[5 marks sum]

Question 15 Marks

Let $A=\left|\begin{array}{ll}3 & 2 \\ 0 & 5\end{array}\right|$ and $B=\left|\begin{array}{ll}1 & 0 \\ 1 & 2\end{array}\right|$, find $(i) \ (A+B)(A-B) \ (ii) \ A^2-B^2$. Is $(i)$ equal to $(ii)$ ?

Answer

$\begin{array}{l}\text { (i) } A=\left|\begin{array}{ll}3 & 2 \\ 0 & 5\end{array}\right|_{2 \times 2}, B=\left|\begin{array}{ll}1 & 0 \\ 1 & 2\end{array}\right|_{2 \times 2} \end{array} $
$ A+B=\left|\begin{array}{ll}3+1 & 2+0 \\ 0+1 & 5+2\end{array}\right|=\left|\begin{array}{ll}4 & 2 \\ 1 & 7\end{array}\right|_{2 \times 2} $
$ A-B=\left|\begin{array}{ll}3-1 & 2-0 \\ 0-1 & 5-2\end{array}\right|=\left|\begin{array}{cc}2 & 2 \\ -1 & 3\end{array}\right|_{2 \times 2} $
$ \left.(A+B)(A-B)=\left|\begin{array}{cc}4 & 2 \\ 1 & 7\end{array}\right| \begin{array}{cc}2 & 2 \\ -1 & 3\end{array} \right\rvert\, $
$ =\left|\begin{array}{ll}8-2 & 8+6 \\ 2-7 & 2+21\end{array}\right|$
$=\left|\begin{array}{cc}6 & 14 \\ -5 & 23\end{array}\right| \ldots \ldots . .(1)$
$(ii) A ^2$
$=\left|\begin{array}{ll}3 & 2 \\ 0 & 5\end{array}\right|\left|\begin{array}{ll}3 & 2 \\ 0 & 5\end{array}\right|$
$=\left|\begin{array}{cc}9+0 & 6+10 \\ 0+0 & 0+25\end{array}\right|$
$=\left|\begin{array}{cc}9 & 16 \\ 0 & 25\end{array}\right|_{2 \times 2}$
$B^2=\left|\begin{array}{ll}1 & 0 \\ 1 & 2\end{array}\right|\left|\begin{array}{ll}1 & 0 \\ 1 & 2\end{array}\right|$
$=\left|\begin{array}{ll}1+0 & 0+0 \\ 1+2 & 0+4\end{array}\right|$
$=\left|\begin{array}{ll}1 & 0 \\ 3 & 4\end{array}\right|_{2 \times 2}$
$ \left|A^2-B^2=\left|\begin{array}{ll} 9 & 16 \\ 0 & 25 \end{array}\right|-\left|\begin{array}{ll} 1 & 0 \\ 3 & 4 \end{array}\right|=\left|\begin{array}{cc} 8 & 16 \\ -3 & 21 \end{array}\right|_{2 \times 2} \ldots\right.$

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Question 25 Marks

If $A =\left|\begin{array}{cc}3 & -2 \\ -1 & 4\end{array}\right|, B =\left|\begin{array}{c}2 a \\ 1\end{array}\right|, C =\left|\begin{array}{c}-4 \\ 5\end{array}\right|, D =\left|\begin{array}{l}2 \\ b \end{array}\right|$ and $AB +2 C =4 D$ then find the values of $a$ and $b$.

Answer

$\begin{array}{l}A=\left|\begin{array}{cc}3 & -2 \\ -1 & 4\end{array}\right|_{2 \times 2}, B=\left|\begin{array}{c}2 a \\ 1\end{array}\right|_{2 \times 1}\end{array} $
$ C =\left|\begin{array}{c}-4 \\ 5\end{array}\right|_{2 \times 1}, D=\left|\begin{array}{l}2 \\ b \end{array}\right|_{2 \times 1} $
$ AB =\left|\begin{array}{cc}3 & -2 \\ -1 & 4\end{array}\right|\left|\begin{array}{c}2 a \\ 1\end{array}\right| $
$ =\left|\begin{array}{c}6 a -2 \\ -2 a +4\end{array}\right|_{2 \times 1} $
$ 2 C =\left|\begin{array}{c}-8 \\ 10\end{array}\right| $
$ AB +2 C =\left|\begin{array}{c}6 a -2 \\ -2 a +4\end{array}\right|\left|\begin{array}{c}-8 \\ 10\end{array}\right| $
$ =\left|\begin{array}{c}6 a -10 \\ -2 a +14\end{array}\right|_{2 \times 1} $
$ 4 D =\left|\begin{array}{c}8 \\ 4 b \end{array}\right|$
Given, $A B+2 C=4 D$
$ \begin{array}{l} \left|\begin{array}{c} 6 a-10 \\ -2 a+14 \end{array}\right|=\left|\begin{array}{c} 8 \\ 4 b \end{array}\right| \\\end{array} $
$6 a-10=8$
$\Rightarrow 6 a=18$
$\Rightarrow a=3$
$-2 a+14=4 b$
$\Rightarrow-2(3)+14=4 b$
$\Rightarrow 8=4 b$
$\Rightarrow 2=b$

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Question 35 Marks

If $X=\left|\begin{array}{cc}1 & -2 \\ 1 & 3\end{array}\right|, Y=\left|\begin{array}{cc}-3 & 0 \\ 4 & 1\end{array}\right|$ and $Z=\left|\begin{array}{cc}5 & -1 \\ 3 & 2\end{array}\right|$, prove that $X(Y+Z)=X Y+X Z$

Answer

$\begin{aligned} & X=\left|\begin{array}{cc}1 & -2 \\ 1 & 3\end{array}\right|_{2 \times 2} Y=\left|\begin{array}{cc}-3 & 0 \\ 4 & 1\end{array}\right|_{2 \times 2} Z=\left|\begin{array}{cc}5 & -1 \\ 3 & 2\end{array}\right|_{2 \times 2} \end{aligned}$
$ (Y+Z)=\left|\begin{array}{cc}-3 & 0 \\ 4 & 1\end{array}\right|+\left|\begin{array}{cc}5 & -1 \\ 3 & 2\end{array}\right|=\left|\begin{array}{cc}2 & -1 \\ 7 & 3\end{array}\right|_{2 \times 2} $
$ X(Y+Z)=\left|\begin{array}{cc}1 & -2 \\ 1 & 3\end{array}\right|\left|\begin{array}{cc}2 & -1 \\ 7 & 3\end{array}\right|$
$ \begin{aligned} & =\left|\begin{array}{cc} 2-14 & -1-6 \\ 2+21 & -1+9 \end{array}\right| \end{aligned} $
$ X(Y+Z)=\left|\begin{array}{cc} -12 & -7 \\ 23 & 8 \end{array}\right|_{2 \times 2} \ldots \ldots \ldots (1)$
$\begin{aligned} & X Y=\left|\begin{array}{cc}1 & -2 \\ 1 & 3\end{array}\right|\left|\begin{array}{cc}-3 & 0 \\ 4 & 1\end{array}\right| \end{aligned} $
$=\left|\begin{array}{cc}-3-8 & 0-2 \\ -3+12 & 0+3\end{array}\right| $
$ =\left|\begin{array}{cc}-11 & -2 \\ 9 & 3\end{array}\right|_{2 \times 2}$
$\begin{aligned} & X Z=\left|\begin{array}{cc}1 & -2 \\ 1 & 3\end{array}\right|\left|\begin{array}{cc}5 & -1 \\ 3 & 2\end{array}\right| \end{aligned} $
$ =\left|\begin{array}{cc}5-6 & -1-4 \\ 5+9 & -1+6\end{array}\right| $
$ =\left|\begin{array}{cc}-1 & -5 \\ 14 & 5\end{array}\right|_{2 \times 2}$
$ X Y+X Z=\left|\begin{array}{cc} -11 & -2 \\ 9 & 3 \end{array}\right|+\left|\begin{array}{cc} -1 & -5 \\ 14 & 5 \end{array}\right|$
$=\left|\begin{array}{cc} -12 & -7 \\ 23 & 8 \end{array}\right|_{2 \times 2} \ldots \ldots \text { (2) }$
from $(1)$ and $(2) X(Y+Z)=X Y+Y Z$

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Question 45 Marks

If $\left|\begin{array}{ll}3 a +2 b & 2 a - b \\ 4 p -3 q & 2 p + q \end{array}\right|=\left|\begin{array}{cc}12 & 1 \\ 16 & 8\end{array}\right|$, find the values of $a , b , p$ and $q.$

Answer

$\begin{array}{l}\left|\begin{array}{ll}3 a+2 b & 2 a-b \\ 4 p-3 q & 2 p+q\end{array}\right|=\left|\begin{array}{ll}12 & 1 \\ 16 & 8\end{array}\right| \end{array}$
$3 a+2 b=12 ........(1)$
$2 a-b=1 ......(2)$
$b=2 a-1$
$4 p-3 q=16 .......(3)$
$2 p+q=8......(4)$
putting the value of $b$ in $(1)$
$3 a+4 a-2=12$
$7 a=14$
$a=2$
from $(2)$
$4-b=1$
$b=3$
putting the value of $q$ in $(3)$
$4 p-24+6 p=16$
$10 p=40$
$p=4$
from$(4)$
$8+q=8$
$q=0$

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Question 55 Marks

If $A =\left|\begin{array}{cc} p & q \\ 8 & 5\end{array}\right|, B =\left|\begin{array}{cc}3 p & 5 q \\ 2 q & 7\end{array}\right|$ and if $A + B =\left|\begin{array}{cc}12 & 6 \\ 2 r & 3 s \end{array}\right|$, find the values of $p , q , r$ and $s$.

Answer

$\begin{array}{l}\left|\begin{array}{ll}3 a+2 b & 2 a-b \\ 4 p-3 q & 2 p+q\end{array}\right|=\left|\begin{array}{ll}12 & 1 \\ 16 & 8\end{array}\right| \end{array}$
$3 a+2 b=12 ........(1)$
$2 a-b=1 ......(2)$
$b=2 a-1$
$4 p-3 q=16 .......(3)$
$2 p+q=8......(4)$
putting the value of $b$ in $(1)$
$3 a+4 a-2=12$
$7 a=14$
$a=2$
from $(2)$
$4-b=1$
$b=3$
putting the value of $q$ in $(3)$
$4 p-24+6 p=16$
$10 p=40$
$p=4$
from$(4)$
$8+q=8$
$q=0$

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Question 65 Marks

If $A=\left|\begin{array}{ll}5 & r \\ p & 7\end{array}\right|, c$ and if $A+B=(9,7),(5,8)$, find the values of $p, q, r$ and $s$.

Answer

$\begin{array}{l}\left|\begin{array}{ll}3 a+2 b & 2 a-b \\ 4 p-3 q & 2 p+q\end{array}\right|=\left|\begin{array}{ll}12 & 1 \\ 16 & 8\end{array}\right| \end{array}$
$3 a+2 b=12 ........(1)$
$2 a-b=1 ......(2)$
$b=2 a-1$
$4 p-3 q=16 .......(3)$
$2 p+q=8......(4)$
putting the value of $b$ in $(1)$
$3 a+4 a-2=12$
$7 a=14$
$a=2$
from $(2)$
$4-b=1$
$b=3$
putting the value of $q$ in $(3)$
$4 p-24+6 p=16$
$10 p=40$
$p=4$
from$(4)$
$8+q=8$
$q=0$

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Mathematics [5 marks sum]

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