Question 12 Marks
If $A=\left[\begin{array}{ll}1 & 2 \\ 2 & 1\end{array}\right]$ and $B=\left[\begin{array}{ll}2 & 1 \\ 1 & 2\end{array}\right]$, fin $A(B A)$
Answer$\begin{aligned} & A =\left[\begin{array}{ll}1 & 2 \\ 2 & 1\end{array}\right] \\ & B =\left[\begin{array}{ll}2 & 1 \\ 1 & 2\end{array}\right] \\ & BA =\left[\begin{array}{ll}2 & 1 \\ 1 & 2\end{array}\right] \times\left[\begin{array}{ll}1 & 2 \\ 2 & 1\end{array}\right] \\ & =\left[\begin{array}{cc}2+2 & 4+1 \\ 1+4 & 2+2\end{array}\right] \\ & =\left[\begin{array}{cc}4 & 5 \\ 5 & 4\end{array}\right] \\ & A ( BA )=\left[\begin{array}{ll}1 & 2 \\ 2 & 1\end{array}\right] \times\left[\begin{array}{ll}4 & 5 \\ 5 & 4\end{array}\right] \\ & =\left[\begin{array}{cc}4+10 & 5+8 \\ 8+5 & 10+4\end{array}\right] \\ & =\left[\begin{array}{cc}14 & 13 \\ 13 & 14\end{array}\right] .\end{aligned}$
View full question & answer→Question 22 Marks
Given $A=\left[\begin{array}{ll}1 & 1 \\ 8 & 3\end{array}\right]$, evaluate $A^2-4 A$
Answer$\begin{aligned} & A=\left[\begin{array}{ll}1 & 1 \\ 8 & 3\end{array}\right] \\ & A^2-4 A=\left[\begin{array}{ll}1 & 1 \\ 8 & 3\end{array}\right]\left[\begin{array}{ll}1 & 1 \\ 8 & 3\end{array}\right]-4\left[\begin{array}{ll}1 & 1 \\ 8 & 3\end{array}\right] \\ & =\left[\begin{array}{cc}1+8 & 1+3 \\ 8+24 & 8+9\end{array}\right]-\left[\begin{array}{cc}4 & 4 \\ 32 & 12\end{array}\right] \\ & =\left[\begin{array}{cc}9 & 4 \\ 32 & 17\end{array}\right]-\left[\begin{array}{cc}4 & 4 \\ 32 & 12\end{array}\right] \\ & =\left[\begin{array}{cc}9-4 & 4-4 \\ 32-32 & 17-12\end{array}\right] \\ & =\left[\begin{array}{cc}5 & 0 \\ 0 & 5\end{array}\right] .\end{aligned}$
View full question & answer→Question 32 Marks
Solve the matrix equation $: \left[\begin{array}{l}4 \\ 1\end{array}\right], X =\left[\begin{array}{ll}-4 & 8 \\ -1 & 2\end{array}\right]$
Answer$\left[\begin{array}{l}4 \\ 1\end{array}\right], X=\left[\begin{array}{ll}-4 & 8 \\ -1 & 2\end{array}\right]$
Let matrix $X=[x y]$
$\therefore\left[\frac{4}{1}\right][x, y]=\left[\begin{array}{ll}-4 & 8 \\ -1 & 2\end{array}\right]$
$\Rightarrow\left[\begin{array}{cc}4 x & 4 y \\ x & y\end{array}\right]=\left[\begin{array}{ll}-4 & 8 \\ -1 & 2\end{array}\right]$
Comparing the corresponding elements
$4 x=-4 \Rightarrow x=-1$
$ 4 y=8 \Rightarrow y=2$
$ \therefore x=[-1 2] .$
View full question & answer→Question 42 Marks
Let $M \times\left[\begin{array}{ll}1 & 1 \\ 0 & 2\end{array}\right]=\left[\begin{array}{ll}1 & 2\end{array}\right]$ where $M$ is a matrix.
(i) State the order of matrix M
(ii) Find the matrix $M$
AnswerGiven
(i) $M$ is the order of $1 \times 2$
$
\begin{aligned}
& \text { let } M=[ x y] \\
& \therefore\left[\begin{array}{ll}
x & y
\end{array}\right] \times\left[\begin{array}{ll}
1 & 1 \\
0 & 2
\end{array}\right]=\left[\begin{array}{ll}
1 & 2
\end{array}\right] \\
& \Rightarrow\left[\begin{array}{ll}
x+0 & x+2 y
\end{array}\right]=\left[\begin{array}{ll}
1 & 2
\end{array}\right]
\end{aligned}
$
Comparing the corresponding elements
$
\begin{aligned}
& x=1 \text { and } x+2 y=2 \\
& \Rightarrow 1+2 y=2 \\
& \Rightarrow 2 y=2-1=1 \\
& \Rightarrow y=\frac{1}{2}
\end{aligned}
$
Hence $x=1, y=\frac{1}{2}$
$
\therefore M=\left[\begin{array}{ll}
1 & \frac{1}{2}
\end{array}\right] .
$
View full question & answer→Question 52 Marks
If $\left[\begin{array}{ll}a & 1 \\ 1 & 0\end{array}\right]\left[\begin{array}{cc}4 & 3 \\ -3 & 2\end{array}\right]=\left[\begin{array}{cc}b & 11 \\ 4 & c\end{array}\right]$ find $a, b$ and $c$
Answer$
\begin{aligned}
& {\left[\begin{array}{ll}
a & 1 \\
1 & 0
\end{array}\right]\left[\begin{array}{cc}
4 & 3 \\
-3 & 2
\end{array}\right]=\left[\begin{array}{cc}
b & 11 \\
4 & c
\end{array}\right]} \\
& \Rightarrow\left[\begin{array}{cc}
4 a-3 & 3 a+2 \\
4+0 & 3+0
\end{array}\right]=\left[\begin{array}{cc}
b & 11 \\
4 & c
\end{array}\right] \\
& \Rightarrow\left[\begin{array}{cc}
4 a-3 & 3 a+2 \\
4 & 3
\end{array}\right]=\left[\begin{array}{cc}
b & 11 \\
4 & c
\end{array}\right]
\end{aligned}
$
Comparing the corresponding elements
$
\begin{aligned}
& 3 a+2=11 \\
& \Rightarrow 3 a=11-2=9 \\
& \therefore a=\frac{9}{3}=3 \\
& 4 a-3=b \\
& \Rightarrow b=4 \times 3-3 \\
& =12-3 \\
& =9 \\
& 3=c
\end{aligned}
$
Hence $a=3, b=9, c=3$.
View full question & answer→Question 62 Marks
If $P=\left[\begin{array}{cc}4 & 6 \\ 2 & -8\end{array}\right], Q=\left[\begin{array}{cc}2 & -3 \\ -1 & 1\end{array}\right]$ Find $2 P Q$
Answer$\begin{aligned} & P=\left[\begin{array}{cc}4 & 6 \\ 2 & -8\end{array}\right] \\ & Q=\left[\begin{array}{cc}2 & -3 \\ -1 & 1\end{array}\right] \\ & 2 P Q=2\left[\begin{array}{cc}4 & 6 \\ 2 & -8\end{array}\right] \times\left[\begin{array}{cc}2 & -3 \\ -1 & 1\end{array}\right] \\ & =2\left[\begin{array}{cc}8-6 & -12+6 \\ 4+8 & -6-8\end{array}\right] \\ & =2\left[\begin{array}{cc}8 & -6 \\ 12 & -14\end{array}\right] \\ & =\left[\begin{array}{cc}4 & -12 \\ 24 & -28\end{array}\right] .\end{aligned}$
View full question & answer→Question 72 Marks
Find $x,y$ if $\left[\begin{array}{cc}-2 & 0 \\ 3 & 1\end{array}\right]\left[\begin{array}{c}-1 \\ 2 x\end{array}\right]+3\left[\begin{array}{c}-2 \\ 1\end{array}\right]=2\left[\begin{array}{l}y \\ 3\end{array}\right].$
Answer$\left[\begin{array}{cc}-2 & 0 \\ 3 & 1\end{array}\right]\left[\begin{array}{c}-1 \\ 2 x\end{array}\right]+3\left[\begin{array}{c}-2 \\ 1\end{array}\right]=2\left[\begin{array}{l}y \\ 3\end{array}\right]$
$\left[\begin{array}{c}2+0 \\ -3+2 x\end{array}\right]+\left[\begin{array}{c}-6 \\ 3\end{array}\right]=\left[\begin{array}{c}2 y \\ 6\end{array}\right]$
$\left[\begin{array}{c}-4 \\ 2 x\end{array}\right]=\left[\begin{array}{c}2 y \\ 6\end{array}\right]$
$\Rightarrow 2 y =-4,2 x =6$
$\Rightarrow y =-2, x =3$
Thus required values is $x=3, y=-2.$
View full question & answer→Question 82 Marks
If $A=\left[\begin{array}{ll}3 & x \\ 0 & 1\end{array}\right]$ and $B=\left[\begin{array}{cc}9 & 16 \\ 0 & -y\end{array}\right]$ find $x$ and $y$ when $A^2=B$
AnswerGiven
$A=\left[\begin{array}{ll}3 & x \\ 0 & 1\end{array}\right]$ and $B=\left[\begin{array}{cc}9 & 16 \\ 0 & -y\end{array}\right]$ find $x$ and $y$ when $A^2=B$
Now, $A^2= A \times A$
$
\begin{aligned}
& =\left[\begin{array}{ll}
3 & x \\
0 & 1
\end{array}\right] \times\left[\begin{array}{ll}
3 & x \\
0 & 1
\end{array}\right] \\
& =\left[\begin{array}{cc}
9 & 3 x+x \\
0 & 1
\end{array}\right] \\
& =\left[\begin{array}{cc}
9 & 4 x \\
0 & 1
\end{array}\right]
\end{aligned}
$
We have $A^2=B$
Two matrices are equal if each and every corresponding element is equal.
Thus $\left[\begin{array}{cc}9 & 4 x \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}9 & 16 \\ 0 & -y\end{array}\right]$
$\Rightarrow 4 x =16$ and $1=- y$
$\Rightarrow x=4$ and $y=-1$
View full question & answer→Question 92 Marks
If $A=\left[\begin{array}{ll}2 & x \\ 0 & 1\end{array}\right]$ and $B=\left[\begin{array}{cc}4 & 36 \\ 0 & 1\end{array}\right]$, find the value of $x$, given that $A^2-B$
AnswerGiven
$
\begin{aligned}
& A ^2=\left[\begin{array}{ll}
2 & x \\
0 & 1
\end{array}\right]\left[\begin{array}{ll}
2 & x \\
0 & 1
\end{array}\right] \\
& A ^2=\left[\begin{array}{ll}
2 & x \\
0 & 1
\end{array}\right]\left[\begin{array}{ll}
2 & x \\
0 & 1
\end{array}\right] \\
& =\left[\begin{array}{cc}
4+0 & 2 x+x \\
0+0 & 0+1
\end{array}\right] \\
& =\left[\begin{array}{cc}
4 & 3 x \\
0 & 1
\end{array}\right] \\
& \because A ^2= B
\end{aligned}
$
$
\therefore\left[\begin{array}{cc}
4 & 3 x \\
0 & 1
\end{array}\right]=\left[\begin{array}{cc}
4 & 36 \\
0 & 1
\end{array}\right]
$
Corresponding the corresponding elements
$
\begin{aligned}
& 3 x =36 \\
& \Rightarrow x =12
\end{aligned}
$
Hence $x=12$.
View full question & answer→Question 102 Marks
If $\left[\begin{array}{ll}3 & 4 \\ 5 & 5\end{array}\right]=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$ write down the values of $a, b, c$ and $d$
AnswerGiven
$
\begin{aligned}
& {\left[\begin{array}{ll}
3 & 4 \\
5 & 5
\end{array}\right]=\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]} \\
& \Rightarrow\left[\begin{array}{ll}
3 & 4 \\
2 & 5
\end{array}\right]=\left[\begin{array}{ll}
a+0 & 0+b \\
c+0 & 0+d
\end{array}\right] \\
& \Rightarrow\left[\begin{array}{ll}
3 & 4 \\
2 & 5
\end{array}\right]=\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]
\end{aligned}
$
Comparing the corresponding elements
$
a=3, b=4, c=2, d=5 \text {. }
$
View full question & answer→Question 112 Marks
$\left[\begin{array}{ll}1 & 2 \\ 3 & 3\end{array}\right]\left[\begin{array}{ll}x & 0 \\ 0 & y\end{array}\right]=\left[\begin{array}{ll}x & 0 \\ 9 & 0\end{array}\right]$ find the values of $x$ and $y$
AnswerGiven
$
\begin{aligned}
& {\left[\begin{array}{ll}
1 & 2 \\
3 & 3
\end{array}\right]\left[\begin{array}{ll}
x & 0 \\
0 & y
\end{array}\right]=\left[\begin{array}{ll}
x & 0 \\
9 & 0
\end{array}\right]} \\
& \Rightarrow\left[\begin{array}{cc}
x+0 & 0 \\
3 x+2 y \\
3 x+0 & 0+3 y
\end{array}\right]=\left[\begin{array}{ll}
x & 0 \\
9 & 0
\end{array}\right] \\
& \Rightarrow\left[\begin{array}{cc}
x & 2 y \\
3 x & 3 y
\end{array}\right]=\left[\begin{array}{ll}
x & 0 \\
9 & 0
\end{array}\right]
\end{aligned}
$
Comparing the corresponding elements
$
\begin{aligned}
& 2 y=0 \\
& \Rightarrow y=0 \\
& 3 x=9 \\
& \Rightarrow x=3
\end{aligned}
$
Hence $x =3, y =0$.
View full question & answer→Question 122 Marks
Find $x$ and $y$ if $\left[\begin{array}{cc}x+y & y \\ 2 x & x-y\end{array}\right]\left[\begin{array}{c}2 \\ -1\end{array}\right]=\left[\begin{array}{l}3 \\ 2\end{array}\right]$
AnswerGiven
$\begin{aligned} & {\left[\begin{array}{cc} x+y & y \\ 2 x & x-y \end{array}\right]\left[\begin{array}{c} 2 \\\ -1 \end{array}\right]=\left[\begin{array}{l} 3 \\ 2 \end{array}\right]} \end{aligned} $
$ \Rightarrow\left[\begin{array}{cc} 2 x+2 y & -y \\ 4 x & -x+y \end{array}\right]=\left[\begin{array}{l} 3 \\ 2 \end{array}\right] $
$ \Rightarrow\left[\begin{array}{ll} 2 x & +y \\ 3 x & +y \end{array}\right]=\left[\begin{array}{l} 3 \\ 2 \end{array}\right] $
Comparing the corresponding elements
$2 x+y=3.....(1)$
$ 3 x+y=2.....(2)$
Subtracting, we get
$- x =1$
$ \Rightarrow x =-1$
Substituting the value of $x$ in $(i)$
$ 2(-1)+y=3$
$ \Rightarrow-2+y=3$
$ \Rightarrow y=3+2=5$
Hence $x=-1, y=5.$
View full question & answer→Question 132 Marks
Find $x$ and $y$ if $\left[\begin{array}{cc}2 x & x \\ y & 3 y\end{array}\right]\left[\begin{array}{l}3 \\ 2\end{array}\right]=\left[\begin{array}{c}16 \\ 9\end{array}\right]$
Answer$
\begin{aligned}
& {\left[\begin{array}{ll}
2 x & x \\
y & 3 y
\end{array}\right]\left[\begin{array}{l}
3 \\
2
\end{array}\right]=\left[\begin{array}{c}
16 \\
9
\end{array}\right]} \\
& \Rightarrow\left[\begin{array}{l}
2 x \times 3+x \times 2 \\
y \times 3+3 y \times 2
\end{array}\right]=\left[\begin{array}{c}
16 \\
9
\end{array}\right] \\
& \Rightarrow\left[\begin{array}{l}
6 x+2 x \\
3 y+6 y
\end{array}\right]=\left[\begin{array}{c}
16 \\
9
\end{array}\right] \\
& \Rightarrow\left[\begin{array}{l}
8 x \\
9 y
\end{array}\right]=\left[\begin{array}{c}
16 \\
9
\end{array}\right]
\end{aligned}
$
Comparing, we get
$
\begin{aligned}
& 8 x =16 \\
& \Rightarrow x =\frac{16}{8}=2
\end{aligned}
$
and
$ 9 y=9 $
$ \Rightarrow y=\frac{9}{9}=1$
Here $x =2, y =1$.
View full question & answer→Question 142 Marks
Find $x$ and $y$ if $\left[\begin{array}{cc}-3 & 2 \\ 0 & -5\end{array}\right]\left[\begin{array}{l}x \\ 2\end{array}\right]=\left[\begin{array}{l}5 \\ y\end{array}\right]$
Answer$\begin{aligned} & {\left[\begin{array}{cc}-3 & 2 \\ 0 & -5\end{array}\right]\left[\begin{array}{l}x \\ 2\end{array}\right]=\left[\begin{array}{l}5 \\ y\end{array}\right]} \end{aligned} $
$ \Rightarrow\left[\begin{array}{cc}-3 x & 4 \\ 0 & -10\end{array}\right]=\left[\begin{array}{c}-5 \\ y\end{array}\right] $
$ \Rightarrow\left[\begin{array}{cc}-3 x & +4 \\ 0 & -10\end{array}\right]=\left[\begin{array}{c}-5 \\ y\end{array}\right]$
Comparing the corresponding elements
$-3 x+4=-5$
$ \Rightarrow-3 x=-5-4=-9$
$ \therefore x=\frac{-9}{-3}=3$
$ -10=y$
$ \Rightarrow y=-10$
Hence $x=3, y=-10$.
View full question & answer→Question 152 Marks
If $A=\left[\begin{array}{ll}1 & 1 \\ x & x\end{array}\right]$, find the value of $x$, so that $A^2-0$
AnswerGiven
$
\begin{aligned}
& A =\left[\begin{array}{ll}
1 & 1 \\
x & x
\end{array}\right] \\
& A ^2=\left[\begin{array}{ll}
1 & 1 \\
x & x
\end{array}\right]\left[\begin{array}{ll}
1 & 1 \\
x & x
\end{array}\right] \\
& =\left[\begin{array}{cc}
1+x & 1+x \\
x+x^2 & x+x^2
\end{array}\right] \\
& \because A ^2=0 \\
& \therefore=\left[\begin{array}{cc}
1+x & 1+x \\
x+x^2 & x+x^2
\end{array}\right]=0 \\
& =\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]
\end{aligned}
$
Comparing
$
\begin{aligned}
& 1+x=0 \\
& \Rightarrow x=-1
\end{aligned}
$
View full question & answer→Question 162 Marks
If $A=\left[\begin{array}{ll}2 & 5 \\ 1 & 3\end{array}\right] B=\left[\begin{array}{cc}1 & -1 \\ -3 & 2\end{array}\right]$, find $A B$ and $B A$, is $A B=B A$ ?
Answer$
\begin{aligned}
& A=\left[\begin{array}{ll}
2 & 5 \\
1 & 3
\end{array}\right] \\
& B=\left[\begin{array}{cc}
1 & -1 \\
-3 & 2
\end{array}\right] \\
& \therefore A \times B=\left[\begin{array}{ll}
2 & 5 \\
1 & 3
\end{array}\right] \times\left[\begin{array}{cc}
1 & -1 \\
-3 & 2
\end{array}\right] \\
& =\left[\begin{array}{cc}
2-15 & -2+10 \\
1-9 & -1+6
\end{array}\right] \\
& =\left[\begin{array}{cc}
-13 & 8 \\
-8 & 5
\end{array}\right]
\end{aligned}
$
and
$
\begin{aligned}
& B \times A =\left[\begin{array}{cc}
1 & -1 \\
-3 & 2
\end{array}\right] \times\left[\begin{array}{ll}
2 & 5 \\
1 & 3
\end{array}\right] \\
& =\left[\begin{array}{cc}
2-1 & 5-3 \\
-6+2 & -15+6
\end{array}\right] \\
& =\left[\begin{array}{cc}
1 & 2 \\
-4 & -9
\end{array}\right]
\end{aligned}
$
Hence $A B \neq B A$.
View full question & answer→Question 172 Marks
If $A=\left[\begin{array}{cc}1 & 0 \\ 0 & -1\end{array}\right]$, find $A 2$ and $A 3$. Also state that which of these is equal to $A$
Answer$
\begin{aligned}
& A=\left[\begin{array}{cc}
1 & 0 \\
0 & -1
\end{array}\right] \\
& A^2=A \times A=\left[\begin{array}{cc}
1 & 0 \\
0 & -1
\end{array}\right]\left[\begin{array}{cc}
1 & 0 \\
0 & -1
\end{array}\right] \\
& =\left[\begin{array}{ll}
1+0 & 0+0 \\
0+0 & 0+1
\end{array}\right] \\
& =\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right] \\
& A^3=A^2+A=\left[\begin{array}{cc}
1 & 0 \\
0 & 1
\end{array}\right] \times\left[\begin{array}{cc}
1 & 0 \\
0 & -1
\end{array}\right] \\
& =\left[\begin{array}{cc}
1+0 & 0+0 \\
0+0 & 0-1
\end{array}\right] \\
& =\left[\begin{array}{cc}
1 & 0 \\
0 & -1
\end{array}\right]
\end{aligned}
$
From above it is clear that $A^3=A$.
View full question & answer→Question 182 Marks
If $A=\left[\begin{array}{ll}1 & 2 \\ 2 & 3\end{array}\right]$ and $B=\left[\begin{array}{ll}2 & 1 \\ 3 & 2\end{array}\right], C=\left[\begin{array}{ll}1 & 3 \\ 3 & 1\end{array}\right]$ find the matrix $C(B-A)$
Answer$\begin{aligned} & A=\left[\begin{array}{ll}1 & 2 \\ 2 & 3\end{array}\right] \\ & B=\left[\begin{array}{ll}2 & 1 \\ 3 & 2\end{array}\right] \\ & C=\left[\begin{array}{ll}1 & 3 \\ 3 & 1\end{array}\right] \\ & B-A=\left[\begin{array}{ll}2 & 1 \\ 3 & 2\end{array}\right]-\left[\begin{array}{ll}1 & 2 \\ 2 & 3\end{array}\right] \\ & =\left[\begin{array}{ll}1 & -1 \\ 1 & -1\end{array}\right] \\ & C(B-A)=\left[\begin{array}{ll}1 & 3 \\ 3 & 1\end{array}\right] \times\left[\begin{array}{ll}1 & -1 \\ 1 & -1\end{array}\right] \\ & =\left[\begin{array}{ll}1+3 & -1-3 \\ 3+1 & -3-1\end{array}\right] \\ & =\left[\begin{array}{ll}4 & -4 \\ 4 & -4\end{array}\right] . \\ & \end{aligned}$
View full question & answer→Question 192 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 $(B+C) A$
Answer$\begin{aligned} & (B+C) A \\ & A=\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right] \\ & B=\left[\begin{array}{ll}2 & 1 \\ 4 & 2\end{array}\right] \\ & C=\left[\begin{array}{ll}5 & 1 \\ 7 & 4\end{array}\right] \\ & (B+C) A=\left[\begin{array}{cc}7 & 2 \\ 11 & 6\end{array}\right] \times\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right] \\ & =\left[\begin{array}{cc}7+6 & 14+8 \\ 11+18 & 22+24\end{array}\right] \\ & =\left[\begin{array}{cc}13 & 22 \\ 29 & 46\end{array}\right] .\end{aligned}$
View full question & answer→Question 202 Marks
$A=\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]$ and $B=\left[\begin{array}{ll}6 & 1 \\ 1 & 1\end{array}\right], C=\left[\begin{array}{cc}-2 & -3 \\ 0 & 1\end{array}\right]$ find each of the following and state if they are equal. $A+C B$
Answer$
\begin{aligned}
& A + CB \\
& =\left[\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right]+\left[\begin{array}{cc}
-2 & -3 \\
0 & 1
\end{array}\right] \times\left[\begin{array}{ll}
6 & 1 \\
1 & 1
\end{array}\right] \\
& =\left[\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right]+\left[\begin{array}{cc}
-12-3 & -2-3 \\
0+1 & 0+1
\end{array}\right] \\
& =\left[\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right]+\left[\begin{array}{cc}
-15 & -5 \\
1 & 1
\end{array}\right] \\
& =\left[\begin{array}{cc}
1-15 & 2-5 \\
3+1 & 4+1
\end{array}\right] \\
& =\left[\begin{array}{cc}
-14 & -3 \\
4 & 5
\end{array}\right]
\end{aligned}
$
We can say that $C A+B \neq A+C B$.
View full question & answer→Question 212 Marks
$A=\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]$ and $B=\left[\begin{array}{ll}6 & 1 \\ 1 & 1\end{array}\right], C=\left[\begin{array}{cc}-2 & -3 \\ 0 & 1\end{array}\right]$ find each of the following and state if they are equal. $C A+B$
Answer$\begin{aligned} & CA + B \\ & CA =\left[\begin{array}{cc}-2 & -3 \\ 0 & 1\end{array}\right]\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right] \\ & =\left[\begin{array}{cc}-2-9 & -4-12 \\ 0+3 & 0+4\end{array}\right] \\ & =\left[\begin{array}{cc}-11 & -16 \\ 3 & 4\end{array}\right] \\ & CA + B =\left[\begin{array}{cc}-11 & -16 \\ 3 & 4\end{array}\right]+\left[\begin{array}{ll}6 & 1 \\ 1 & 1\end{array}\right] \\ & =\left[\begin{array}{cc}-11+6 & -16+1 \\ 3+1 & 4+1\end{array}\right] \\ & =\left[\begin{array}{cc}-5 & -15 \\ 4 & 5\end{array}\right] .\end{aligned}$
View full question & answer→Question 222 Marks
If $A=\left[\begin{array}{cc}3 & 5 \\ 4 & -2\end{array}\right]$ and $B=\left[\begin{array}{l}2 \\ 4\end{array}\right]$, is the product $A B$ possible? Give a reason. If yes, find $A B.$
AnswerYes, the product is possible because of
number of column in $A=$ number of row in $B$
i.e. $(2 \times 2) .(2 \times 1)=(2 \times 1)$ is the order of the matrix.
$ \begin{aligned} & AB =\left[\begin{array}{cc} 3 & 5 \\ 4 & -2 \end{array}\right]\left[\begin{array}{l} 2 \\ 4 \end{array}\right] \end{aligned} $
$ =\left[\begin{array}{c} 3 \times 2+5 \times 4 \\ 4 \times 2+(-2) \times 4 \end{array}\right] $
$ =\left[\begin{array}{c} 6+20 \\ 8-8 \end{array}\right] $
$ =\left[\begin{array}{c} 26 \\ 0 \end{array}\right] .$
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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$\begin{aligned} & A+2 X=2 B+C \\ & \Rightarrow\left[\begin{array}{cc}2 & -6 \\ 2 & 0\end{array}\right]+2 X=2\left[\begin{array}{cc}-3 & 2 \\ 4 & 0\end{array}\right]+\left[\begin{array}{ll}4 & 0 \\ 0 & 2\end{array}\right] \\ & \Rightarrow\left[\begin{array}{cc}2 & -6 \\ 2 & 0\end{array}\right]+2 X=\left[\begin{array}{cc}-6+4 & 4+0 \\ 8+0 & 0+2\end{array}\right] \\ & \Rightarrow\left[\begin{array}{cc}2 & -6 \\ 2 & 0\end{array}\right]+2 X=\left[\begin{array}{cc}-2 & 4 \\ 8 & 2\end{array}\right] \\ & \Rightarrow 2 X=\left[\begin{array}{cc}-2 & 4 \\ 8 & 2\end{array}\right]-\left[\begin{array}{cc}2 & -6 \\ 2 & 0\end{array}\right]=\left[\begin{array}{cc}-4 & 10 \\ 6 & 2\end{array}\right] \\ & \Rightarrow X=\frac{1}{2}\left[\begin{array}{cc}-4 & 10 \\ 6 & 2\end{array}\right] \\ & \Rightarrow X=\left[\begin{array}{cc}-2 & 5 \\ 3 & 1\end{array}\right]\end{aligned}$
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If $\left[\begin{array}{cc}1 & 4 \\ -2 & 3\end{array}\right]+2 M=3\left[\begin{array}{cc}3 & 2 \\ 0 & -3\end{array}\right]$, find the matrix $M$
Answer$\begin{aligned} & {\left[\begin{array}{cc}1 & 4 \\ -2 & 3\end{array}\right]+2 M =3\left[\begin{array}{cc}3 & 2 \\ 0 & -3\end{array}\right]} \\ & 2 M =3\left[\begin{array}{cc}3 & 2 \\ -2 & 3\end{array}\right]-\left[\begin{array}{cc}1 & 4 \\ -2 & 3\end{array}\right] \\ & =\left[\begin{array}{cc}9 & 6 \\ 0 & -9\end{array}\right]-\left[\begin{array}{cc}1 & 4 \\ -2 & 3\end{array}\right] \\ & =\left[\begin{array}{cc}9-1 & 6-4 \\ 0-(-2) & -9-3\end{array}\right] \\ & =\left[\begin{array}{cc}8 & 2 \\ 2 & -12\end{array}\right] \\ & \therefore M =\frac{1}{2}\left[\begin{array}{cc}8 & 2 \\ 2 & -12\end{array}\right] \\ & =\left[\begin{array}{cc}4 & 1 \\ 1 & -6\end{array}\right] . . . \text { (Dividing by } 2 \text { ) }\end{aligned}$
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Solve the matrix equation $\left[\begin{array}{ll}2 & 1 \\ 5 & 0\end{array}\right]-3 X =\left[\begin{array}{cc}-7 & 4 \\ 2 & 6\end{array}\right]$
Answer$\begin{aligned} & {\left[\begin{array}{ll}2 & 1 \\ 5 & 0\end{array}\right]-3 X =\left[\begin{array}{cc}-7 & 4 \\ 2 & 6\end{array}\right]} \\ & {\left[\begin{array}{ll}2 & 1 \\ 5 & 0\end{array}\right]-\left[\begin{array}{cc}-7 & 4 \\ 2 & 6\end{array}\right]=3 X } \\ & \therefore X =\frac{1}{3}\left[\begin{array}{ll}9 & -3 \\ 3 & -6\end{array}\right] \\ & =\left[\begin{array}{ll}3 & -1 \\ 1 & -2\end{array}\right] .\end{aligned}$
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If $A=\left[\begin{array}{cc}0 & -1 \\ 1 & 2\end{array}\right]$ and $B=\left[\begin{array}{cc}1 & 2 \\ -1 & 1\end{array}\right]$ Find the matrix $X$ if : $X-3 B=2 A$
Answer$\begin{aligned} & A=\left[\begin{array}{cc}0 & -1 \\ 1 & 2\end{array}\right] \\ & B=\left[\begin{array}{cc}1 & 2 \\ -1 & 1\end{array}\right] \\ & X-3 B=2 A \\ & \Rightarrow x=2 A+3 B \\ & X=2\left[\begin{array}{cc}0 & -1 \\ 1 & 2\end{array}\right]+3\left[\begin{array}{cc}1 & 2 \\ -1 & 1\end{array}\right] \\ & =\left[\begin{array}{cc}0 & -2 \\ 2 & 4\end{array}\right]+\left[\begin{array}{cc}3 & 6 \\ -3 & 3\end{array}\right] \\ & =\left[\begin{array}{cc}0+3 & -2+6 \\ 2-3 & 4+3\end{array}\right] \\ & =\left[\begin{array}{cc}3 & 4 \\ -1 & 7\end{array}\right] .\end{aligned}$
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If $A=\left[\begin{array}{cc}0 & -1 \\ 1 & 2\end{array}\right]$ and $B=\left[\begin{array}{cc}1 & 2 \\ -1 & 1\end{array}\right]$ Find the matrix $X$ if : $3 A+X=B$
Answer$\begin{aligned} & A=\left[\begin{array}{cc}0 & -1 \\ 1 & 2\end{array}\right] \\ & B=\left[\begin{array}{cc}1 & 2 \\ -1 & 1\end{array}\right] \\ & 3 A + X = B \\ & \Rightarrow X = B -3 A \\ & X=\left[\begin{array}{cc}1 & 2 \\ -1 & 1\end{array}\right]-3\left[\begin{array}{cc}0 & -1 \\ 1 & 2\end{array}\right] \\ & =\left[\begin{array}{cc}1 & 2 \\ -1 & 1\end{array}\right]-\left[\begin{array}{cc}0 & -3 \\ 3 & 6\end{array}\right] \\ & =\left[\begin{array}{cc}1-0 & 2+3 \\ -1-3 & 1-6\end{array}\right] \\ & =\left[\begin{array}{cc}1 & 5 \\ -4 & -5\end{array}\right] \text {. } \\ & \end{aligned}$
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$A=\left[\begin{array}{cc}1 & 2 \\ -2 & 3\end{array}\right]$ and $B=\left[\begin{array}{cc}-2 & -1 \\ 1 & 2\end{array}\right], C\left[\begin{array}{cc}0 & 3 \\ 2 & -1\end{array}\right]$ Find $A+2 B-3 C$
Answer$\begin{aligned} & A=\left[\begin{array}{cc}1 & 2 \\ -2 & 3\end{array}\right] \text { and } B=\left[\begin{array}{cc}-2 & -1 \\ 1 & 2\end{array}\right], C\left[\begin{array}{cc}0 & 3 \\ 2 & -1\end{array}\right] \\ & \therefore A+2 B-3 C \\ & =\left[\begin{array}{cc}1 & 2 \\ -2 & 3\end{array}\right]+2\left[\begin{array}{cc}-2 & -1 \\ 1 & 2\end{array}\right]-3\left[\begin{array}{cc}0 & 3 \\ 2 & -1\end{array}\right] \\ & =\left[\begin{array}{cc}1 & 2 \\ -2 & 3\end{array}\right]+\left[\begin{array}{cc}-4 & -2 \\ 2 & 4\end{array}\right]-\left[\begin{array}{cc}0 & 9 \\ 6 & -3\end{array}\right] \\ & =\left[\begin{array}{cc}1-4-0 & 2-2-9 \\ -2+2-6 & 3+4+3\end{array}\right] \\ & =\left[\begin{array}{cc}-3 & -9 \\ -6 & 10\end{array}\right] .\end{aligned}$
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Given $A=\left[\begin{array}{ll}1 & 4 \\ 2 & 3\end{array}\right]$ and $B=\left[\begin{array}{ll}-4 & -1 \\ -3 & -2\end{array}\right]$ find a matrix $C$ such that $C+B=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]$
Answer$\begin{aligned} & A=\left[\begin{array}{ll}1 & 4 \\ 2 & 3\end{array}\right] \\ & B=\left[\begin{array}{ll}-4 & -1 \\ -3 & -2\end{array}\right] \\ & C+B=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right] \\ & C=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]-B=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]-\left[\begin{array}{ll}-4 & -1 \\ -3 & -2\end{array}\right] \\ & =\left[\begin{array}{ll}0-(-4) & 0-(-1) \\ 0-(-3) & 0-(-2)\end{array}\right] \\ & =\left[\begin{array}{ll}4 & 1 \\ 3 & 2\end{array}\right] .\end{aligned}$
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Given $A=\left[\begin{array}{ll}1 & 4 \\ 2 & 3\end{array}\right]$ and $B=\left[\begin{array}{rr}-4 & -1 \\ -3 & -2\end{array}\right]$ find the matrix $2 A+B$
Answer$\begin{aligned} & A=\left[\begin{array}{ll}1 & 4 \\ 2 & 3\end{array}\right] \\ & B=\left[\begin{array}{ll}-4 & -1 \\ -3 & -2\end{array}\right] \\ & 2 A+B=2\left[\begin{array}{ll}1 & 4 \\ 2 & 3\end{array}\right]+\left[\begin{array}{ll}-4 & -1 \\ -3 & -2\end{array}\right] \\ & =\left[\begin{array}{ll}2 & 8 \\ 4 & 6\end{array}\right]+\left[\begin{array}{ll}-4 & -1 \\ -3 & -2\end{array}\right] \\ & =\left[\begin{array}{ll}2-4 & 8-1 \\ 4-3 & 6-2\end{array}\right] \\ & =\left[\begin{array}{cc}-2 & 7 \\ 1 & 4\end{array}\right] \text {. } \\ & \end{aligned}$
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If $A=\left[\begin{array}{ll}1 & 4 \\ 2 & 3\end{array}\right]$ and $B=\left[\begin{array}{ll}1 & 2 \\ 3 & 1\end{array}\right]$ Compute $3 A+4 B$
Answer$\begin{aligned} & A=\left[\begin{array}{ll}1 & 4 \\ 2 & 3\end{array}\right] \\ & B=\left[\begin{array}{ll}1 & 2 \\ 3 & 1\end{array}\right] \\ & 3 A+4 B=3\left[\begin{array}{ll}1 & 4 \\ 2 & 3\end{array}\right]+4\left[\begin{array}{ll}1 & 2 \\ 3 & 1\end{array}\right] \\ & =\left[\begin{array}{cc}3 & 12 \\ 6 & 9\end{array}\right]+\left[\begin{array}{cc}4 & 8 \\ 12 & 4\end{array}\right] \\ & =\left[\begin{array}{cc}3+4 & 12+8 \\ 6+12 & 9+4\end{array}\right] \\ & =\left[\begin{array}{cc}7 & 20 \\ 18 & 13\end{array}\right] .\end{aligned}$
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If $A=\left[\begin{array}{cc}2 & 0 \\ -3 & 1\end{array}\right]$ and $B=\left[\begin{array}{cc}0 & 1 \\ -2 & 3\end{array}\right]$ find $2 A-3 B$
Answer$\begin{aligned} & A=\left[\begin{array}{cc}2 & 0 \\ -3 & 1\end{array}\right] \\ & B=\left[\begin{array}{cc}0 & 1 \\ -2 & 3\end{array}\right] \\ & \therefore 2 A-3 B=2\left[\begin{array}{cc}2 & 0 \\ -3 & 1\end{array}\right]-3\left[\begin{array}{cc}0 & 1 \\ -2 & 3\end{array}\right] \\ & =\left[\begin{array}{cc}4 & 0 \\ -6 & 2\end{array}\right]-\left[\begin{array}{cc}0 & 3 \\ -6 & 9\end{array}\right] \\ & =\left[\begin{array}{cc}4-0 & 0-3 \\ -6+6 & 2-9\end{array}\right] \\ & =\left[\begin{array}{cc}4 & -3 \\ 0 & -7\end{array}\right] .\end{aligned}$
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Given that $M=\left[\begin{array}{ll}2 & 0 \\ 1 & 2\end{array}\right]$ and $N=\left[\begin{array}{cc}2 & 0 \\ -1 & 2\end{array}\right]$, find $M+2 N$
Answer$\begin{aligned} & M=\left[\begin{array}{ll}2 & 0 \\ 1 & 2\end{array}\right] \\ & N=\left[\begin{array}{cc}2 & 0 \\ -1 & 2\end{array}\right] \\ & \therefore M+2 N=\left[\begin{array}{ll}2 & 0 \\ 1 & 2\end{array}\right]+2\left[\begin{array}{cc}2 & 0 \\ -1 & 2\end{array}\right] \\ & =\left[\begin{array}{ll}2 & 0 \\ 1 & 2\end{array}\right]+\left[\begin{array}{cc}4 & 0 \\ -2 & 4\end{array}\right] \\ & =\left[\begin{array}{ll}2+4 & 0+0 \\ 1-2 & 2+4\end{array}\right] \\ & =\left[\begin{array}{cc}6 & 0 \\ -1 & 6\end{array}\right] . \\ & \end{aligned}$
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Find the values of $x_t y_1 a$ and $b$ if $\left[\begin{array}{cc}x-2 & y \\ a+2 b & 3 a-b\end{array}\right]=\left[\begin{array}{ll}3 & 1 \\ 5 & 1\end{array}\right]$
AnswerComparing corresponding elements
x – 2 = 3, y = 1
x = 3 + 2 = 5
a + 2b = 5 ……(i)
3a – b = 1 ……..(ii)
Multiplying (i) by 1 and (ii) by 2
a + 2b = 5,
6a – 2b = 2
Adding, we get,
7a = 7
⇒ a = 1
Substituting the value of a in ...(i)
1 + 2b = 5
⇒ 2b = 5 – 1 = 4
⇒ b = 2
Hence x = 5, y = 1, a = 1, b = 2.
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Find the values of $x_1 y$ and $z$ if $\left[\begin{array}{cc}x+2 & 6 \\ 3 & 5 z\end{array}\right]=\left[\begin{array}{cc}-5 & y^2+y \\ 3 & 20\end{array}\right]$
AnswerComparing the corresponding elements of equal determinents,
$
\begin{aligned}
& x +2=-5 \\
& \Rightarrow x =-5-2=-7 \\
& \therefore x =-7,5 z =-20 \\
& \Rightarrow z =-\frac{20}{5} \\
& =-4 \\
& \Rightarrow y ^2+ y =6 \\
& \Rightarrow y ^2+ y -6=0 \\
& \Rightarrow y ^2+3 y -2 y -6=0 \\
& \Rightarrow y ( y +3)-2( y +3)=0 \\
& \Rightarrow( y +3)( y -2)=0
\end{aligned}
$
Either $y+3=0$,
then $y=-3$ or $y-2=0$,
then $y =2$
Hence $x=-7, y=-3,2, z=-4$.
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If $\left[\begin{array}{cc}x+3 & 4 \\ y-4 & x+y\end{array}\right]=\left[\begin{array}{ll}5 & 4 \\ 3 & 9\end{array}\right]$, find values of $x$ and $y$
Answer$
\left[\begin{array}{cc}
x+3 & 4 \\
y-4 & x+y
\end{array}\right]=\left[\begin{array}{ll}
5 & 4 \\
3 & 9
\end{array}\right]
$
Comparing the corresponding terms, we get.
$
\begin{aligned}
& x+3=5 \\
& \Rightarrow x=5-3=2 \\
& \Rightarrow y-4=3 \\
& \Rightarrow y=3+4=7 \\
& x=2, y=7
\end{aligned}
$
View full question & answer→Question 372 Marks
Find the value of $x$ if $\left[\begin{array}{cc}3 x+y & -y \\ 2 y-x & 3\end{array}\right]=\left[\begin{array}{cc}1 & 2 \\ -5 & 3\end{array}\right]$
Answer$
\left[\begin{array}{cc}
3 x+y & -y \\
2 y-x & 3
\end{array}\right]=\left[\begin{array}{cc}
1 & 2 \\
-5 & 3
\end{array}\right]
$
Comparing the corresponding terms, we get.
$
\begin{aligned}
& -y=2 \\
& \Rightarrow y=-2 \\
& 3 x+y=1 \\
& \Rightarrow 3 x=1-y \\
& \Rightarrow 3 x=1-(-2) \\
& =1+2 \\
& =3 \\
& \Rightarrow x=\frac{3}{3} \\
& =1
\end{aligned}
$
Hence $x=1, y=-2$.
View full question & answer→Question 382 Marks
Find the values of $x$ and $y$ if $:\left[\begin{array}{c}2 x+y \\ 3 x-2 y\end{array}\right]=\left[\begin{array}{l}5 \\ 4\end{array}\right]$
AnswerComparing corresponding elements,
$2x + y = 5 …(i)$
$3x – 2y = 4 …(ii)$
Multiply $(i)$ by $2$ and $(ii)$ by $'1\ ’$ we get
$4x + 2y = 10,$
$3x – 2y = 4$
Adding we get,
$7x = 14$
$\Rightarrow x = 2$
Substituting the value of $x$ in $(i)$
$2 x^ 2 + y = 5$
$\Rightarrow 4 + y = 5$
$y = 5 – 4$
$= 1$
Hence $x = 2, y = 1.$
View full question & answer→Question 392 Marks
Find $X$ if $Y=\left[\begin{array}{ll}3 & 2 \\ 1 & 4\end{array}\right]$ and $2 X+Y=\left[\begin{array}{cc}1 & 0 \\ -3 & 2\end{array}\right]$
AnswerGiven
$
\begin{aligned}
& 2 X+Y=\left[\begin{array}{cc}
1 & 0 \\
-3 & 2
\end{array}\right] \\
& \Rightarrow 2 X=2 X+Y=\left[\begin{array}{cc}
1 & 0 \\
-3 & 2
\end{array}\right]-Y \\
& \Rightarrow 2 X=\left[\begin{array}{cc}
1 & 0 \\
-3 & 2
\end{array}\right]-\left[\begin{array}{ll}
3 & 2 \\
1 & 4
\end{array}\right] \\
& =\left[\begin{array}{cc}
1-3 & 0-2 \\
-3-1 & 2-4
\end{array}\right] \\
& =\left[\begin{array}{ll}
-2 & -2 \\
-4 & -2
\end{array}\right] \\
& X =\frac{1}{2}\left[\begin{array}{ll}
-2 & -2 \\
-4 & -2
\end{array}\right] \\
& =\left[\begin{array}{ll}
-1 & -1 \\
-2 & -1
\end{array}\right] . \\
&
\end{aligned}
$
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If $\left[\begin{array}{cc}-1 & 0 \\ 0 & 1\end{array}\right]\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]=\left[\begin{array}{cc}1 & 0 \\ 0 & -1\end{array}\right]$ find $a, b, c$ and $d$
AnswerGiven
$\begin{array}{l}{\left(\begin{array}{cc}-1 & 0 \\ 0 & 1\end{array}\right)\left(\begin{array}{ll}a & b \\ c & d\end{array}\right)=\left(\begin{array}{cc}1 & 0 \\ 0 & -1\end{array}\right)} \end{array} $
$ \Rightarrow\left(\begin{array}{cc}a+0 & -b+0 \\ 0+c & 0+d\end{array}\right)=\left(\begin{array}{cc}1 & 0 \\ 0 & -1\end{array}\right) $
$ \Rightarrow\left(\begin{array}{cc}-a & -b \\ c & d\end{array}\right)=\left(\begin{array}{cc}1 & 0 \\ 0 & -1\end{array}\right)$
Comparing the corresponding elements
$–a = 1$
$\Rightarrow a = –1$
$–b = 0$
$\Rightarrow b = 0$
$c = 0$ and $d = –1$
Hence $a = –1, b = 0, c = 0, d = –1.$
View full question & answer→Question 412 Marks
If $\left[\begin{array}{cc}-1 & 0 \\ 0 & 1\end{array}\right]\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]=\left[\begin{array}{cc}1 & 0 \\ 0 & -1\end{array}\right]$ find $a, b, c$ and $d$
AnswerGiven
$
\begin{aligned}
& {\left[\begin{array}{ll}
-1 & 0 \\
0 & 1
\end{array}\right]\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]=\left[\begin{array}{cc}
1 & 0 \\
0 & -1
\end{array}\right]} \\
& \Rightarrow\left[\begin{array}{ll}
a+0 & -b+0 \\
0+c & 0+d
\end{array}\right]=\left[\begin{array}{cc}
1 & 0 \\
0 & -1
\end{array}\right] \\
& \Rightarrow\left[\begin{array}{cc}
-a & -b \\
c & d
\end{array}\right]=\left[\begin{array}{cc}
1 & 0 \\
0 & -1
\end{array}\right]
\end{aligned}
$
Comparing the corresponding elements
$
\begin{aligned}
& -a=1 \\
& \Rightarrow a=-1 \\
& -b=0 \\
& \Rightarrow b=0 \\
& c=0 \text { and } d=-1
\end{aligned}
$
Hence $a=-1, b=0, c=0, d=-1$.
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