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Question 74 Marks

Let $A=\left[\begin{array}{cc}2 & 1 \\ 0 & -2\end{array}\right], B=\left[\begin{array}{cc}4 & 1 \\ -3 & -2\end{array}\right]$ and $C=\left[\begin{array}{ll}-3 & 2 \\ -1 & 4\end{array}\right]$. Find $A^2 + AC -5B$.

Answer

Given : $A =\left[\begin{array}{cc}2 & 1 \\ 0 & -2\end{array}\right], B =\left[\begin{array}{cc}4 & 1 \\ -3 & -2\end{array}\right]$ and $C =\left[\begin{array}{ll}-3 & 2 \\ -1 & 4\end{array}\right]$
Now,
$ A^2=\left[\begin{array}{cc} 2 & 1 \\ 0 & -2 \end{array}\right]\left[\begin{array}{cc} 2 & 1 \\
0 & -2 \end{array}\right]=\left[\begin{array}{ll} 4+0 & 2-2 \\ 0+0 & 0+4 \end{array}\right]=\left[\begin{array}{ll}
4 & 0 \\ 0 & 4 \end{array}\right] $
$5 B=\left[\begin{array}{cc} 20 & 5 \\ -15 & -10 \end{array}\right]$
$A C=\left[\begin{array}{cc}2 & 1 \\ 0 & -2\end{array}\right]\left[\begin{array}{ll}-3 & 2 \\ -1 & 4\end{array}\right]=\left[\begin{array}{cc}-6-1 & 4+4 \\ 0+2 & 0-8\end{array}\right]=\left[\begin{array}{cc}-7 & 8 \\ 2 & -8\end{array}\right] $
$ \therefore A^2+A C-5 B=\left[\begin{array}{ll}4 & 0 \\ 0 & 4\end{array}\right]+\left[\begin{array}{cc}-7 & 8 \\ 2 & -8\end{array}\right]-\left[\begin{array}{cc}20 & 5 \\ -15 & -10\end{array}\right] $
$ =\left[\begin{array}{cc}4-7-20 & 0+8-5 \\ 0+2+15 & 4-8+10\end{array}\right] $
$ =\left[\begin{array}{cc}-23 & 3 \\ 17 & 6\end{array}\right] .$

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Question 84 Marks

$A=\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]$, show that $A^2 - 5A + 7 I_2 = 0$.

Answer

We have
$A=\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right] $
$ A=\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right] $
$ =\left[\begin{array}{cc}9-1 & 3+2 \\ -3-2 & -1+4\end{array}\right] $
$ =\left[\begin{array}{cc}8 & 5 \\ -5 & 3\end{array}\right] \\ -5 A =\left[\begin{array}{cc}(-5) \cdot 3 & (-5) \cdot 1 \\ (-5) \cdot(-1) & (-5) \cdot 2\end{array}\right] $
$ =\left[\begin{array}{cc}-5 & -5 \\ 5 & -10\end{array}\right]$
$ -5 A =\left[\begin{array}{cc} (-5) \cdot 3 & (-5) \cdot 1 \\ (-5) \cdot(-1) & (-5) \cdot 2
\end{array}\right] $
$=\left[\begin{array}{cc} -5 & -5 \\ 5 & -10 \end{array}\right] 7 I _2 =7\left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right]=\left[\begin{array}{ll} 7 & 0 \\ 0 & 7 \end{array}\right]$
So $A^2-5 A+71_2$
$=\left[\begin{array}{cr} 8 & 5 \\ -5 & 3 \end{array}\right]+\left[\begin{array}{cc} -15 & -5 \\ 5 & -10 \end{array}\right]+\left[\begin{array}{ll} 7 & 0 \\ 0 & 7 \end{array}\right] $
$=\left[\begin{array}{cc} 8-15+7 & 5-5+0 \\ -5+5+0 & 3-10+7 \end{array}\right]=\left[\begin{array}{ll} 0 & 0 \\
0 & 0 \end{array}\right]$
So $A^2 - 5A + 7I_2 = 0$.
Hence proved.

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Question 94 Marks

If $A=\left[\begin{array}{ll}9 & 1 \\ 7 & 8\end{array}\right], B=\left[\begin{array}{cc}1 & 5 \\ 7 & 12\end{array}\right]$ find matrix $C$ such that $5A + 5B + 2C$ is a null matrix.

Answer

Let $C=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$
We have $A =\left[\begin{array}{ll}9 & 1 \\ 7 & 8\end{array}\right], B =\left[\begin{array}{cc}1 & 5 \\ 7 & 12\end{array}\right]$
Now $5 A +3 B +2 C =0$
$ \Rightarrow 5\left[\begin{array}{ll} 9 & 1 \\ 7 & 8 \end{array}\right]+3\left[\begin{array}{cc}
1 & 5 \\ 7 & 12 \end{array}\right]+2\left[\begin{array}{ll} a & b \\ c & d \end{array}\right]=\left[\begin{array}{ll}
0 & 0 \\ 0 & 0\end{array}\right] $
$\Rightarrow\left[\begin{array}{cc} 45 & 5 \\ 35 & 40 \end{array}\right]+\left[\begin{array}{cc} 3 & 15 \\ 21 & 36
\end{array}\right]+\left[\begin{array}{ll} 2 a & 2 b \\ 2 c & 2 d \end{array}\right]=\left[\begin{array}{ll} 0 & 0 \\
0 & 0 \end{array}\right]$
$\Rightarrow\left[\begin{array}{cc} 45+3+2 a & 5+15+2 b \\ 35+21+2 c & 40+36+2 d \end{array}\right]=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right] $
$\Rightarrow\left[\begin{array}{ll} 48+2 a & 20+2 b \\ 56+2 c & 76+2 d \end{array}\right]=\left[\begin{array}{ll}
0 & 0 \\ 0 & 0 \end{array}\right] $
$\Rightarrow 48+2 a =0$
$ \Rightarrow 2 a =-48$
$ \Rightarrow a =-24= 20+2 b =0 $
$\Rightarrow 2 b =20 $
$\Rightarrow b =-10 = 56+2 c =0$
$ \Rightarrow 2 c =56 $
$\Rightarrow c =-28 = 76+2 d =0 $
$\Rightarrow 2 d =-76 $
$\Rightarrow d =-38$
Thus $C =\left[\begin{array}{ll} -24 & -10 \\ -28 & -38\end{array}\right] .$

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Question 104 Marks

If $A=\left[\begin{array}{ll}3 & 1 \\ 2 & 1\end{array}\right]$ and $B=\left[\begin{array}{cc}1 & -2 \\ 5 & 3\end{array}\right],$ then show that $(A - B)^2 \neq A2 - 2AB + B^2.$

Answer

$\begin{array}{l}A-B=\left[\begin{array}{ll}3 & 1 \\ 2 & 1\end{array}\right]-\left[\begin{array}{cc}1 & -2 \\ 5 & 3\end{array}\right] \\ =\left[\begin{array}{ll}3-1 & 1+2 \\ 2-5 & 1-3\end{array}\right]=\left[\begin{array}{cc}2 & 3 \\ -3 & -2\end{array}\right]\end{array}$
$(A - B)^2 = (A - B)(A - B)$
$\begin{array}{l}\Rightarrow( A - B ) 2=\left[\begin{array}{cc}2 & 3 \\ -3 & -2\end{array}\right]\left[\begin{array}{cc}2 & 3 \\ -3 & -2\end{array}\right] \\ =\left[\begin{array}{cc}4-9 & 6-6 \\ -6+6 & -9+4\end{array}\right] \\ =\left[\begin{array}{cc}-5 & 0 \\ 0 & -5\end{array}\right]\end{array}$
$\begin{array}{l}\text { and } A^2=\left[\begin{array}{ll}3 & 1 \\ 2 & 1\end{array}\right]\left[\begin{array}{ll}3 & 1 \\ 2 & 1\end{array}\right] \\ =\left[\begin{array}{ll}9+2 & 3+1 \\ 6+2 & 2+1\end{array}\right]=\left[\begin{array}{cc}11 & 4 \\ 8 & 3\end{array}\right]\end{array}$
$\begin{array}{l}\text { and } B^2=\left[\begin{array}{cc}1 & -2 \\ 5 & 3\end{array}\right]\left[\begin{array}{cc}1 & -2 \\ 5 & 3\end{array}\right] \\ =\left[\begin{array}{cc}1-10 & -2-6 \\ 5+15 & -10+9\end{array}\right] \\ =\left[\begin{array}{cc}-9 & -8 \\ 20 & -1\end{array}\right]\end{array}$
$\begin{array}{l}\text { and } A B=\left[\begin{array}{ll}3 & 1 \\ 2 & 1\end{array}\right]\left[\begin{array}{cc}1 & -2 \\ 5 & 3\end{array}\right] \\ =\left[\begin{array}{ll}3+5 & -6+3 \\ 2+5 & -4+3\end{array}\right]=\left[\begin{array}{ll}8 & -3 \\ 7 & -1\end{array}\right]\end{array}$
$\begin{array}{l}\text { Now } A^2-2 A B+B^2 \\ =\left[\begin{array}{cc}11 & 4 \\ 8 & 3\end{array}\right]-2\left[\begin{array}{cc}8 & -3 \\ 7 & -1\end{array}\right]+\left[\begin{array}{cc}-9 & -8 \\ 20 & -1\end{array}\right] \\ =\left[\begin{array}{cc}11 & 4 \\ 8 & 3\end{array}\right]-\left[\begin{array}{ll}16 & -6 \\ 14 & -2\end{array}\right]+\left[\begin{array}{cc}-9 & -8 \\ 20 & -1\end{array}\right] \\ =\left[\begin{array}{cc}11-6-9 & 4+6-8 \\ 8-14+20 & 3+2-1\end{array}\right] \\ =\left[\begin{array}{cc}-14 & 2 \\ 14 & 4\end{array}\right]\end{array}$
Hence$,$ from above calculations, we get
$(A - B)^2 \neq A^2 - 2AB + B^2.$

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Mathematics [4 marks sum]

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