Question 13 Marks
If $A =\left[\begin{array}{cc}-1 & 3 \\ 2 & 4\end{array}\right], B =\left[\begin{array}{cc}2 & -3 \\ -4 & -6\end{array}\right]$ find the matrix $A B+B A$
Answer$\begin{aligned} & A=\left[\begin{array}{cc}-1 & 3 \\ 2 & 4\end{array}\right] \\ & B=\left[\begin{array}{cc}2 & -3 \\ -4 & -6\end{array}\right] \\ & AB =\left[\begin{array}{cc}-1 & 3 \\ 2 & 4\end{array}\right] \times\left[\begin{array}{cc}2 & -3 \\ -4 & -6\end{array}\right] \\ & =\left[\begin{array}{cc}-2-12 & 3-18 \\ 4-16 & -6-24\end{array}\right] \\ & =\left[\begin{array}{ll}-14 & -15 \\ -12 & -30\end{array}\right] \\ & BA =\left[\begin{array}{cc}2 & -3 \\ -4 & -6\end{array}\right] \times\left[\begin{array}{cc}-1 & 3 \\ 2 & 4\end{array}\right] \\ & =\left[\begin{array}{cc}-2-6 & 6-12 \\ 4-12 & -12-24\end{array}\right] \\ & =\left[\begin{array}{cc}-8 & -6 \\ -8 & -36\end{array}\right] \\ & \therefore A B+B A \\ & =\left[\begin{array}{ll}-14 & -15 \\ -12 & -30\end{array}\right]+\left[\begin{array}{cc}-8 & -6 \\ -8 & -36\end{array}\right] \\ & =\left[\begin{array}{cc}-14-8 & 15-6 \\ -12-8 & -30-36\end{array}\right] \\ & =\left[\begin{array}{ll}-22 & -21 \\ -20 & -66\end{array}\right] \text {. } \\ & \end{aligned}$
View full question & answer→Question 23 Marks
Evaluate: $\left[\begin{array}{rr}4 \sin 30^{\circ} & 2 \cos 60^{\circ} \\ \sin 90^{\circ} & 2 \cos 0^{\circ}\end{array}\right]\left[\begin{array}{ll}4 & 5 \\ 5 & 4\end{array}\right]$
Answer$\begin{aligned} & {\left[\begin{array}{cc}4 \sin 30^{\circ} & 2 \cos 60^{\circ} \\ \sin 90^{\circ} & 2 \cos 0^{\circ}\end{array}\right]\left[\begin{array}{ll}4 & 5 \\ 5 & 4\end{array}\right]} \\ & \sin 30^{\circ}=\frac{1}{2}, \cos 60^{\circ}=\frac{1}{2} \\ & \sin 90^{\circ}=1 \text { and } \cos 0^{\circ}=1 \\ & \therefore\left[\begin{array}{cc}4 \times \frac{1}{2} & 2 \times \frac{1}{2} \\ 1 & 2 \times 1\end{array}\right]\left[\begin{array}{ll}4 & 5 \\ 5 & 4\end{array}\right] \\ & =\left[\begin{array}{ll}2 & 1 \\ 1 & 2\end{array}\right]\left[\begin{array}{cc}4 & 5 \\ 5 & 4\end{array}\right] \\ & =\left[\begin{array}{ll}2 \times 4+1 \times 5 & 2 \times 5+1 \times 4 \\ 1 \times 4+2 \times 5 & 1 \times 5+2 \times 4\end{array}\right] \\ & =\left[\begin{array}{cc}8+5 & 10+4 \\ 4+10 & 5+8\end{array}\right] \\ & =\left[\begin{array}{ll}13 & 14 \\ 14 & 13\end{array}\right] .\end{aligned}$
View full question & answer→Question 33 Marks
Given martices $A=\left[\begin{array}{ll}2 & 1 \\ 4 & 2\end{array}\right]$ and $B=\left[\begin{array}{cc}3 & 4 \\ -1 & -2\end{array}\right], C=\left[\begin{array}{cc}-3 & 1 \\ 0 & -2\end{array}\right]$ Find the products of (i) $A B C$ (ii) $A C B$ and state whether they are equal.
Answer$\begin{aligned} & A=\left[\begin{array}{ll}2 & 1 \\ 4 & 2\end{array}\right] \\ & B=\left[\begin{array}{cc}3 & 4 \\ -1 & -2\end{array}\right] \\ & C=\left[\begin{array}{cc}-3 & 1 \\ 0 & -2\end{array}\right] \\ & A B C=\left[\begin{array}{ll}2 & 1 \\ 4 & 2\end{array}\right] \times\left[\begin{array}{cc}3 & 4 \\ -1 & -2\end{array}\right] \times\left[\begin{array}{cc}-3 & 1 \\ 0 & -2\end{array}\right] \\ & =\left[\begin{array}{cc}6-1 & 8-2 \\ 12-2 & 16-4\end{array}\right]\left[\begin{array}{cc}-3 & 1 \\ 0 & -2\end{array}\right] \\ & =\left[\begin{array}{cc}5 & 6 \\ 10 & 12\end{array}\right] \times\left[\begin{array}{cc}-3 & 1 \\ 0 & -2\end{array}\right] \\ & =\left[\begin{array}{cc}-15+0 & 5-12 \\ -30+0 & 10-24\end{array}\right] \\ & =\left[\begin{array}{cc}-15 & -7 \\ -30 & -14\end{array}\right] \\ & A C B=\left[\begin{array}{cc}2 & 1 \\ 4 & 2\end{array}\right]\left[\begin{array}{cc}-3 & 1 \\ 0 & -2\end{array}\right] \times\left[\begin{array}{cc}3 & 4 \\ -1 & -2\end{array}\right] \\ & =\left[\begin{array}{cc}-6+0 & 2-2 \\ -12+10 & 4-4\end{array}\right] \times\left[\begin{array}{cc}3 \\ -1 & -2\end{array}\right] \\ & =\left[\begin{array}{cc}-6 & 0 \\ -12 & 0\end{array}\right] \times\left[\begin{array}{cc}3 & 4 \\ -1 & -2\end{array}\right] \\ & =\left[\begin{array}{cc}-18+0 & -24+0 \\ -36+0 & -48+0\end{array}\right] \\ & =\left[\begin{array}{ll}-18 & -24 \\ -36 & -48\end{array}\right] \\ & \therefore A B C \neq A C B\end{aligned}$
View full question & answer→Question 43 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 $AB -5 C$
Answer$\begin{aligned} & A =\left[\begin{array}{ll}3 & 7 \\ 2 & 4\end{array}\right], B =\left[\begin{array}{ll}0 & 2 \\ 5 & 3\end{array}\right] \text { and } C =\left[\begin{array}{cc}1 & -5 \\ -4 & 6\end{array}\right] \\ & AB =\left[\begin{array}{ll}3 & 7 \\ 2 & 4\end{array}\right]\left[\begin{array}{ll}0 & 2 \\ 5 & 3\end{array}\right] \\ & =\left[\begin{array}{ll}3 \times 0+7 \times 5 & 3 \times 2+7 \times 3 \\ 2 \times 0+4 \times 5 & 2 \times 2+4 \times 3\end{array}\right] \\ & =\left[\begin{array}{ll}0+35 & 6+21 \\ 0+20 & 4+12\end{array}\right] \\ & =\left[\begin{array}{ll}35 & 27 \\ 20 & 16\end{array}\right] \\ & 5 C =5\left[\begin{array}{cc}1 & -5 \\ -4 & 6\end{array}\right] \\ & =\left[\begin{array}{cc}5 & -25 \\ -20 & 30\end{array}\right] \\ & AB -5 C =\left[\begin{array}{ll}35 & 27 \\ 20 & 16\end{array}\right]-\left[\begin{array}{cc}5 & -25 \\ -20 & 30\end{array}\right] \\ & =\left[\begin{array}{cc}30 & 52 \\ 40 & -14\end{array}\right] .\end{aligned}$
View full question & answer→Question 53 Marks
If $A=\left[\begin{array}{cc}2 & -1 \\ -4 & 5\end{array}\right]$ and $B=\left[\begin{array}{ll}0 & -3\end{array}\right]$ find the matrix $C$ such that $C A=B$
Answer$
A=\left[\begin{array}{cc}
2 & -1 \\
-4 & 5
\end{array}\right] \text { and } B=\left[\begin{array}{cc}
0 & -3
\end{array}\right]
$
Let matrix $C=\left[\begin{array}{ll}x & y\end{array}\right]$
Since the matrix $A$ is $2 \times 2$ and $B=7 \times 2$
$
\begin{aligned}
& \because C A=B \\
& \therefore(x y)\left[\begin{array}{cc}
2 & -1 \\
-4 & 5
\end{array}\right]=[0-3] \\
& =(2 x-4 y-x+5 y) \\
& =[0-3]
\end{aligned}
$
Comparing,
$
\begin{aligned}
& 2 x-4 y=0 \\
& \Rightarrow x-2 y=0 \\
& \therefore x=2 y
\end{aligned}
$
and
$
\begin{aligned}
& -x+5 y=-3 \\
& \Rightarrow-2 y+5 y=-3 \\
& \Rightarrow 3 y=-3 \\
& \Rightarrow y=-1 \\
& \therefore x=2 y \\
& =2 x(-1) \\
& =-2
\end{aligned}
$
Hence C
$
\begin{aligned}
& =\left[\begin{array}{ll}
x & y
\end{array}\right] \\
& =\left[\begin{array}{ll}
-2 & -1
\end{array}\right] .
\end{aligned}
$
View full question & answer→Question 63 Marks
Given $\left[\begin{array}{cc}2 & 1 \\ -3 & 4\end{array}\right], X=\left[\begin{array}{l}7 \\ 6\end{array}\right]$ the matrix $X$
AnswerWe have
$ \left[\begin{array}{cc} 2 & 1 \\ -3 & 4 \end{array}\right], X =\left[\begin{array}{l} 7 \\ 6 \end{array}\right]$ Let $X =\left[\begin{array}{l}x \\ y\end{array}\right]$
So, $\left[\begin{array}{cc}2 & 1 \\ -3 & 4\end{array}\right]\left[\begin{array}{l}x \\ y\end{array}\right]=\left[\begin{array}{l}7 \\ 6\end{array}\right]$
$ \Rightarrow\left[\begin{array}{c} 2 x+y \\ -3 x+4 y \end{array}\right]=\left[\begin{array}{l} 7 \\ 6 \end{array}\right] $
Multiplying $(i)$ by $3$ and $(ii)$ by $2,$ and adding we get:
$6 x+33 y=21$
$ -6 x+8 y=12$
$ 11 y=33$
$ \Rightarrow y =3$
From $(i),$
$2 x=7-3=4$
$ \Rightarrow x=2$
So, $X=\left[\begin{array}{l}2 \\ 2\end{array}\right]$.
View full question & answer→Question 73 Marks
If $P =\left[\begin{array}{ll}2 & 6 \\ 3 & 9\end{array}\right], Q =\left[\begin{array}{ll}3 & x \\ y & 2\end{array}\right]$ find $x$ and $y$ such that $PQ =0$
AnswerGiven
$
\begin{aligned}
& P=\left[\begin{array}{ll}
2 & 6 \\
3 & 9
\end{array}\right] \\
& Q=\left[\begin{array}{ll}
3 & x \\
y & 2
\end{array}\right] \\
& P Q=\left[\begin{array}{ll}
2 & 6 \\
3 & 9
\end{array}\right]\left[\begin{array}{ll}
3 & x \\
y & 2
\end{array}\right] \\
& =\left[\begin{array}{ll}
6+6 y & 2 x+12 \\
9+9 y & 3 x+18
\end{array}\right] \\
& \because P Q=0 \\
& \therefore\left[\begin{array}{ll}
6+6 y & 2 x+12 \\
9+9 y & 3 x+8
\end{array}\right]=\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]
\end{aligned}
$
Comparing the corresponding elements
$
\begin{aligned}
& 6+6 y=0 \\
& \Rightarrow 6 y=-6 \\
& \Rightarrow y=-1 \\
& 2 x+12=0 \\
& \Rightarrow 2 x=-12 \\
& \Rightarrow x=-6
\end{aligned}
$
Hence $x=-6, y=-1$.
View full question & answer→Question 83 Marks
If $A=\left[\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right]$ find $x$ and $y$ so that $A^2-x A+y \mid$
AnswerGiven
$
\begin{aligned}
& A^2=\left[\begin{array}{ll}
2 & 3 \\
1 & 2
\end{array}\right]\left[\begin{array}{ll}
2 & 3 \\
1 & 2
\end{array}\right] \\
& =\left[\begin{array}{ll}
4+3 & 6+6 \\
2+2 & 3+4
\end{array}\right] \\
& =\left[\begin{array}{ll}
7 & 12 \\
4 & 7
\end{array}\right] \\
& \because A ^2= xA + yl \\
& \Rightarrow\left[\begin{array}{cc}
7 & 12 \\
4 & 7
\end{array}\right]=x\left[\begin{array}{ll}
2 & 3 \\
1 & 2
\end{array}\right]+y\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right] \\
& \Rightarrow\left[\begin{array}{cc}
7 & 12 \\
4 & 7
\end{array}\right]=\left[\begin{array}{cc}
2 x & 3 x \\
x & 2 x
\end{array}\right]+\left[\begin{array}{ll}
y & 0 \\
0 & y
\end{array}\right] \\
& =\left[\begin{array}{cc}
2 x+y & 3 x \\
x & 2 x+y
\end{array}\right]
\end{aligned}
$
Comparing the corresponding elements
$
\begin{aligned}
& 3 x=12 \\
& \Rightarrow x=4 \\
& 2 x+y=7 \\
& \Rightarrow 2 x 4+y=7 \\
& \Rightarrow 8+y=7 \\
& \Rightarrow y=7-8=-1
\end{aligned}
$
Hence $x=4, y=-1$.
View full question & answer→Question 93 Marks
If $A=\left[\begin{array}{cc}1 & 4 \\ 0 & -1\end{array}\right], B=\left[\begin{array}{cc}2 & x \\ 0 & -\frac{1}{2}\end{array}\right]$ find the value of $x$ if $A B=B A$
AnswerGiven
$
\begin{aligned}
& A =\left[\begin{array}{cc}
1 & 4 \\
0 & -1
\end{array}\right]\left[\begin{array}{cc}
2 & x \\
0 & -\frac{1}{2}
\end{array}\right] \\
& =\left[\begin{array}{cc}
2+0 & x-2 \\
0+0 & 0+\frac{1}{2}
\end{array}\right] \\
& =\left[\begin{array}{cc}
2 & x-2 \\
0 & \frac{1}{2}
\end{array}\right] \\
& BA =\left[\begin{array}{cc}
2^2 & x \\
0 & -\frac{1}{2}
\end{array}\right]\left[\begin{array}{cc}
1 & 4 \\
0 & -1
\end{array}\right] \\
& =\left[\begin{array}{cc}
2+0 & 8-x \\
0+0 & 0+\frac{1}{2}
\end{array}\right] \\
& =\left[\begin{array}{cc}
2 & 8-x \\
0 & \frac{1}{2}
\end{array}\right] \\
& \because AB = BA \\
& \therefore\left[\begin{array}{ll}
2 & x-2 \\
0 & \frac{1}{2}
\end{array}\right]=\left[\begin{array}{cc}
2 & 8-x \\
0 & \frac{1}{2}
\end{array}\right]
\end{aligned}
$
Comparing the corresponding elements
$
\begin{aligned}
& x-2=8-x \\
& \Rightarrow x+x=8+2 \\
& \Rightarrow 2 x=10 \\
& \therefore x=\frac{10}{2}=5 .
\end{aligned}
$
View full question & answer→Question 103 Marks
Find the value of $x$ given that $A^2=B$ Where $A=\left[\begin{array}{ll}2 & 12 \\ 0 & 1\end{array}\right]$ and $B=\left[\begin{array}{ll}4 & x \\ 0 & 1\end{array}\right]$
Answer$
\begin{aligned}
& A=\left[\begin{array}{cc}
2 & 12 \\
0 & 1
\end{array}\right] \text { and } \\
& B=\left[\begin{array}{ll}
4 & x \\
0 & 1
\end{array}\right] \\
& A^2=B \\
& \Rightarrow A \times A=B \\
& \Rightarrow\left[\begin{array}{cc}
2 & 12 \\
0 & 1
\end{array}\right]\left[\begin{array}{cc}
2 & 12 \\
0 & 1
\end{array}\right]=\left[\begin{array}{ll}
4 & x \\
0 & 1
\end{array}\right] \\
& \Rightarrow\left[\begin{array}{cc}
2 \times 2+12 \times 0 & 2 \times 12+12 \times 1 \\
0 \times 2+1 \times 0 & 0 \times 12+1 \times 1
\end{array}\right]=\left[\begin{array}{ll}
4 & x \\
0 & 1
\end{array}\right] \\
& \Rightarrow\left[\begin{array}{cc}
4+0 & 24+12 \\
0+0 & 0+1
\end{array}\right]=\left[\begin{array}{cc}
4 & x \\
0 & 1
\end{array}\right] \\
& \Rightarrow\left[\begin{array}{cc}
4 & 36 \\
0 & 1
\end{array}\right]=\left[\begin{array}{ll}
4 & x \\
0 & 1
\end{array}\right]
\end{aligned}
$
Comparing the corresponding elements of two equal matrices, $x=36$.
View full question & answer→Question 113 Marks
Find the matrix $X$ of order $2 \times 2$ 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$\begin{aligned} & \text { Given } \\ & {\left[\begin{array}{ll}3 & 7 \\ 2 & 4\end{array}\right]\left[\begin{array}{ll}0 & 2 \\ 5 & 3\end{array}\right]+2 X =\left[\begin{array}{cc}1 & -5 \\ -4 & 6\end{array}\right]} \\ & \Rightarrow\left[\begin{array}{ll}0+35 & 6+21 \\ 0+20 & 4+12\end{array}\right]+2 X =\left[\begin{array}{cc}1 & -5 \\ -4 & 6\end{array}\right] \\ & \Rightarrow\left[\begin{array}{ll}35 & 27 \\ 20 & 16\end{array}\right]+2 X =\left[\begin{array}{cc}1 & -5 \\ -4 & 6\end{array}\right] \\ & 2 X =-\left[\begin{array}{ll}35 & 27 \\ 20 & 16\end{array}\right]+\left[\begin{array}{cc}1 & -5 \\ -4 & 6\end{array}\right] \\ & =\left[\begin{array}{ll}-34 & -32 \\ -24 & -10\end{array}\right] \\ & X =\frac{1}{2}\left[\begin{array}{ll}-34 & -32 \\ -24 & -10\end{array}\right] \\ & =\left[\begin{array}{cc}-17 & -16 \\ -12 & -5\end{array}\right] .\end{aligned}$
View full question & answer→Question 123 Marks
Show that $\left[\begin{array}{ll}1 & 2 \\ 2 & 1\end{array}\right]$ is a solution of the matrix equation $X^2-2 X-3 I=0$, Where $I$ is the unit matrix of order 2
AnswerGiven
$
x^2-2 x-31=0
$
Solution $=\left[\begin{array}{ll}1 & 2 \\ 2 & 1\end{array}\right]$
or
$
\begin{aligned}
& X=\left[\begin{array}{ll}
1 & 2 \\
2 & 1
\end{array}\right] \\
& \therefore X^2=\left[\begin{array}{ll}
1 & 2 \\
2 & 1
\end{array}\right]\left[\begin{array}{ll}
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 & 4 \\
4 & 5
\end{array}\right]
\end{aligned}
$
Now $X^2-2 X-31$
$
\begin{aligned}
& =\left[\begin{array}{ll}
5 & 4 \\
4 & 5
\end{array}\right]-2\left[\begin{array}{ll}
1 & 2 \\
2 & 1
\end{array}\right]-3\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right] \\
& =\left[\begin{array}{ll}
5 & 4 \\
4 & 5
\end{array}\right]-\left[\begin{array}{ll}
2 & 4 \\
4 & 2
\end{array}\right]-\left[\begin{array}{ll}
3 & 0 \\
0 & 3
\end{array}\right] \\
& =\left[\begin{array}{ll}
5-2-3 & 4-4+0 \\
4-4-0 & 5-2-3
\end{array}\right] \\
& =\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right] \\
& \therefore X ^2=2 X -31=0 \\
&
\end{aligned}
$
Hence proved.
View full question & answer→Question 133 Marks
If $X=\left[\begin{array}{cc}4 & 1 \\ -1 & 2\end{array}\right]$, show that $6 X-X^2=91$ Where $I$ is the unit matrix.
Answer$
\begin{aligned}
& X=\left[\begin{array}{cc}
4 & 1 \\
-1 & 2
\end{array}\right] \\
& x^2=x \times x=\left[\begin{array}{cc}
4 & 1 \\
-1 & 2
\end{array}\right]\left[\begin{array}{cc}
4 & 1 \\
-1 & 2
\end{array}\right] \\
& =\left[\begin{array}{cc}
16-1 & 4+2 \\
-4-2 & -1+4
\end{array}\right] \\
& =\left[\begin{array}{ll}
15 & 6 \\
-6 & 3
\end{array}\right] \\
& \text { L.H.S. } 6 \text { X- } x^2=6\left[\begin{array}{cc}
4 & 1 \\
-1 & 2
\end{array}\right]-\left[\begin{array}{cc}
15 & 6 \\
-6 & 3
\end{array}\right] \\
& =\left[\begin{array}{cc}
24 & 6 \\
-6 & 12
\end{array}\right]-\left[\begin{array}{cc}
15 & 6 \\
-6 & 3
\end{array}\right] \\
& =\left[\begin{array}{cc}
24-15 & 6-6 \\
-6-6 & 12-3
\end{array}\right] \\
& =\left[\begin{array}{ll}
9 & 0 \\
0 & 9
\end{array}\right] \\
& =9\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right] \\
& =91 \\
& =\text { R.H.S. }
\end{aligned}
$
Hence proved.
View full question & answer→Question 143 Marks
If $A=\left[\begin{array}{cc}2 & 1 \\ 0 & -2\end{array}\right]$ and $B=\left[\begin{array}{cc}4 & 1 \\ -3 & -2\end{array}\right], C=\left[\begin{array}{ll}-3 & 2 \\ -1 & 4\end{array}\right]$ Find $A^2+A C-5 B$
Answer$
\begin{aligned}
& A=\left[\begin{array}{cc}
2 & 1 \\
0 & -2
\end{array}\right] \\
& B=\left[\begin{array}{cc}
4 & 1 \\
-3 & -2
\end{array}\right] \\
& C=\left[\begin{array}{ll}
-3 & 2 \\
-1 & 4
\end{array}\right] \\
& A^2+A C-5 B \\
& =\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}{cc}
2 & 1 \\
0 & -2
\end{array}\right]\left[\begin{array}{ll}
-3 & 2 \\
-1 & 4
\end{array}\right]-5\left[\begin{array}{cc}
4 & 1 \\
-3 & -2
\end{array}\right]
\end{aligned}
$
(Substituting the values from questions)
$
\begin{aligned}
& =\left[\begin{array}{ll}
4+0 & 2-2 \\
0+0 & 0+4
\end{array}\right]+\left[\begin{array}{cc}
-6-1 & 0+4 \\
0+2 & -8
\end{array}\right]-5\left[\begin{array}{cc}
4 & 1 \\
-3 & -2
\end{array}\right] \\
& =\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] .
\end{aligned}
$
View full question & answer→Question 153 Marks
If $A=\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]$ and $B=\left[\begin{array}{ll}2 & 1 \\ 4 & 2\end{array}\right], C=\left[\begin{array}{ll}5 & 1 \\ 7 & 4\end{array}\right]$, compute $A(B+C)$
Answer$\begin{aligned} & A(B+C) \\ & A=\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right] \\ & B=\left[\begin{array}{ll}2 & 1 \\ 4 & 2\end{array}\right] \\ & C=\left[\begin{array}{ll}5 & 1 \\ 7 & 4\end{array}\right] \\ & A(B+C)=\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]\left[\begin{array}{ll}2 & 1 \\ 4 & 2\end{array}\right]+\left[\begin{array}{ll}5 & 1 \\ 7 & 4\end{array}\right] \\ & =\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]\left[\begin{array}{cc}2+5 & 1+1 \\ 4+7 & 2+4\end{array}\right] \\ & =\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]\left[\begin{array}{cc}7 & 2 \\ 11 & 6\end{array}\right] \\ & =\left[\begin{array}{cc}7+22 & 2+12 \\ 21+44 & 6+24\end{array}\right] \\ & =\left[\begin{array}{ll}29 & 14 \\ 65 & 30\end{array}\right] .\end{aligned}$
View full question & answer→Question 163 Marks
If $A=\left[\begin{array}{cc}1 & -2 \\ 2 & -1\end{array}\right]$ and $B=\left[\begin{array}{cc}3 & 2 \\ -2 & 1\end{array}\right]$ Find $2 B-A^2$
Answer$\begin{aligned} & A=\left[\begin{array}{cr}1 & -2 \\ 2 & -1\end{array}\right] \\ & B=\left[\begin{array}{cr}3 & 2 \\ -2 & 1\end{array}\right] \\ & 2 B=2\left[\begin{array}{cr}3 & 2 \\ -2 & 1\end{array}\right] \\ & =\left[\begin{array}{cr}6 & 4 \\ -4 & 2\end{array}\right] \\ & A^2= A \times A =\left[\begin{array}{ll}1 & -2 \\ 2 & -1\end{array}\right]\left[\begin{array}{cc}1 & -2 \\ 2 & -1\end{array}\right] \\ & =\left[\begin{array}{cc}1-4 & -2+2 \\ 2-2 & -4+1\end{array}\right] \\ & =\left[\begin{array}{cc}-3 & 0 \\ 0 & -3\end{array}\right] \\ & \therefore 2 B - A ^2=\left[\begin{array}{cc}6 & 4 \\ -4 & 2\end{array}\right]-\left[\begin{array}{cc}3- & 0 \\ 0 & -3\end{array}\right] \\ & =\left[\begin{array}{cc}6-(-3) & 4-0 \\ -4-0 & 2-(-3)\end{array}\right] \\ & =\left[\begin{array}{cc}6+3 & 4 \\ -4 & 2+3\end{array}\right] \\ & =\left[\begin{array}{cc}9 & 4 \\ -4 & 5\end{array}\right] .\end{aligned}$
View full question & answer→Question 173 Marks
If $A=\left[\begin{array}{cc}2 & a \\ -3 & 5\end{array}\right]$ and $B=\left[\begin{array}{cc}-2 & 3 \\ 7 & b\end{array}\right], C=\left[\begin{array}{cc}c & 9 \\ -1 & -11\end{array}\right]$ and $5 A+2 B=C$, find the values of $a, b, c$
Answer$
\begin{aligned}
& A =\left[\begin{array}{cc}
2 & a \\
-3 & 5
\end{array}\right] \text { and } B =\left[\begin{array}{cc}
-2 & 3 \\
7 & b
\end{array}\right], C =\left[\begin{array}{cc}
c & 9 \\
-1 & -11
\end{array}\right] \text { and } 5 A +2 B = C \\
& \Rightarrow 5\left[\begin{array}{cc}
2 & a \\
-3 & 5
\end{array}\right]+2\left[\begin{array}{cc}
-2 & 3 \\
7 & b
\end{array}\right]=\left[\begin{array}{cc}
c & 9 \\
-1 & -11
\end{array}\right] \\
& \Rightarrow\left[\begin{array}{cc}
10 & 5 a \\
-15 & 25
\end{array}\right]+\left[\begin{array}{cc}
-4 & 6 \\
14 & 2 b
\end{array}\right]=\left[\begin{array}{cc}
c & 9 \\
-1 & -11
\end{array}\right] \\
& \Rightarrow\left[\begin{array}{cc}
10-4 & 5 a+6 \\
15+14 & 25+2 b
\end{array}\right]=\left[\begin{array}{cc}
c & 9 \\
-1 & -11
\end{array}\right] \\
& \Rightarrow\left[\begin{array}{cc}
6 & 5 a+6 \\
-1 & 25+2 b
\end{array}\right]=\left[\begin{array}{cc}
c & 9 \\
-1 & -11
\end{array}\right]
\end{aligned}
$
Comparing each term
$
\begin{aligned}
& 5 a+6=9 \\
& \Rightarrow 5 a=9-6=3 \\
& \Rightarrow a=\frac{3}{5} \\
& \Rightarrow 25+2 b=-11 \\
& \Rightarrow 2 b=-11-25=-36 \\
& \Rightarrow b=-\frac{36}{2}=-18 \\
& c=6
\end{aligned}
$
Hence $a=\frac{3}{5}, b=-18$ and $c=6$.
View full question & answer→Question 183 Marks
If $\left[\begin{array}{ll}a & 3 \\ 4 & 2\end{array}\right]+\left[\begin{array}{cc}2 & b \\ 1 & -2\end{array}\right]-\left[\begin{array}{cc}1 & 1 \\ -2 & c\end{array}\right]=\left[\begin{array}{ll}5 & 0 \\ 7 & 3\end{array}\right]$ Find the value of $a, b$ and $c$
Answer$
\begin{aligned}
& {\left[\begin{array}{ll}
a & 3 \\
4 & 2
\end{array}\right]+\left[\begin{array}{cc}
2 & b \\
1 & -2
\end{array}\right]-\left[\begin{array}{cc}
1 & 1 \\
-2 & c
\end{array}\right]=\left[\begin{array}{ll}
5 & 0 \\
7 & 3
\end{array}\right]} \\
& \Rightarrow\left[\begin{array}{cc}
a+2-1 & 3+b-1 \\
4+1+2 & 2-2-c
\end{array}\right]=\left[\begin{array}{ll}
5 & 0 \\
7 & 3
\end{array}\right] \\
& \Rightarrow\left[\begin{array}{cc}
a+1 & b+2 \\
7 & -c
\end{array}\right]=\left[\begin{array}{ll}
5 & 0 \\
7 & 3
\end{array}\right]
\end{aligned}
$
Comparing the corresponding elements:
$
\begin{aligned}
& a+1=5 \\
& \Rightarrow a-4 \\
& b+2=0 \\
& \Rightarrow b=-2 \\
& -c=3 \\
& \Rightarrow c=-3 .
\end{aligned}
$
View full question & answer→Question 193 Marks
If $\left[\begin{array}{cc}5 & 2 \\ -1 & y+1\end{array}\right]-2\left[\begin{array}{cc}1 & 2 x-1 \\ 3 & -2\end{array}\right]=\left[\begin{array}{cc}3 & -8 \\ -7 & 2\end{array}\right]$ Find the values of $x$ and $y$
Answer$
\begin{aligned}
& {\left[\begin{array}{cc}
5 & 2 \\
-1 & y+1
\end{array}\right]-2\left[\begin{array}{cc}
1 & 2 x-1 \\
3 & -2
\end{array}\right]=\left[\begin{array}{cc}
3 & -8 \\
-7 & 2
\end{array}\right]} \\
& \Rightarrow\left[\begin{array}{cc}
5 & 2 \\
-1 & y+1
\end{array}\right]-\left[\begin{array}{cc}
2 & 4 x-2 \\
6 & -4
\end{array}\right]=\left[\begin{array}{cc}
3 & -8 \\
-7 & 2
\end{array}\right] \\
& \Rightarrow\left[\begin{array}{cc}
5-2 & 2-4 x+2 \\
-1-6 & y+1+4
\end{array}\right]=\left[\begin{array}{cc}
3 & -8 \\
-7 & 2
\end{array}\right] \\
& \Rightarrow\left[\begin{array}{cc}
3 & 4-4 x \\
-7 & y+5
\end{array}\right]=\left[\begin{array}{cc}
3 & -8 \\
-7 & 2
\end{array}\right]
\end{aligned}
$
Comparing the corresponding terms, we get
$
\begin{aligned}
& 4-4 x =-8 \\
& \Rightarrow-4 x =-8-4 \\
& \Rightarrow-4 x =-12 \\
& \Rightarrow x =\frac{-12}{-4}=3
\end{aligned}
$
and
$
\begin{aligned}
& y+5=2 \\
& \Rightarrow y=2-5=-3 \\
& \therefore x=3, y=-3 .
\end{aligned}
$
View full question & answer→Question 203 Marks
If $2\left[\begin{array}{ll}3 & 4 \\ 5 & x\end{array}\right]+\left[\begin{array}{ll}1 & y \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}z & 0 \\ 10 & 5\end{array}\right]$ Find the values of $x$ and $y$
Answer$
\begin{aligned}
& 2\left[\begin{array}{ll}
3 & 4 \\
5 & x
\end{array}\right]+\left[\begin{array}{ll}
1 & y \\
0 & 1
\end{array}\right]=\left[\begin{array}{cc}
z & 0 \\
10 & 5
\end{array}\right] \\
& {\left[\begin{array}{cc}
6 & 8 \\
10 & 2 x
\end{array}\right]+\left[\begin{array}{ll}
1 & y \\
0 & 1
\end{array}\right]=\left[\begin{array}{cc}
z & 0 \\
10 & 5
\end{array}\right]} \\
& \Rightarrow\left[\begin{array}{cc}
6+1 & 8+y \\
10+0 & 2 x+1
\end{array}\right]=\left[\begin{array}{cc}
z & 0 \\
10 & 5
\end{array}\right] \\
& \Rightarrow\left[\begin{array}{cc}
7 & 8+y \\
10 & 2 x+1
\end{array}\right]=\left[\begin{array}{cc}
z & 0 \\
10 & 5
\end{array}\right]
\end{aligned}
$
Comparing,
$
\begin{aligned}
& 2 x+1=5 \\
& \Rightarrow 2 x=5-1=4 \\
& \therefore x=\frac{4}{2}=2 \\
& 8+y=0 \\
& \Rightarrow y=-8 \\
& z=7
\end{aligned}
$
Hence $x=2, y=-8, z=7$.
View full question & answer→Question 213 Marks
If $2\left[\begin{array}{ll}3 & 4 \\ 5 & x\end{array}\right]+a\left[\begin{array}{ll}1 & y \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}7 & 0 \\ 10 & 5\end{array}\right]$ Find the values of $x$ and $y$
Answer$
\begin{aligned}
& 2\left[\begin{array}{ll}
3 & 4 \\
5 & x
\end{array}\right]+\left[\begin{array}{ll}
1 & y \\
0 & 1
\end{array}\right]=\left[\begin{array}{cc}
7 & 0 \\
10 & 5
\end{array}\right] \\
& {\left[\begin{array}{cc}
6 & 8 \\
10 & 2 x
\end{array}\right]+\left[\begin{array}{cc}
1 & y \\
0 & 1
\end{array}\right]=\left[\begin{array}{cc}
7 & 0 \\
10 & 5
\end{array}\right]} \\
& \Rightarrow\left[\begin{array}{cc}
6+1 & 8+y \\
10+0 & 2 x+1
\end{array}\right] \\
& =\left[\begin{array}{cc}
7 & 0 \\
10 & 5
\end{array}\right]
\end{aligned}
$
Comparing the corresponding elements,
$
\begin{aligned}
& 8+y=0 \\
& \text { then } y=-8 \\
& 2 x+1=5
\end{aligned}
$
then $2 x=5-1=4$
$\Rightarrow x =2$
Hence $x=2, y=-8$.
View full question & answer→Question 223 Marks
Find $X$ and $Y$ If $X+Y=\left[\begin{array}{ll}7 & 0 \\ 2 & 5\end{array}\right]$ and $X-Y=\left[\begin{array}{ll}3 & 0 \\ 0 & 3\end{array}\right]$
Answer$
\begin{aligned}
& X+Y=\left[\begin{array}{ll}
7 & 0 \\
2 & 5
\end{array}\right] \ldots \text {....(i) } \\
& X-Y=\left[\begin{array}{ll}
3 & 0 \\
0 & 3
\end{array}\right] \ldots \text { (ii) }
\end{aligned}
$
Adding (i) and (ii) we get,
$
\begin{aligned}
& 2 x=\left[\begin{array}{ll}
7 & 0 \\
2 & 5
\end{array}\right]+\left[\begin{array}{ll}
3 & 0 \\
0 & 3
\end{array}\right] \\
& =\left[\begin{array}{ll}
7+3 & 0+0 \\
2+0 & 5+3
\end{array}\right] \\
& =\left[\begin{array}{cc}
10 & 0 \\
2 & 8
\end{array}\right] \\
& \therefore x=\frac{1}{2}\left[\begin{array}{cc}
10 & 0 \\
2 & 8
\end{array}\right] \\
& =\left[\begin{array}{ll}
5 & 0 \\
1 & 4
\end{array}\right]
\end{aligned}
$
Subtracting (ii) from (i),
$
\begin{aligned}
& 2 y=\left[\begin{array}{ll}
7 & 0 \\
2 & 5
\end{array}\right]-\left[\begin{array}{ll}
3 & 0 \\
0 & 3
\end{array}\right] \\
& \Rightarrow 2 y=\left[\begin{array}{ll}
7-3 & 0-0 \\
2-0 & 5-3
\end{array}\right] \\
& =\left[\begin{array}{ll}
4 & 0 \\
2 & 2
\end{array}\right] \\
& \therefore y=\frac{1}{2}\left[\begin{array}{ll}
4 & 0 \\
2 & 2
\end{array}\right] \\
& =\left[\begin{array}{ll}
2 & 0 \\
1 & 1
\end{array}\right]
\end{aligned}
$
Hence $x=\left[\begin{array}{ll}5 & 0 \\ 1 & 4\end{array}\right], y=\left[\begin{array}{ll}2 & 0 \\ 1 & 1\end{array}\right]$.
View full question & answer→Question 233 Marks
Find the values of $x, y,$ a and $b,$ if $\left[\begin{array}{ccc}3 x+4 y & 2 & x-2 y \\ a+b & 2 a-b & -1\end{array}\right]=\left[\begin{array}{ccc}2 & 2 & 4 \\ 5 & 5 & 1\end{array}\right]$
AnswerComparing the corresponding terms, we get.
3x + 4y = 2 ……(i)
x – 2y = 4 …….(ii)
Multiplying (i) by 1 and (ii) by 2
3x + 4y = 2,
2x – 4y = 8
Adding we get,
5x = 10
⇒ x = 2
Substituting the value of x in (i)
3 x 2 + 4y = 2,
6 + 4y = 2,
4y = 2 – 6
= –4
y = –1
∴ x = 2, y = –1
a + b = 5 ...(iii)
2a – b = –5. ...(iv)
View full question & answer→Question 243 Marks
If $A =\left[\begin{array}{ll}3 & 3 \\ p & q\end{array}\right]$ and $A ^2=0$ find $p$ and $q$
AnswerGiven
$
\begin{aligned}
& A =\left[\begin{array}{ll}
3 & 3 \\
p & q
\end{array}\right] \\
& A ^2= A \times A =\left[\begin{array}{ll}
3 & 3 \\
p & q
\end{array}\right]\left[\begin{array}{ll}
3 & 3 \\
p & q
\end{array}\right] \\
& =\left[\begin{array}{cc}
9+3 p & 9+3 q \\
3 p+p q & 3 p+q^2
\end{array}\right] \\
& \text { But } A ^2=0 \\
& \therefore\left[\begin{array}{cc}
9+3 p & 9+3 q \\
3 p+p q & 3 p+q^2
\end{array}\right]=\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]
\end{aligned}
$
Comparing the corresponding elements
$
\begin{aligned}
& 9+3 p=0 \\
& \Rightarrow 3 p=-9 \\
& \Rightarrow p=-3 \\
& 9+3 q=0 \\
& \Rightarrow 3 q=-9 \\
& \Rightarrow q=-3
\end{aligned}
$
Hence $p=-3, q=-3$.
View full question & answer→Question 253 Marks
If $A=\left[\begin{array}{cc}3 & -5 \\ -4 & 2\end{array}\right]$ find $A^2-5 A-14$ I
Where $I$ is unit matrix of order $2 \times 2$
AnswerGiven
$
\begin{aligned}
& A=\left[\begin{array}{cc}
3 & -5 \\
-4 & 2
\end{array}\right] \\
& A^2=A \times A=\left[\begin{array}{cc}
3 & -5 \\
-4 & 2
\end{array}\right]\left[\begin{array}{cc}
3 & -5 \\
-4 & 2
\end{array}\right] \\
& =\left[\begin{array}{cc}
9+20 & -15-10 \\
-12-8 & 20+4
\end{array}\right] \\
& =\left[\begin{array}{cc}
29 & -25 \\
-20 & 24
\end{array}\right] \\
& 5 A=5\left[\begin{array}{cc}
3 & -5 \\
-4 & 2
\end{array}\right] \\
& =\left[\begin{array}{cc}
15 & -25 \\
-20 & 10
\end{array}\right] \\
& \therefore A^2-5 A-14 I=\left[\begin{array}{cc}
29 & -25 \\
-20 & 24
\end{array}\right] \\
& -\left[\begin{array}{cc}
15 & -25 \\
-20 & 10
\end{array}\right]-14\left[\begin{array}{cc}
1 & 0 \\
0 & 1
\end{array}\right] \\
& =\left[\begin{array}{cc}
29 & -25 \\
-20 & 24
\end{array}\right]-\left[\begin{array}{cc}
15 & -20 \\
-20 & 10
\end{array}\right]-\left[\begin{array}{cc}
14 & 0 \\
0 & 1
\end{array}\right] \\
& =\left[\begin{array}{cc}
29-15-14 & -25+25-0 \\
-20+20+0 & 24-10-14
\end{array}\right] \\
& =\left[\begin{array}{cc}
0 & 0 \\
0 & 0
\end{array}\right]
\end{aligned}
$
View full question & answer→Question 263 Marks
If $A=\left[\begin{array}{ll}3 & 2 \\ 0 & 5\end{array}\right]$ and $B=\left[\begin{array}{ll}1 & 0 \\ 1 & 2\end{array}\right]$ find the each of the following and state it they are equal: $A^2-B^2$
AnswerGiven
$
A=\left[\begin{array}{ll}
3 & 2 \\
0 & 5
\end{array}\right]
$
and
$
\begin{aligned}
& B=\left[\begin{array}{ll}
1 & 0 \\
1 & 2
\end{array}\right] \\
& A^2-B^2
\end{aligned}
$
$
=\left[\begin{array}{ll}
3 & 2 \\
0 & 5
\end{array}\right] \times\left[\begin{array}{ll}
3 & 2 \\
0 & 5
\end{array}\right]-\left[\begin{array}{ll}
1 & 0 \\
1 & 2
\end{array}\right] \times\left[\begin{array}{ll}
1 & 0 \\
1 & 2
\end{array}\right]
$
$=\left[\begin{array}{ll}9+0 & 6+10 \\ 0+0 & 0+25\end{array}\right]-\left[\begin{array}{ll}1+0 & 0+0 \\ 1+2 & 0+4\end{array}\right]$
$=\left[\begin{array}{ll}9 & 16 \\ 0 & 25\end{array}\right]-\left[\begin{array}{ll}1 & 0 \\ 3 & 4\end{array}\right]$
$=\left[\begin{array}{ll}9-1 & 16-0 \\ 0-3 & 25-4\end{array}\right]$
$=\left[\begin{array}{cc}8 & 16 \\ -3 & 21\end{array}\right]$
We see that $(A+B)(A-B) \neq A^2-B^2$.
View full question & answer→Question 273 Marks
If $A=\left[\begin{array}{ll}3 & 2 \\ 0 & 5\end{array}\right]$ and $B=\left[\begin{array}{ll}1 & 0 \\ 1 & 2\end{array}\right]$ find the each of the following and state it they are equal: $(A+B)(A-B)$
AnswerGiven
$
A=\left[\begin{array}{ll}
3 & 2 \\
0 & 5
\end{array}\right]
$
and
$
\begin{aligned}
& B=\left[\begin{array}{ll}
1 & 0 \\
1 & 2
\end{array}\right] \\
& (A+B)(A-B) \\
& =\left\{\left[\begin{array}{ll}
3 & 2 \\
0 & 5
\end{array}\right]+\left[\begin{array}{ll}
1 & 0 \\
1 & 2
\end{array}\right]\right\} \times\left\{\left[\begin{array}{ll}
3 & 2 \\
0 & 5
\end{array}\right]-\left[\begin{array}{ll}
1 & 0 \\
1 & 2
\end{array}\right]\right\} \\
& =\left[\begin{array}{ll}
3+1 & 2+0 \\
0+1 & 5+2
\end{array}\right] \times\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] \times\left[\begin{array}{cc}
2 & 2 \\
-1 & 3
\end{array}\right] \\
& =\left[\begin{array}{cc}
8-2 & 8+6 \\
2-7 & 2+21
\end{array}\right] \\
& =\left[\begin{array}{cc}
6 & 14 \\
-5 & 23
\end{array}\right] . \\
&
\end{aligned}
$
View full question & answer→Question 283 Marks
If $A =\left[\begin{array}{ll}1 & 4 \\ 1 & 0\end{array}\right], B =\left[\begin{array}{cc}2 & 1 \\ 3 & -1\end{array}\right]$ and $C =\left[\begin{array}{ll}2 & 3 \\ 0 & 5\end{array}\right]$ compute $( AB ) C$ $=(C B) A$ ?
AnswerGiven
$
\begin{aligned}
& A=\left[\begin{array}{ll}
1 & 4 \\
1 & 0
\end{array}\right] \\
& B=\left[\begin{array}{cc}
2 & 1 \\
3 & -1
\end{array}\right] \\
& C=\left[\begin{array}{ll}
2 & 3 \\
0 & 5
\end{array}\right] \\
& (A B) C=\left[\left[\begin{array}{ll}
1 & 4 \\
1 & 0
\end{array}\right] \times\left[\begin{array}{cc}
2 & 1 \\
3 & -1
\end{array}\right]\right]\left[\begin{array}{ll}
2 & 3 \\
0 & 5
\end{array}\right] \\
& =\left[\begin{array}{cc}
2+12 & 1-14 \\
2+0 & 1+0
\end{array}\right]\left[\begin{array}{ll}
2 & 3 \\
0 & 5
\end{array}\right] \\
& =\left[\begin{array}{cc}
14 & -13 \\
2 & 1
\end{array}\right]\left[\begin{array}{ll}
2 & 3 \\
0 & 5
\end{array}\right] \\
& =\left[\begin{array}{cc}
28+0 & 42-15 \\
4+0 & 6+5
\end{array}\right] \\
& =\left[\begin{array}{cc}
28 & 27 \\
4 & 11
\end{array}\right] \\
& \text { (CB) } A=\left[\left[\begin{array}{ll}
2 & 3 \\
0 & 5
\end{array}\right]\left[\begin{array}{cc}
2 & 1 \\
3 & -1
\end{array}\right]\right]\left[\begin{array}{ll}
1 & 4 \\
1 & 0
\end{array}\right] \\
& =\left[\begin{array}{cc}
4+9 & 2-3 \\
0+15 & 0-5
\end{array}\right]\left[\begin{array}{ll}
1 & 4 \\
1 & 0
\end{array}\right] \\
& =\left[\begin{array}{ll}
13 & 1 \\
15 & 5
\end{array}\right]\left[\begin{array}{ll}
1 & 4 \\
1 & 0
\end{array}\right] \\
& =\left[\begin{array}{cc}
13-1 & 52+0 \\
15-5 & 60+0
\end{array}\right] \\
& =\left[\begin{array}{ll}
12 & 52 \\
10 & 60
\end{array}\right] \\
&
\end{aligned}
$
It is clear from above that
$
(A B) C \neq (C B) A .
$
View full question & answer→Question 293 Marks
If $A=\left[\begin{array}{cc}1 & 2 \\ -3 & 4\end{array}\right], B=\left[\begin{array}{cc}0 & 1 \\ -2 & 5\end{array}\right]$ and $C=\left[\begin{array}{cc}-2 & 0 \\ -1 & 1\end{array}\right]$ find $A(4 B-$ (3C)
Answer$
\begin{aligned}
& A=\left[\begin{array}{cc}
1 & 2 \\
-3 & 4
\end{array}\right], B=\left[\begin{array}{cc}
0 & 1 \\
-2 & 5
\end{array}\right] \text { and } C=\left[\begin{array}{ll}
-2 & 0 \\
-1 & 1
\end{array}\right] \\
& 4 B-3 C=4\left[\begin{array}{cc}
0 & 1 \\
-2 & 5
\end{array}\right]-3\left[\begin{array}{ll}
-2 & 0 \\
-1 & 1
\end{array}\right] \\
& =\left[\begin{array}{cc}
0 & 4 \\
-8 & 20
\end{array}\right]-\left[\begin{array}{ll}
-6 & 0 \\
-3 & 3
\end{array}\right] \\
& =\left[\begin{array}{cc}
0-(-6) & 4-0 \\
-8-(-3) & 20-3
\end{array}\right] \\
& =\left[\begin{array}{cc}
0+6 & 4-0 \\
-8+3 & 20-3
\end{array}\right] \\
& =\left[\begin{array}{cc}
6 & 4 \\
-5 & 17
\end{array}\right]
\end{aligned}
$
Now $A(4 B-3 C)=\left[\begin{array}{cc}1 & 2 \\ -3 & 4\end{array}\right]\left[\begin{array}{cc}6 & 4 \\ -5 & 17\end{array}\right]$
$
\begin{aligned}
& =\left[\begin{array}{cc}
1 \times 6+2(-5) & 1 \times 4+2 \times 17 \\
-3 \times+6+4 x(-5) & -3 \times 4+4 \times 17
\end{array}\right] \\
& =\left[\begin{array}{cc}
6-10 & 4+34 \\
-18-20 & -12+68
\end{array}\right] \\
& =\left[\begin{array}{cr}
-4 & 38 \\
-38 & 56
\end{array}\right] .
\end{aligned}
$
View full question & answer→Question 303 Marks
Determine the matrices $A$ and $B$ when $A+2 B=$
$
\left[\begin{array}{cc}
1 & 2 \\
6 & -3
\end{array}\right] \text { and } 2 A - B =\left[\begin{array}{cc}
2 & -1 \\
2 & -1
\end{array}\right]
$
Answer$
\begin{aligned}
& A+2 B=\left[\begin{array}{cc}
1 & 2 \\
6 & -3
\end{array}\right] \ldots \ldots(\text { (i) } \\
& 2 A-B=\left[\begin{array}{ll}
2 & -1 \\
2 & -1
\end{array}\right] \ldots \ldots \text { (iii) }
\end{aligned}
$
Multiplying (i) by 1 and (ii) by 2
$
\begin{aligned}
& A+2 B=\left[\begin{array}{cc}
1 & 2 \\
6 & -3
\end{array}\right] \\
& 4 A-2 B=2\left[\begin{array}{ll}
2 & -1 \\
2 & -1
\end{array}\right]=\left[\begin{array}{ll}
4 & -2 \\
4 & -2
\end{array}\right]
\end{aligned}
$
Adding, we get
$
\begin{aligned}
& 5 A=\left[\begin{array}{cc}
1 & 2 \\
6 & -3
\end{array}\right]+\left[\begin{array}{cc}
4 & -2 \\
4 & -2
\end{array}\right]=\left[\begin{array}{cc}
5 & 0 \\
10 & -5
\end{array}\right] \\
& A=\frac{1}{5}\left[\begin{array}{cc}
5 & 0 \\
10 & 5
\end{array}\right]=\left[\begin{array}{cc}
1 & 0 \\
2 & -1
\end{array}\right]
\end{aligned}
$
From (i) $A+2 B=\left[\begin{array}{cc}1 & 2 \\ 6 & -3\end{array}\right]$
$
\begin{aligned}
& =\left[\begin{array}{cc}
1 & 0 \\
2 & -1
\end{array}\right]+2 B =\left[\begin{array}{cc}
1 & 2 \\
6 & -3
\end{array}\right] \\
& 2 B =\left[\begin{array}{cc}
1 & 2 \\
6 & -3
\end{array}\right]-\left[\begin{array}{cc}
1 & 0 \\
2 & -1
\end{array}\right]=\left[\begin{array}{cc}
0 & 2 \\
4 & -2
\end{array}\right] \\
& \therefore B =\frac{1}{2}\left[\begin{array}{cc}
0 & 2 \\
4 & -2
\end{array}\right]=\left[\begin{array}{cc}
0 & 1 \\
2 & -1
\end{array}\right]
\end{aligned}
$
Hence $A=\left[\begin{array}{cc}1 & 0 \\ 2 & -1\end{array}\right]$ and $B=\left[\begin{array}{cc}0 & 1 \\ 2 & -1\end{array}\right]$.
View full question & answer→Question 313 Marks
Find $a, b, c$ and $d$ if $3\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]=\left[\begin{array}{cc}4 & a+b \\ c+d & 3\end{array}\right]+\left[\begin{array}{cc}a & 6 \\ -1 & 2 d\end{array}\right]$
AnswerGiven
$
\begin{aligned}
& 3\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]=\left[\begin{array}{cc}
4 & a+b \\
c+d & 3
\end{array}\right]+\left[\begin{array}{cc}
a & 6 \\
-1 & 2 d
\end{array}\right] \\
& \Rightarrow\left[\begin{array}{ll}
3 a & 3 b \\
3 c & 3 d
\end{array}\right]=\left[\begin{array}{cc}
4+a & a+b+6 \\
c+d-1 & 3+2 d
\end{array}\right]
\end{aligned}
$
Comparing the corresponding elements:
$
\begin{aligned}
& 3 a=4+a \\
& \Rightarrow 3 a-a=4 \\
& \Rightarrow 2 a=4 \\
& \therefore a=2 \\
& 3 b=a+b+6 \\
& \Rightarrow 3 b-b=2+6 \\
& \Rightarrow 2 b=8 \\
& \therefore b=4 \\
& 3 d=3+2 d \\
& \Rightarrow 3 d-2 d=3 \\
& \therefore d=3 \\
& 3 c=c+d-1 \\
& \Rightarrow 3 c-c=3-1 \\
& 2 c=2 \\
& \Rightarrow c=1
\end{aligned}
$
Hence $a=2, b=4, c=1, d=3$.
View full question & answer→Question 323 Marks
$ \text { If } A=\left[\begin{array}{cc} \sec 60^{\circ} & \cos 90^{\circ} \\ -3 \tan 45^{\circ} & \sin 90^{\circ} \end{array}\right]$ and $B=\left[\begin{array}{cc} 0 & \cos 45^{\circ} \\ -2 & 3 \sin 90^{\circ} \end{array}\right] $ Find $: BA$
AnswerGiven
$A=\left[\begin{array}{cc} \sec 60^{\circ} & \cos 90^{\circ} \\ -3 \tan 45^{\circ} & \sin 90^{\circ} \end{array}\right]$
and
$ B=\left[\begin{array}{cc} 0 & \cos 45^{\circ} \\ -2 & 3 \sin 90^{\circ} \end{array}\right]$
$\begin{aligned} & A=\left[\begin{array}{cc}\sec 60^{\circ} & \cos 90^{\circ} \\ -3 \tan 45^{\circ} & \sin 90^{\circ}\end{array}\right]=\left[\begin{array}{ll}2 & 0 \\ 3 & 1\end{array}\right] \ldots\left(\because \sec 60^{\circ}=2, \cos 90^{\circ}=0, \tan 45^{\circ}=1, \sin 90^{\circ}=1\right)\end{aligned} $
$ B=\left[\begin{array}{cc}0 & \cos 45^{\circ} \\ -2 & 3 \sin 90^{\circ}\end{array}\right]=\left[\begin{array}{cc}0 & 1 \\ -2 & 3\end{array}\right] \ldots\left(\because \cot 45^{\circ}=1\right)$
$\begin{aligned} & B A=\left[\begin{array}{cc}0 & 1 \\ -2 & 3\end{array}\right]\left[\begin{array}{ll}2 & 0 \\ 3 & 1\end{array}\right] \end{aligned} $
$=\left[\begin{array}{cc}0-3 & 0+1 \\ -4-9 & 0+3\end{array}\right] $
$ =\left[\begin{array}{cc}-3 & 1 \\ -13 & 3\end{array}\right] .$
View full question & answer→Question 333 Marks
If $A =\left[\begin{array}{cc}\sec 60^{\circ} & \cos 90^{\circ} \\ -3 \tan 45^{\circ} & \sin 90^{\circ}\end{array}\right]$ and $B =\left[\begin{array}{cc}0 & \cos 45^{\circ} \\ -2 & 3 \sin 90^{\circ}\end{array}\right]$ Find $A^2$
Answer$\begin{aligned} & \text { Given } \\ & A=\left[\begin{array}{cc}\sec 60^{\circ} & \cos 90^{\circ} \\ -3 \tan 45^{\circ} & \sin 90^{\circ}\end{array}\right] \\ & \text { and } \\ & B=\left[\begin{array}{cc}0 & \cos 45^{\circ} \\ -2 & 3 \sin 90^{\circ}\end{array}\right] \\ & A=\left[\begin{array}{cc}\sec 60^{\circ} & \cos 90^{\circ} \\ -3 \tan 45^{\circ} & \sin 90^{\circ}\end{array}\right]=\left[\begin{array}{ll}2 & 0 \\ 3 & 1\end{array}\right] \ldots\left(\because \sec 60^{\circ}=2, \cos 90^{\circ}=0, \tan 45^{\circ}=1, \sin 90^{\circ}=1\right) \\ & B=\left[\begin{array}{cc}0 & \cos 45^{\circ} \\ -2 & 3 \sin 90^{\circ}\end{array}\right]=\left[\begin{array}{cc}0 & 1 \\ -2 & 3\end{array}\right] \ldots\left(\because \cot 45^{\circ}=1\right) \\ & A^2=A \times A=\left[\begin{array}{cc}2 & 0 \\ -3 & 1\end{array}\right]\left[\begin{array}{cc}2 & 0 \\ -3 & 1\end{array}\right] \\ & =\left[\begin{array}{cc}4+0 & 0+0 \\ -6-3 & 0+1\end{array}\right] \\ & =\left[\begin{array}{cc}4 & 0 \\ -9 & 1\end{array}\right] . \\ & \end{aligned}$
View full question & answer→Question 343 Marks
Find $a$ and $b$ if $\left[\begin{array}{ll}a-b & b-4 \\ b+4 & a-2\end{array}\right]\left[\begin{array}{ll}2 & 0 \\ 0 & 2\end{array}\right]=\left[\begin{array}{cc}2 & -2 \\ 14 & 0\end{array}\right]$
AnswerGiven
$
\begin{aligned}
& {\left[\begin{array}{ll}
a-b & b-4 \\
b+4 & a-2
\end{array}\right]\left[\begin{array}{ll}
2 & 0 \\
0 & 2
\end{array}\right]=\left[\begin{array}{cc}
2 & -2 \\
14 & 0
\end{array}\right]} \\
& \Rightarrow\left[\begin{array}{cc}
2 a-2 b+0 & 0+2 a-2 b \\
2 b+8+0 & 0+2 a-4
\end{array}\right]=\left[\begin{array}{cc}
-2 & -2 \\
14 & 0
\end{array}\right] \\
& \Rightarrow\left[\begin{array}{cc}
2 a-2 b & 2 a-2 b \\
2 b+8 & 2 a-4
\end{array}\right]=\left[\begin{array}{cc}
-2 & -2 \\
14 & 0
\end{array}\right]
\end{aligned}
$
Comparing the corresponding elements
$
\begin{aligned}
& 2 a-4=0 \\
& \Rightarrow 2 a=4 \\
& \Rightarrow a=2 \\
& 2 a-2 b=-2 \\
& \Rightarrow 2 \times 2-2 b=-2 \\
& \Rightarrow 4-2 b=-2 \\
& \Rightarrow-2 b=-2-4=-6 \\
& \Rightarrow b=3
\end{aligned}
$
Hence $a =2, b =3$.
View full question & answer→Question 353 Marks
If $A =\left[\begin{array}{cc}\frac{3}{5} & \frac{2}{5} \\ x & y\end{array}\right]$ and $A ^2= I$, find $x , y$
AnswerGiven
$A =\left(\begin{array}{ll}\frac{3}{5} & \frac{2}{5} \\ x & y\end{array}\right)$
$A ^2= A \times A =\left(\begin{array}{cc}\frac{3}{5} & \frac{2}{5} \\ x & y\end{array}\right)\left(\begin{array}{cc}\frac{3}{5} & \frac{2}{5} \\ x & y\end{array}\right)$
$
\left[\begin{array}{cc}
\frac{9}{25}+\frac{2}{5} x & \frac{6}{25} \frac{2}{5} y \\
\frac{3}{5} x+x y & \frac{2}{5} x+y^2
\end{array}\right]=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]
$
Comparing the corresponding elements,
$
\begin{aligned}
& \frac{9}{25}+\frac{2}{5} x=1 \\
& \Rightarrow \frac{2}{5} x=1-\frac{9}{25}=\frac{16}{25} \\
& x=\frac{16}{25} \times \frac{5}{2}=\frac{8}{5} \\
& \frac{6}{25}+\frac{2}{5} y=0 \\
& \Rightarrow \frac{2}{5} y=\frac{-6}{25} \\
& y=\frac{-6}{25} \times \frac{5}{2}=\frac{-3}{5}
\end{aligned}
$
Hence $x=\frac{8}{5}, y=\frac{-3}{5}$.
View full question & answer→Question 363 Marks
Find the values of $a$ and below
$
\left[\begin{array}{cc}
a+3 & b^2+2 \\
0 & -6
\end{array}\right]=\left[\begin{array}{cc}
2 a+1 & 3 b \\
0 & b^2-5 b
\end{array}\right]
$
Answer$
\left[\begin{array}{cc}
a+3 & b^2+2 \\
0 & -6
\end{array}\right]=\left[\begin{array}{cc}
2 a+1 & 3 b \\
0 & b^2-5 b
\end{array}\right]
$
comparing the corresponding elements
$
\begin{aligned}
& a+3=2 a+1 \\
& \Rightarrow 2 a-a=3-1 \\
& \Rightarrow a=2 \\
& b^2+2=3 b \\
& \Rightarrow b^2-3 b+2=0 \\
& \Rightarrow b^2-b-2 b+2=0 \\
& \Rightarrow b(b-1)-2(b-1)=0 \\
& \Rightarrow(b-1)(b-2)=0 .
\end{aligned}
$
Either $b-1=0$,
then $b=1$
or
$
b-2=0 \text {, }
$
then $b=2$
Hence $a=2, b=2$ or 1 .
View full question & answer→