Question 14 Marks
If $A =\left[\begin{array}{ll}1 & 3 \\ 3 & 4\end{array}\right] B =\left[\begin{array}{ll}-2 & 1 \\ -3 & 2\end{array}\right]$ and $A^2-5 B^2=5 C$ Find the matrix $C$ where $C$ is a $2$ by $2$ matrix.
Answer$A^2=A \times A=p\left[\begin{array}{ll}1 & 3 \\ 3 & 4\end{array}\right]\left[\begin{array}{ll}1 & 3 \\ 3 & 4\end{array}\right]$
$\left[\begin{array}{cc}1 \times 1+3 \times 3 & 1 \times 3+3 \times 4 \\ 3 \times 1+4 \times 3 & 3 \times 3+34 \times 40\end{array}\right]$
$=\left[\begin{array}{cc}10 & 15 \\ 15 & 25\end{array}\right]$
$\begin{array}{l}B^2=B \times B=\left[\begin{array}{ll}-2 & 1 \\ -3 & 2\end{array}\right]\left[\begin{array}{ll}-2 & 1 \\ -3 & 2\end{array}\right]\end{array}$
$=\left[\begin{array}{ll}-2 \times-2+1 \times-3 & -2 \times 1+1 \times 2 \\ -3 \times-2+2 \times-3 & -3 \times 1+2 \times 2\end{array}\right]$
$=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
Given: $A^2-5 B^2=5 C$
$\begin{array}{l}\Rightarrow\left[\begin{array}{ll}10 & 15 \\ 15 & 25\end{array}\right]-5\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=5 C\end{array} $
$\Rightarrow\left[\begin{array}{ll}10 & 15 \\ 15 & 25\end{array}\right]-\left[\begin{array}{ll}5 & 0 \\ 0 & 5\end{array}\right]=5 C$
$\Rightarrow\left[\begin{array}{cc}5 & 15 \\ 15 & 20\end{array}\right]=5 C $
$ \Rightarrow 5\left[\begin{array}{ll}1 & 3 \\ 3 & 4\end{array}\right]=5 C$
$\Rightarrow C=\left[\begin{array}{ll}1 & 3 \\ 3 & 4\end{array}\right]$
View full question & answer→Question 24 Marks
Given $A=\left[\begin{array}{cc}2 & 0 \\ -1 & 7\end{array}\right]$ and $1=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$ and $A^2=9 A+m l$. Find $m$
Answer$A^2=9 A+M I$
$\Rightarrow A^2-9 A=m l$
Now, $A^2=A A$
$=\left[\begin{array}{cc}2 & 0 \\ -1 & 7\end{array}\right]\left[\begin{array}{cc}2 & 0 \\ -1 & 7\end{array}\right] \\ =\left[\begin{array}{cc}4 & 0 \\ -9 & 49\end{array}\right]$
Substituting $A^2$ in $(1),$ we have
$A^2-9 A=m l$
$\Rightarrow\left[\begin{array}{cc}4 & 0 \\ -9 & 49\end{array}\right]-9\left[\begin{array}{cc}2 & 0 \\ -1 & 7\end{array}\right]=m\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] $
$ \Rightarrow\left[\begin{array}{cc}4 & 0 \\ -9 & 49\end{array}\right]-\left[\begin{array}{cc}18 & 0 \\ -9 & 63\end{array}\right]=\left[\begin{array}{cc}m & 0 \\ 0 & m\end{array}\right] $
$ \Rightarrow\left[\begin{array}{cc}-14 & 0 \\ 0 & -14\end{array}\right]=\left[\begin{array}{cc}m & 0 \\ 0 & m\end{array}\right]$
$\Rightarrow m=-14$
View full question & answer→Question 34 Marks
if $A=\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]$ and $I=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$, find $A^2-5 A+7 I$
Answer$A=\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right], I=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
$A^2=\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]=\left[\begin{array}{cc}9-1 & 3+2 \\ -3-2 & -1+4\end{array}\right]=\left[\begin{array}{cc}8 & 5 \\ -5 & 3\end{array}\right]$
$A^2-5 A+7 I=\left[\begin{array}{cc}8 & 5 \\ -5 & 3\end{array}\right]-5\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]+7\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
$\begin{array}{l}=\left[\begin{array}{cc}8 & 5 \\ -5 & 3\end{array}\right]-\left[\begin{array}{cc}15 & 5 \\ -5 & 10\end{array}\right]+\left[\begin{array}{ll}7 & 0 \\ 0 & 7\end{array}\right] \end{array}$
$ =\left[\begin{array}{cc}-7 & 0 \\ 0 & -7\end{array}\right]+\left[\begin{array}{ll}7 & 0 \\ 0 & 7\end{array}\right] $
$ =\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right] $
$ =0$
View full question & answer→Question 44 Marks
Find $x$ and $y$, if $\left(\begin{array}{cc}3 & -2 \\ -1 & 4\end{array}\right)\left(\begin{array}{c}2 x \\ 1\end{array}\right)+2\left(\begin{array}{c}-4 \\ 5\end{array}\right)=4\left(\begin{array}{l}2 \\ y\end{array}\right)$
Answer$\left(\begin{array}{cc}3 & -2 \\ -1 & 4\end{array}\right)\left(\begin{array}{c}2 x \\ 1\end{array}\right)+2\left(\begin{array}{c}-4 \\ 5\end{array}\right)=4\left(\begin{array}{l}2 \\ y\end{array}\right)$
$\left(\begin{array}{c}6 x-2 \\ -2 x+4\end{array}\right)+\left(\begin{array}{c}-8 \\ 10\end{array}\right)=\left(\begin{array}{c}8 \\ 4 y\end{array}\right)$
$\left(\begin{array}{c}6 x-10-8 \\ -2 x+14+10\end{array}\right)=\left(\begin{array}{c}8 \\ 4 y\end{array}\right)$
$\text { Now } 6 x-10=8$
$\therefore 6 x=18$
$\therefore x=\frac{18}{6}=3$
and
$-x + 14 = 4y$
$-2 x^3 + 14 = 4y$
or
$4y = 14 - 6 = 8$
$\therefore y=\frac{8}{4}=2$
$\therefore x=3, y=2$
View full question & answer→Question 54 Marks
If $A=\left[\begin{array}{cc}3 & a \\ -4 & 8\end{array}\right], B=\left[\begin{array}{cc}c & 4 \\ -3 & 0\end{array}\right], C=\left[\begin{array}{cc}-1 & 4 \\ 3 & b\end{array}\right]$ and $3 A -2 C =6 B$, find the values of $a, b, c$.
Answer$3A - 2C = 6B$
$\begin{array}{l}3\left[\begin{array}{cc}3 & a \\ -4 & 8\end{array}\right]-2\left[\begin{array}{cc}-1 & 4 \\ 3 & b\end{array}\right]=6\left[\begin{array}{cc}c & 4 \\ -3 & 0\end{array}\right] \end{array} $
$ {\left[\begin{array}{cc}9 & 3 a \\ -12 & 24\end{array}\right]-\left[\begin{array}{cc}-2 & 8 \\ 6 & 2 b\end{array}\right]=\left[\begin{array}{cc}6 c & 24 \\ -18 & 0\end{array}\right]} $
$ {\left[\begin{array}{cc}11 & 3 a-8 \\ -18 & 24-2 b\end{array}\right]=\left[\begin{array}{cc}6 c & 24 \\ -18 & 0\end{array}\right]}$
Comparing the corresponding elements we get
$3 a-8=24 $
$\Rightarrow 3 a=32 $
$\Rightarrow a=\frac{32}{3}=10 \frac{2}{3}$
$24-2 b=0 $
$\Rightarrow 2 b=24$
$\Rightarrow b=12$
$11=6 c $
$\Rightarrow c=\frac{11}{6}=1 \frac{5}{6}$
View full question & answer→Question 64 Marks
Given $A=\left[\begin{array}{cc}2 & -6 \\ 2 & 0\end{array}\right], B=\left[\begin{array}{cc}-3 & 2 \\ 4 & 0\end{array}\right], C=\left[\begin{array}{ll}4 & 0 \\ 0 & 2\end{array}\right]$ Find the matrix $X$ such that $A+2 X=2 B+C$.
Answer$A + 2X = 2B + C$
$\begin{array}{l}\Rightarrow\left[\begin{array}{cc}2 & -6 \\ 2 & 0\end{array}\right]+2 X=2\left[\begin{array}{cc}-3 & 2 \\ 4 & 0\end{array}\right]+\left[\begin{array}{ll}4 & 0 \\ 0 & 2\end{array}\right] \end{array} $
$ \Rightarrow\left[\begin{array}{cc}2 & -6 \\ 2 & 0\end{array}\right]+2 X=\left[\begin{array}{cc}-6+4 & 4+0 \\ 8+0 & 0+2\end{array}\right]$
$ \Rightarrow\left[\begin{array}{cc}2 & -6 \\ 2 & 0\end{array}\right]+2 X=\left[\begin{array}{cc}-2 & 4 \\ 8 & 2\end{array}\right]$
$\Rightarrow 2 X=\left[\begin{array}{cc}-2 & 4 \\ 8 & 2\end{array}\right]-\left[\begin{array}{cc}2 & -6 \\ 2 & 0\end{array}\right]=\left[\begin{array}{cc}-4 & 10 \\ 6 & 2\end{array}\right] $
$ \Rightarrow X=\frac{1}{2}\left[\begin{array}{cc}-4 & 10 \\ 6 & 2\end{array}\right] $
$ \Rightarrow X=\left[\begin{array}{cc}-2 & 5 \\ 3 & 1\end{array}\right]$
View full question & answer→Question 74 Marks
If $A=\left[\begin{array}{ll}2 & 5 \\ 1 & 3\end{array}\right], B=\left[\begin{array}{cc}4 & -2 \\ -1 & 3\end{array}\right]$ and $I$ is Identity matrix of same order and $A^t$ is the transpose of matrix $A$ find $A^t \cdot B+B I$
Answer$A=\left[\begin{array}{ll}2 & 5 \\ 1 & 3\end{array}\right]$
$\therefore A^t=\left[\begin{array}{ll}2 & 1 \\ 5 & 3\end{array}\right]$
$\begin{array}{l}A^t . B=\left[\begin{array}{ll}2 & 1 \\ 5 & 3\end{array}\right]\left[\begin{array}{cc}4 & -2 \\ -1 & 3\end{array}\right] \end{array} $
$ =\left[\begin{array}{ll}2 \times 4+1 \times(-1) & 2 \times(-2)+1 \times 3 \\ 3 \times 4+3 \times(-1) & 5 \times(-2)+3 \times 3\end{array}\right] $
$ =\left[\begin{array}{cc}7 & -1 \\ 17 & -1\end{array}\right]$
$B .I=\left[\begin{array}{cc}4 & -2 \\ -1 & 3\end{array}\right]\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
$=\left[\begin{array}{cc}4 & -2 \\ -1 & 3\end{array}\right]$
$\therefore A^t \cdot B+B I=\left[\begin{array}{cc}7 & -1 \\ 17 & -1\end{array}\right]+\left[\begin{array}{cc}4 & -2 \\ -1 & 3\end{array}\right] $
$ =\left[\begin{array}{cc}11 & -3 \\ 16 & 2\end{array}\right]$
View full question & answer→Question 84 Marks
If $A=\left[\begin{array}{ll}1 & 3 \\ 2 & 4\end{array}\right], B=\left[\begin{array}{ll}1 & 2 \\ 4 & 3\end{array}\right]$ and $C=\left[\begin{array}{ll}4 & 3 \\ 1 & 2\end{array}\right]$ find $1) \ce{(AB)C , 2) A(BC)}$ Is $\ce{A(BC) = (AB)C} ?$
Answer$1)$
$\begin{array}{l}( AB ) C =\left[\begin{array}{ll}1 & 3 \\ 2 & 4\end{array}\right]\left[\begin{array}{ll}1 & 2 \\ 4 & 3\end{array}\right]\left[\begin{array}{ll}4 & 3 \\ 1 & 2\end{array}\right] \end{array} $
$ =\left[\begin{array}{ll}1(1)+3(4) & 1(2)+3(3) \\ 2(1)+4(4) & 2(2)+4(3)\end{array}\right]$
$=\left[\begin{array}{ll}4 & 3 \\ 1 & 2\end{array}\right] $
$ =\left[\begin{array}{ll}13 & 11 \\ 18 & 16\end{array}\right]\left[\begin{array}{ll}4 & 3 \\ 1 & 2\end{array}\right] $
$ =\left[\begin{array}{ll}13(4)+11(1) & 13(3)+11(2) \\ 18(4)+18(3) & 18(3)+16(2)\end{array}\right] $
$ =\left[\begin{array}{ll}52+11 & 39+22 \\ 72+16 & 54+32\end{array}\right] \\ =\left[\begin{array}{ll}63 & 61 \\ 88 & 86\end{array}\right]$
$2)$
$\begin{array}{l} A(B C)=\left[\begin{array}{ll}1 & 3 \\ 2 & 4\end{array}\right]\left[\begin{array}{ll}1 & 2 \\ 4 & 3\end{array}\right]\left[\begin{array}{ll}4 & 3 \\ 1 & 2\end{array}\right] \end{array} $
$ =\left[\begin{array}{ll}1 & 3 \\ 2 & 4\end{array}\right]\left[\begin{array}{ll}1(4)+2(1) & 1(3)+2(2) \\ 4(4)+3(1) & 4(3)+3(2)\end{array}\right] $
$ =\left[\begin{array}{ll}1 & 3 \\ 2 & 4\end{array}\right]\left[\begin{array}{cc}4+2 & 3+4 \\ 16+3 & 12+6\end{array}\right] $
$ =\left[\begin{array}{ll}1 & 3 \\ 2 & 4\end{array}\right]\left[\begin{array}{cc}6 & 7 \\ 19 & 18\end{array}\right] $
$ =\left[\begin{array}{cc}6+57 & 7+54 \\ 12+76 & 14+72\end{array}\right] $
$ =\left[\begin{array}{ll}63 & 61 \\ 88 & 86\end{array}\right] $
$ \therefore A(B C)=( AB ) C $
View full question & answer→Question 94 Marks
If $A=\left[\begin{array}{lll}2 & 1 & -1 \\ 0 & 1 & -2\end{array}\right]$ Find $A^t . A$ Where $A^t . A$ is the transpose of matrix $A$
Answer$A=\left[\begin{array}{lll}2 & 1 & -1 \\ 0 & 1 & -2\end{array}\right]$
$A^t=\left[\begin{array}{cc}2 & 0 \\ 1 & 1 \\ -1 & -2\end{array}\right]$
$A^t \cdot A=\left[\begin{array}{cc}2 & 0 \\ 1 & 1 \\ -1 & -2\end{array}\right]\left[\begin{array}{ccc}2 & 1 & -1 \\ 0 & 1 & -2\end{array}\right] $
$ =\left[\begin{array}{ccc}4+0 & 2+0 & -2-0 \\ 2+0 & 1+1 & -1-2 \\ -2-0 & -1-2 & 1+4\end{array}\right] $
$ =\left[\begin{array}{ccc}4 & 2 & -2 \\ 2 & 2 & -3 \\ -2 & -3 & 5\end{array}\right]$
View full question & answer→Question 104 Marks
If $P=\left[\begin{array}{cc}1 & 2 \\ 2 & -1\end{array}\right]$ and $Q=\left[\begin{array}{ll}1 & 0 \\ 2 & 1\end{array}\right]$ then computer:
$(1) P^2-Q^2$
$(2) (P+Q)(P-Q)$
is $(P+Q)(P-Q)=P^2-Q^2$ true for matrix algebra?
Answer$P^2=\left[\begin{array}{cc}1 & 2 \\ 2 & -1\end{array}\right]\left[\begin{array}{cc}1 & 2 \\ 2 & -1\end{array}\right]=\left[\begin{array}{ll}1+4 & 2-2 \\ 2-2 & 4+1\end{array}\right]=\left[\begin{array}{ll}5 & 0 \\ 0 & 5\end{array}\right]$
$Q^2=\left[\begin{array}{ll}1 & 0 \\ 2 & 1\end{array}\right]\left[\begin{array}{ll}1 & 0 \\ 2 & 1\end{array}\right]=\left[\begin{array}{ll}1+0 & 0+0 \\ 2+2 & 0+1\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 4 & 1\end{array}\right]$
$p^2-Q^2=\left[\begin{array}{ll}5 & 0 \\ 0 & 5\end{array}\right]-\left[\begin{array}{ll}1 & 0 \\ 4 & 1\end{array}\right]=\left[\begin{array}{cc}4 & 0 \\ -4 & 4\end{array}\right] \\ P+Q=\left[\begin{array}{cc}1 & 2 \\ 2 & -1\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 2 & 1\end{array}\right]=\left[\begin{array}{ll}2 & 2 \\ 4 & 0\end{array}\right] $
$ P-Q=\left[\begin{array}{cc}1 & 2 \\ 2 & -1\end{array}\right]-\left[\begin{array}{ll}1 & 0 \\ 2 & 1\end{array}\right]=\left[\begin{array}{cc}0 & 2 \\ 0 & -2\end{array}\right] $
$ (P+Q)(P-Q)=\left[\begin{array}{ll}2 & 2 \\ 4 & 0\end{array}\right]\left[\begin{array}{cc}0 & 2 \\ 0 & -2\end{array}\right]$
$=\left[\begin{array}{ll}0+0 & 4-4 \\ 0+0 & 8-0\end{array}\right]=\left[\begin{array}{ll}0 & 0 \\ 0 & 8\end{array}\right]$
Clearly it can be said that $(P+Q)(P-Q)=P^2-Q^2$ not true for matrix algebra.
View full question & answer→Question 114 Marks
If $A =\left[\begin{array}{ll}3 & 7 \\ 2 & 4\end{array}\right], B=\left[\begin{array}{ll}0 & 2 \\ 5 & 3\end{array}\right]$ and $C=\left[\begin{array}{cc}1 & -5 \\ -4 & 6\end{array}\right]$ Find $A B - 5C$
AnswerGiven : $A=\left[\begin{array}{ll}3 & 7 \\ 2 & 4\end{array}\right], B=\left[\begin{array}{ll}0 & 2 \\ 5 & 3\end{array}\right]$ and $C=\left[\begin{array}{cc}1 & -5 \\ -4 & 6\end{array}\right]$
Now,
$\begin{array}{l} AB =\left[\begin{array}{ll}3 & 7 \\ 2 & 4\end{array}\right]\left[\begin{array}{ll}0 & 2 \\ 5 & 3\end{array}\right] \\ =\left[\begin{array}{ll}3 \times 0+7 \times 5 & 3 \times 2+7 \times 3 \\ 2 \times 0+4 \times 5 & 2 \times 2+4 \times 3\end{array}\right]\end{array}$
$\begin{array}{l}=\left[\begin{array}{ll}0+35 & 6+21 \\ 0+20 & 4+12\end{array}\right] \\ =\left[\begin{array}{ll}35 & 27 \\ 20 & 16\end{array}\right]\end{array}$
$5 C=5\left[\begin{array}{cc}1 & -5 \\ -4 & 6\end{array}\right]=\left[\begin{array}{cc}5 & -25 \\ -20 & 30\end{array}\right]$
$A B-5 C=\left[\begin{array}{ll}35 & 27 \\ 20 & 16\end{array}\right]-\left[\begin{array}{cc}5 & -25 \\ -20 & 30\end{array}\right]$
$=\left[\begin{array}{cc}30 & 52 \\ 40 & -14\end{array}\right]$
View full question & answer→Question 124 Marks
In the given case below, Find $:$
$a)$ The order of matrix $M$
$b)$ The matrix $M \left[\begin{array}{ll}1 & 4 \\ 2 & 1\end{array}\right] \times M=\left[\begin{array}{c}13 \\ 5\end{array}\right]$
AnswerLet the order of matrix $M$ be $a \times b$
$\left[\begin{array}{ll}1 & 4 \\ 2 & 1\end{array}\right]_{2 \times 2} \times M_{a \times b}=\left[\begin{array}{c}13 \\ 5\end{array}\right]_{2 \times 1}$
Clearly the order of matrix $M$ is $2 \times 1$
Let $M=\left[\begin{array}{l}a \\ b\end{array}\right]$
$\left[\begin{array}{ll}1 & 4 \\ 2 & 1\end{array}\right] \times M=\left[\begin{array}{c}13 \\ 5\end{array}\right]$
$\left[\begin{array}{ll}1 & 4 \\ 2 & 1\end{array}\right] \times\left[\begin{array}{l}a \\ b\end{array}\right]=\left[\begin{array}{c}13 \\ 5\end{array}\right]$
$\left[\begin{array}{l}a+4 b \\ 2 a+b\end{array}\right]=\left[\begin{array}{c}13 \\ 5\end{array}\right]$
Comparing the corresponding elements we get
$a + 4b = 13 .....(1)$
$2a + b = 5....(2)$
Multiplying $(2)$ by $4$ we get
$8a + 4b = 20 ....(3)$
Substracting $(1)$ from $(3)$ we get
$a=7=>a=1$
From $(2)$ we get
$b = 5 - 2a = 5- 2 = 3$
$\therefore M=\left[\begin{array}{l}a \\ b\end{array}\right]=\left[\begin{array}{l}1 \\ 3\end{array}\right]$
View full question & answer→Question 134 Marks
In the given case below find
$(a)$ The order of matrix $M$.
$(b)$ The matrix $M$
$M \times\left[\begin{array}{ll}1 & 1 \\ 0 & 2\end{array}\right]=[1,2]$
AnswerWe know, the product of two matrices is defined only when the number of columns of the first matrix is equal to the number of rows of the second matrix
Let the order of matrix $M$ be $a \times b.$
$M_{a \times b} \times\left[\begin{array}{ll}1 & 1 \\ 0 & 2\end{array}\right]_{2 \times 2}=\left[\begin{array}{ll}1 & 2\end{array}\right]_{1 \times 2}$
Clearly, the order of matrix $M$ is $1 \times 2$
Let $M=[a, b]$
$\begin{array}{l}M \times\left[\begin{array}{ll}1 & 1 \\ 0 & 2\end{array}\right]=\left[\begin{array}{ll}1 & 2\end{array}\right] \\ {[a, b] \times\left[\begin{array}{ll}1 & 1 \\ 0 & 2\end{array}\right]=\left[\begin{array}{ll}1 & 2\end{array}\right]} \\ {[a+0 \quad a+2 b]=[1,2]}\end{array}$
Comparing the corresponding elements we get
$a=1 $ and $ a+2 b=2$
$ \Rightarrow 2 b=2-1=1 $
$\Rightarrow b=\frac{1}{2}$
$\therefore M=\left[a b\right]=\left[1 \frac{1}{2}\right]$
View full question & answer→Question 144 Marks
Solve for $x$ and $y : \left[\begin{array}{ll}2 & 5 \\ 5 & 2\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{c}-7 \\ 14\end{array}\right]$
Answer$\begin{array}{l}{\left[\begin{array}{ll}2 & 5 \\ 5 & 2\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{l}-7 \\ 14\end{array}\right]} \end{array} $
$ {\left[\begin{array}{l}2 x+5 y \\ 5 x+2 y\end{array}\right]=\left[\begin{array}{l}-7 \\ 14\end{array}\right]}$
Comparing the corresponding elements, we get
$2x + 5y = -7 ....(1)$
$5x + 2y = 14 ....(2)$
Multiplying $(1)$ and $2$ and $(2)$ with $5$ we get
$4x + 10y = -14...(3)$
$25x + 10y = 70 ....(4)$
Substracting $(3)$ from $(4),$ we get
$21x = 84 $
$\Rightarrow x = 4$
From $(2), 2y = 14 - 5x $
$= 14 - 20 $
$= -6 $
$\Rightarrow y = -3$
View full question & answer→Question 154 Marks
If $A=\left[\begin{array}{cc}1 & 4 \\ 1 & -3\end{array}\right]$ and $B=\left[\begin{array}{cc}1 & 2 \\ -1 & -1\end{array}\right]$ Find:
$i. (A+B)^2$
$ii. A^2+B^2$
$iii.$ Is $(A+B)^2=A^2+B^2$ ?
Answer$A+B=\left[\begin{array}{cc}1 & 4 \\ 1 & -3\end{array}\right]+\left[\begin{array}{cc}1 & 2 \\ -1 & -1\end{array}\right]$
$=\left[\begin{array}{cc}1+1 & 4+2 \\ 1-1 & -3-1\end{array}\right] =\left[\begin{array}{cc}2 & 6 \\ 0 & -4\end{array}\right]$
Now, $( A + B )^2=( A + B )( A + B )$
$\begin{array}{l}=\left[\begin{array}{cc}2 & 6 \\ 0 & -4\end{array}\right]\left[\begin{array}{cc}2 & 6 \\ 0 & -4\end{array}\right] \end{array} $
$=\left[\begin{array}{cc}(2)(2)+(6)(0) & (2)(6)+(6)(-4) \\ (0)(2)+(-4)(0) & (0)(6)+(-4)(-4)\end{array}\right] $
$ =\left[\begin{array}{cc}4 & -12 \\ 0 & 16\end{array}\right]$
$(ii)\ A^2=\left[\begin{array}{cc}1 & 4 \\ 1 & -3\end{array}\right]\left[\begin{array}{cc}1 & 4 \\ 1 & -3\end{array}\right] $
$=\left[\begin{array}{cc}1+4 & 4-12 \\ 1-3 & 4+9\end{array}\right] $
$ =\left[\begin{array}{cc}5 & -8 \\ -2 & 13\end{array}\right]$
$B ^2=\left[\begin{array}{cc}1 & 2 \\ -1 & -1\end{array}\right]\left[\begin{array}{cc}1 & 2 \\ -1 & -1\end{array}\right]$
$\left[\begin{array}{cc}1-2 & 2-2 \\ -1+1 & -2+1\end{array}\right]$
$\left[\begin{array}{cc}-1 & 0 \\ 0 & -1\end{array}\right] A^2+B^2=\left[\begin{array}{cc}5 & -8 \\ -2 & 13\end{array}\right]+\left[\begin{array}{cc}-1 & 0 \\ 0 & -1\end{array}\right]$
$A ^2+ B ^2=\left[\begin{array}{cc}4 & -8 \\ -2 & 12\end{array}\right]$
$(iii)$ No $,(A+B)^2 \neq A^2+B^2$.
View full question & answer→Question 164 Marks
If $A=\left[\begin{array}{ll}0 & -1 \\ 4 & -3\end{array}\right], B =\left[\begin{array}{c}-5 \\ 6\end{array}\right]$ and $3 A \times M =2 B$; Find matrix $M$
AnswerLet the order of matrix $M$ be $a$ x $b.$
$3A \ x m = 2B$
$3\left[\begin{array}{ll}0 & -1 \\ 4 & -3\end{array}\right]_{2 \times 2} \times M_{a \times b}=2\left[\begin{array}{c}-5 \\ 6\end{array}\right]_{2 \times 1}$
Clearly the order of matrix $M$ is $2 x 1$
Let $M=\left[\begin{array}{l}x \\ y\end{array}\right]$
Then
$3\left[\begin{array}{ll}0 & -1 \\ 4 & -3\end{array}\right] \times\left[\begin{array}{l}x \\ y\end{array}\right]=2\left[\begin{array}{c}-5 \\ 6\end{array}\right] $
$ {\left[\begin{array}{cc}0 & -3 \\ 12 & -9\end{array}\right] \times\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{c}-10 \\ 2\end{array}\right]} $
$ {\left[\begin{array}{c}-3 y \\ 12 x-9 y\end{array}\right]=\left[\begin{array}{c}-10 12\end{array}\right]}$
Comparing the corressponding elements we get
$-3y = - 10$
$\Rightarrow y=\frac{10}{3}$
$12 x-9 y=12$
$=12 x=42$
$\Rightarrow x=\frac{7}{2} $
$ \therefore M=\left[\begin{array}{c}\frac{7}{2} \\ \frac{10}{3}\end{array}\right]$
View full question & answer→Question 174 Marks
Given $\left[\begin{array}{cc}2 & 1 \\ -3 & 4\end{array}\right] x=\left[\begin{array}{l}7 \\ 6\end{array}\right]$ Write the matrix $x$
AnswerLet $ x=\left[\begin{array}{l} x \\ y \end{array}\right] $
$ \therefore\left[\begin{array}{cc} 2 & 1 \\ -3 & 4 \end{array}\right] \times\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{l} 7 \\ 6 \end{array}\right] $
$ \Rightarrow\left[\begin{array}{c} 2 x+y \\ -3 x+4 y \end{array}\right]=\left[\begin{array}{l} 7 \\ 6 \end{array}\right] $
$ \Rightarrow 2 x+y=7 \ldots(1) \\ -3 x+4 y=6 \ldots(2) $
Multiplying by $4$ in equation $(1)$ and solving with equation $(2)$
$ 8 x+4 y=28$
$-3 x+4 y=6$
$( +) \_(-) \_(-)$
$11 x=22$
$x=2 $
Putting the value of $x$ in equation $(1)$, we get
$ \therefore 2 \times 2+y=7$
$y=7-4=3 $
$\therefore$ The matrix $x =\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{l}2 \\ 3\end{array}\right]$
View full question & answer→Question 184 Marks
Given matrix $B=\left[\begin{array}{ll}1 & 1 \\ 8 & 3\end{array}\right]$ Find the matrix $X$ if, $X=B^2-4 B$. Hence, solve for $a$ and $b$ given $X\left[\begin{array}{l}a \\ b\end{array}\right]=\left[\begin{array}{c}5 \\ 50\end{array}\right]$
Answer$B^2=B \times B=\left[\begin{array}{ll}1 & 1 \\ 8 & 3\end{array}\right]\left[\begin{array}{ll}1 & 1 \\ 8 & 3\end{array}\right]=\left[\begin{array}{ll}1 \times 1+1 \times 8 & 1 \times 1+1 \times 3 \\ 8 \times 1+3 \times 8 & 8 \times 1+3 \times 3\end{array}\right]=\left[\begin{array}{cc}9 & 4 \\ 32 & 17\end{array}\right]$
$4 B=4\left[\begin{array}{ll}1 & 1 \\ 8 & 3\end{array}\right]=\left[\begin{array}{cc}4 & 4 \\ 32 & 12\end{array}\right]$
Given : $X=B^2-4 B$
$\Rightarrow X=\left[\begin{array}{cc}9 & 4 \\ 32 & 17\end{array}\right]-\left[\begin{array}{cc}4 & 4 \\ 32 & 12\end{array}\right]=\left[\begin{array}{ll}5 & 0 \\ 0 & 5\end{array}\right]$
To find: $a$ and $b$
$\times\left[\begin{array}{l}a \\ b\end{array}\right]=\left[\begin{array}{c}5 \\ 50\end{array}\right] ....... $ given
$\Rightarrow\left[\begin{array}{ll}5 & 0 \\ 0 & 5\end{array}\right]\left[\begin{array}{l}a \\ b\end{array}\right]=\left[\begin{array}{c}5 \\ 50\end{array}\right] $
$ \Rightarrow\left[\begin{array}{l}5 a \\ 5 b\end{array}\right]=\left[\begin{array}{c}5 \\ 50\end{array}\right] $
$ \Rightarrow 5\left[\begin{array}{l}a \\ b\end{array}\right]=5\left[\begin{array}{c}1 \\ 10\end{array}\right]$
$\Rightarrow a = 1$ and $b = 10$
View full question & answer→Question 194 Marks
Given matrix $A =\left[\begin{array}{l}4 \sin 30^{\circ}, \cos 0^{\circ} \\ \cos 0^{\circ}, 4 \sin 30^{\circ}\end{array}\right]$ and $B=\left[\begin{array}{l}4 \\ 5\end{array}\right]$ If $AX = B$. Find the matrix $' X \ '$
AnswerLet the matrix $X =\left[\begin{array}{l}x \\ y\end{array}\right]$
$AX = B$
$\begin{array}{l}\Rightarrow\left[\begin{array}{c}4 \sin 30^{\circ}, \cos 0^{\circ} \\ \cos 0^{\circ}, 4 \sin 30^{\circ}\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{l}4 \\ 5\end{array}\right] \\ \Rightarrow\left[\begin{array}{cc}4\left(\frac{1}{2}\right) & 1 \\ 1 & 4\left(\frac{1}{2}\right)\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{l}4 \\ 5\end{array}\right] \\ \Rightarrow\left[\begin{array}{ll}2 & 1 \\ 1 & 2\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{l}4 \\ 5\end{array}\right] \\ \Rightarrow\left[\begin{array}{l}2 x+y \\ x+2 y\end{array}\right]=\left[\begin{array}{l}4 \\ 5\end{array}\right]\end{array}$
$\Rightarrow 2x + y = 4 .............(1)$
$x + 2y = 5 .........(2)$
Multiplying $(1)$ by $2,$ we get
$4x + 2y = 8 ….(3)$
Subtracting $(2)$ from $(3)$, we get
$3x = 3$
$\Rightarrow x = 1$
Substituting the value of $x$ in $(1),$ we get
$2(1) + y = 4$
$\Rightarrow 2 + y = 4$
$\Rightarrow y = 2$
Hence, the matrix $X =\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{l}1 \\ 2\end{array}\right]$
View full question & answer→Question 204 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 $A^2+A B+B^2$
Answer$A=\left[\begin{array}{ll}1 & 0 \\ 2 & 1\end{array}\right], B=\left[\begin{array}{cc}2 & 3 \\ -1 & 0\end{array}\right]$
$A^2=A \times A=\left[\begin{array}{ll}1 & 0 \\ 2 & 1\end{array}\right] \times\left[\begin{array}{ll}1 & 0 \\ 2 & 1\end{array}\right]$
$=\left[\begin{array}{ll}1 \times 1+0 \times 2 & 1 \times 0+0 \times 1 \\ 2 \times 1+1 \times 2 & 2 \times 0+1 \times 1\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 4 & 1\end{array}\right]$
$\begin{array}{l}A B=A \times B=\left[\begin{array}{ll}1 & 0 \\ 2 & 1\end{array}\right] \times\left[\begin{array}{cc}2 & 3 \\ -1 & 0\end{array}\right] \\ =\left[\begin{array}{ll}1 \times 2+0 \times(-1) & 1 \times 3+0 \times 0 \\ 2 \times 2+1 \times(-1) & 2 \times 3+1 \times 0\end{array}\right] \\ =\left[\begin{array}{ll}2 & 3 \\ 3 & 6\end{array}\right]\end{array}$
$\begin{array}{l}B^2=B \times B=\left[\begin{array}{cc}2 & 3 \\ -1 & 0\end{array}\right] \times\left[\begin{array}{cc}2 & 3 \\ -1 & 0\end{array}\right] \\ =\left[\begin{array}{cc}2 \times 2+3 \times(-1) & 2 \times 3+3 \times 0 \\ (-1) \times 2+0 \times(-1) & -1 \times 3+0 \times 0\end{array}\right] \\ =\left[\begin{array}{cc}1 & 6 \\ -2 & -3\end{array}\right] \\\end{array}$
$\therefore A^2+A B+B^2=\left[\begin{array}{ll}1 & 0 \\ 4 & 1\end{array}\right]+\left[\begin{array}{ll}2 & 3 \\ 3 & 6\end{array}\right]+\left[\begin{array}{cc}1 & 6 \\ -2 & -3\end{array}\right]$
$=\left[\begin{array}{ll}4 & 9 \\ 5 & 4\end{array}\right]$
View full question & answer→Question 214 Marks
if $A=\left[\begin{array}{ll}2 & 1 \\ 1 & 3\end{array}\right]$ and $B=\left[\begin{array}{c}3 \\ -11\end{array}\right]$ Find the matrix $X$ such that $AX = B$
AnswerLet the order of the matrix $X$ be a $x b.$
$AX = B$
$\left[\begin{array}{ll}2 & 1 \\ 1 & 3\end{array}\right]_{2 \times 2} \times X_{a \times b}=\left[\begin{array}{c}3 \\ -11\end{array}\right]_{2 \times 1}$
Clearly the order of the matrix $X$ is $2 \times 1$
Let $X=[(x),(y)]$
$\left[\begin{array}{ll}2 & 1 \\ 1 & 3\end{array}\right] \times\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{c}3 \\ -11\end{array}\right]$
Comparing the two matrices we get
$2x = y = 3 ...(1)$
$x = 3y = -11 ...(2)$
Multiplying $(1)$ with $3$ we get
$6x + 3y = 9 ...(3)$
Substracting $(2)$ from $(3)$ we get
$5x = 20$
$x = 4$
From $(1)$ we have
$y = 3 - 2x = 3 - 8 = -5$
$\therefore X=\left[\begin{array}{c}4 \\ -5\end{array}\right]$
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