Question 12 Marks
If $A=\left[\begin{array}{ll}2 & 5 \\ 3 & 7\end{array}\right], B=\left[\begin{array}{cc}1 & 7 \\ -3 & 0\end{array}\right]$, find the matrix $A-4 B+7 I$, where I is the unit matrix of order 2 .
Answer$
\begin{aligned}
A-4 B +7 I & =\left(\begin{array}{ll}
2 & 5 \\
3 & 7
\end{array}\right]-4\left[\begin{array}{rr}
1 & 7 \\
-3 & 0
\end{array}\right]+7\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right] \\
& =\left[\begin{array}{rr}
2 & 5 \\
3 & 7
\end{array}\right]-\left[\begin{array}{rr}
4 & 28 \\
-12 & 0
\end{array}\right]+\left[\begin{array}{ll}
7 & 0 \\
0 & 7
\end{array}\right] \\
& =\left[\begin{array}{rr}
2-4+7 & 5-28+0 \\
3-(-12)+0 & 7-0+7
\end{array}\right] \\
& =\left[\begin{array}{rr}
5 & -23 \\
15 & 14
\end{array}\right] .
\end{aligned}
$
View full question & answer→Question 22 Marks
Check whether following matrices are invertible or not: $\left[\begin{array}{lll}1 & 2 & 3 \\ 2 & 4 & 5 \\ 2 & 4 & 6\end{array}\right]$
AnswerLet $A=\left[\begin{array}{lll}1 & 2 & 3 \\ 2 & 4 & 5 \\ 2 & 4 & 6\end{array}\right]$
Then $|A|=\left|\begin{array}{lll}1 & 2 & 3 \\ 2 & 4 & 5 \\ 2 & 4 & 6\end{array}\right|$
$
\begin{aligned}
& =1(24-20)-2(12-10)+3(8-8) \\
& =4-4+0 \\
& =0
\end{aligned}
$
$\therefore A$ is a singular matrix.
Hence, A-1 does not exist.
View full question & answer→Question 32 Marks
Check whether following matrices are invertible or not: $\left[\begin{array}{lll}3 & 4 & 3 \\ 1 & 1 & 0 \\ 1 & 4 & 5\end{array}\right]$
AnswerLet $A=\left[\begin{array}{lll}3 & 4 & 3 \\ 1 & 1 & 0 \\ 1 & 4 & 5\end{array}\right]$
Then $|A|=\left|\begin{array}{lll}3 & 4 & 3 \\ 1 & 1 & 0 \\ 1 & 4 & 5\end{array}\right|$
$
\begin{aligned}
& =3(5-0)-4(5-0)+3(4-1) \\
& =15-20+9 \\
& =4 \neq 0
\end{aligned}
$
$\therefore A$ is a non-singular matrix.
Hence, $A ^{-1}$ exists.
View full question & answer→Question 42 Marks
Check whether following matrices are invertible or not: $\left[\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right]$
AnswerLet $A=\left[\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right]$
Then $|A|=\left|\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right|$
$=1-1$
$=0$
$\therefore A$ is a singular matrix.
Hence, $A ^{-1}$ does not exist.
View full question & answer→Question 52 Marks
Check whether following matrices are invertible or not: $\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
AnswerLet $A=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
Then $|A|=\left|\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right|$
$=1-0$
$=1 \neq 0$
$\therefore A$ is a non-singular matrix.
Hence, $A^{-1}$ exists.
View full question & answer→Question 62 Marks
If $A=\left[\begin{array}{cc}1 & 2 \\ -1 & 3\end{array}\right]$, then find $A^3$.
Answer$
\begin{aligned}
& A^2=A \cdot A=\left(\begin{array}{rr}
1 & 2 \\
-1 & 3
\end{array}\right)\left(\begin{array}{rr}
1 & 2 \\
-1 & 3
\end{array}\right) \\
& =\left[\begin{array}{rr}
1-2 & 2+6 \\
-1-3 & -2+9
\end{array}\right]=\left[\begin{array}{ll}
-1 & 8 \\
-4 & 7
\end{array}\right] \\
& \therefore A^3=A^2 \cdot A=\left[\begin{array}{ll}
-1 & 8 \\
-4 & 7
\end{array}\right]\left[\begin{array}{rr}
1 & 2 \\
-1 & 3
\end{array}\right] \\
& =\left(\begin{array}{ll}
-1-8 & -2+24 \\
-4-7 & -8+21
\end{array}\right) \\
& \therefore A^3=\left(\begin{array}{rr}
-9 & 22 \\
-11 & 13
\end{array}\right)\\
&
\end{aligned}
$
View full question & answer→Question 72 Marks
Find the adjoint of the following matrices : $\left[\begin{array}{cc}2 & -3 \\ 3 & 5\end{array}\right]$
Answer$
A=\left[\begin{array}{rr}
2 & -3 \\
3 & 5
\end{array}\right]
$
Here, $a_{11}=2, M _{11}=5$
$
\begin{aligned}
& \therefore A _{11}=(-1)^{1+1}(5)=5 \\
& a_{12}=-3, M _{12}=3 \\
& \therefore A _{12}=(-1)^{1+2}(3)=-3
\end{aligned}
$
$
\begin{aligned}
& a_{21}=3, M _{21}=-3 \\
& \therefore A _{21}=(-1)^{2+1}(-3)=3 \\
& a_{22}=5, M _{22}=2 \\
& \therefore A _{22}=(-1)^{2+2}=2
\end{aligned}
$
$\therefore$ the cofactor matrix $=\left(\begin{array}{ll} A _{11} & A _{12} \\ A _{21} & A _{22}\end{array}\right]=\left[\begin{array}{rr}5 & -3 \\ 3 & 2\end{array}\right)$
$\therefore \operatorname{adj} A=\left[\begin{array}{rr}5 & 3 \\ -3 & 2\end{array}\right] \cdot$
View full question & answer→Question 82 Marks
Find the cofactor matrix of the following matrices : $\left[\begin{array}{cc}1 & 2 \\ 5 & -8\end{array}\right]$
AnswerLet $A =\left(\begin{array}{rr}1 & 2 \\ 5 & -8\end{array}\right)$
Here, $a_{11}=1, M _{11}=-8$
$
\begin{aligned}
& \therefore A _{11}=(-1)^{1+1} M _{11}=-8 \\
& a_{12}=2, M _{12}=5 \\
& \therefore A _{12}=(-1)^{1+2} M _{12}=-1(5)=-5 \\
& a_{21}=5, M _{21}=2 \\
& \therefore A _{21}=(-1)^{2+1} M _{21}=-1(2)=-2 \\
& a_{22}=-8, M _{22}=1 \\
& \therefore A _{22}=(-1)^{2+2} M _{22}=1 . \\
& \therefore \text { cofactor matrix }=\left[\begin{array}{ll}
A _{11} & A _{12} \\
A _{21} & A _{22}
\end{array}\right] \\
& \quad=\left[\begin{array}{rr}
-8 & -5 \\
-2 . & 1
\end{array}\right]
\end{aligned}
$
View full question & answer→Question 92 Marks
If $A=\left[\begin{array}{cc}5 & 4 \\ -2 & 3\end{array}\right]$ and $B=\left[\begin{array}{cc}-1 & 3 \\ 4 & -1\end{array}\right]$, then find $C^{\top}$, such that $3 A-$ $2 B + C = I$, where $I$ is the unit matrix of order 2 .
Answer$
\begin{aligned}
& 3 A -2 B + C = I \\
& \therefore C = I -3 A +2 B \\
& =\left(\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right)-3\left(\begin{array}{rr}
5 & 4 \\
-2 & 3
\end{array}\right)+2\left(\begin{array}{rr}
-1 & 3 \\
4 & -1
\end{array}\right) \\
& =\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]-\left[\begin{array}{rr}
15 & 12 \\
-6 & 9
\end{array}\right]+\left[\begin{array}{rr}
-2 & 6 \\
8 & -2
\end{array}\right] \\
& =\left(\begin{array}{rr}
1-15+(-2) & 0-12+6 \\
0-(-6)+8 & 1-9-2
\end{array}\right) \\
& \therefore C=\left[\begin{array}{rr}
-16 & -6 \\
14 & -10
\end{array}\right] \\
& \therefore C^T=\left(\begin{array}{rr}
-16 & 14 \\
-6 & -10
\end{array}\right) \text {. } \\
&
\end{aligned}
$
View full question & answer→Question 102 Marks
If $A =\left[\begin{array}{ccc}1 & 2 & -5 \\ 2 & -3 & 4 \\ -5 & 4 & 9\end{array}\right]$, prove that $A ^{\top}= A$.
Answer$
\begin{gathered}
A=\left[\begin{array}{rrr}
1 & 2 & -5 \\
2 & -3 & 4 \\
-5 & 4 & 9
\end{array}\right]....(1) \\
\therefore A^{ T }=\left(\begin{array}{rrr}
1 & 2 & -5 \\
2 & -3 & 4 \\
-5 & 4 & 9
\end{array}\right)\ldots(2)
\end{gathered}
$
From (1) and (2), $A ^{ T }= A$.
View full question & answer→Question 112 Marks
If $A=\left[\begin{array}{cc}5 & -3 \\ 4 & -3 \\ -2 & 1\end{array}\right]$, prove that $\left(A^{\top}\right)^{\top}=A$.
Answer$
\begin{array}{r}
A=\left[\begin{array}{rr}
5 & -3 \\
4 & -3 \\
-2 & 1
\end{array}\right] \\
\therefore A^{ T }=\left[\begin{array}{rrr}
5 & 4 & -2 \\
-3 & -3 & 1
\end{array}\right]
\end{array}
$
$
\therefore\left( A ^{ T }\right)^{ T }=\left(\begin{array}{rr}
5 & -3 \\
4 & -3 \\
-2 & 1
\end{array}\right)= A \text {. }
$
View full question & answer→Question 122 Marks
If $A=\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]$, prove that $A^2-5 A+7 I=0$, where I is a $2 \times 2$ unit matrix.
Answer$
\begin{aligned}
A^2=A \cdot A & =\left[\begin{array}{rr}
3 & 1 \\
-1 & 2
\end{array}\right]\left[\begin{array}{rr}
3 & 1 \\
-1 & 2
\end{array}\right] \\
= & \left[\begin{array}{rr}
9-1 & 3+2 \\
-3-2 & -1+4
\end{array}\right]=\left[\begin{array}{rr}
8 & 5 \\
-5 & 3
\end{array}\right] \\
\therefore A ^2-5 A +7 I & =\left[\begin{array}{rr}
8 & 5 \\
-5 & 3
\end{array}\right]-5\left[\begin{array}{rr}
3 & 1 \\
-1 & 2
\end{array}\right]+7\left(\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right] \\
& =\left[\begin{array}{rr}
8 & 5 \\
-5 & 3
\end{array}\right]-\left[\begin{array}{rr}
15 & 5 \\
-5 & 10
\end{array}\right]+\left[\begin{array}{ll}
7 & 0 \\
0 & 7
\end{array}\right] \\
& =\left[\begin{array}{rr}
8-15+7 & 5-5+0 \\
-5+5+0 & 3-10+7
\end{array}\right]=\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]
\end{aligned}
$
$\therefore A ^2-5 A +7 I =0$.
View full question & answer→Question 132 Marks
For each of the following matrices, find its transpose and state whether it is symmetric, skew-symmetric or neither:
$\left[\begin{array}{ccc}0 & 1+2 i & i-2 \\ -1-2 i & 0 & -7 \\ 2-i & 7 & 0\end{array}\right]$
AnswerLet $C=\left(\begin{array}{rrr}0 & 1+2 i & i-2 \\ -1-2 i & 0 & -7 \\ 2-i & 7 & 0\end{array}\right)$
Then $C^{ T }=\left(\begin{array}{rrr}0 & -1-2 i & 2-i \\ 1+2 i & 0 & 7 \\ i-2 & -7 & 0\end{array}\right)$
$\therefore- C ^{ T }=\left(\begin{array}{rrr}0 & 1+2 i & i-2 \\ -1-2 i & 0 & -7 \\ 2-i & 7 & 0\end{array}\right)$
$\therefore C =- C ^{ T }$
Hence, $C$ is a skew-symmetric matrix.
View full question & answer→Question 142 Marks
For each of the following matrices, find its transpose and state whether it is symmetric, skew-symmetric or neither:
$\left[\begin{array}{ccc}2 & 5 & 1 \\ -5 & 4 & 6 \\ -1 & -6 & 3\end{array}\right]$
AnswerLet $B =\left[\begin{array}{ccc}2 & 5 & 1 \\ -5 & 4 & 6 \\ -1 & -6 & 3\end{array}\right]$
Then $B ^{\top}=\left(\begin{array}{rrr}2 & -5 & -1 \\ 5 & 4 & -6 \\ 1 & 6 & 3\end{array}\right)$
$
\therefore B \neq B ^{\top}
$
Also,
$
\begin{aligned}
& - B ^{\top}=\left(\begin{array}{rrr}
2 & -5 & -1 \\
5 & 4 & -6 \\
1 & 6 & 3
\end{array}\right)=\left(\begin{array}{rrr}
-2 & 5 & 1 \\
-5 & -4 & 6 \\
-1 & -6 & -3
\end{array}\right) \\
& \therefore B \neq- B ^{\top}
\end{aligned}
$
Hence, $B$ is neither symmetric nor skew-symmetric matrix.
View full question & answer→Question 152 Marks
For each of the following matrices, find its transpose and state whether it is symmetric, skew-symmetric or neither:
$\left[\begin{array}{ccc}1 & 2 & -5 \\ 2 & -3 & 4 \\ -5 & 4 & 9\end{array}\right]$
AnswerLet $A =\left[\begin{array}{ccc}1 & 2 & -5 \\ 2 & -3 & 4 \\ -5 & 4 & 9\end{array}\right]$
Then $A^{\top}=\left[\begin{array}{rrr}1 & 2 & -5 \\ 2 & -3 & 4 \\ -5 & 4 & 9\end{array}\right]$
Since, $A=A^{\top}, A$ is a symmetric matrix.
View full question & answer→Question 162 Marks
If $A=\left[\begin{array}{ccc}1 & 2 & -3 \\ -3 & 7 & -8 \\ 0 & -6 & 1\end{array}\right], B=\left[\begin{array}{ccc}9 & -1 & 2 \\ -4 & 2 & 5 \\ 4 & 0 & -3\end{array}\right]$, then find the matrix $C$ such that $A + B + C$ is a zero matrix.
Answer$
\begin{aligned}
& A+B+C=0 \\
& \therefore C=-A-B \\
& =-\left[\begin{array}{rrr}
1 & 2 & -3 \\
-3 & 7 & -8 \\
0 & -6 & 1
\end{array}\right]-\left[\begin{array}{rrr}
9 & -1 & 2 \\
-4 & 2 & 5 \\
4 & 0 & -3
\end{array}\right] \\
& =\left[\begin{array}{rrr}
-1 & -2 & 3 \\
3 & -7 & 8 \\
0 & 6 & -1
\end{array}\right]-\left[\begin{array}{rrr}
9 & -1 & 2 \\
-4 & 2 & 5 \\
4 & 0 & -3
\end{array}\right] \\
& =\left[\begin{array}{rrr}
-1-9 & -2-(-1) & 3-2 \\
3-(-4) & -7-2 & 8-5 \\
0-4 & 6-0 & -1-(-3)
\end{array}\right] \\
& \therefore C=\left[\begin{array}{rrr}
-10 & -1 & 1 \\
7 & -9 & 3 \\
-4 & 6 & 2
\end{array}\right] \\
\end{aligned}
$
View full question & answer→Question 172 Marks
If $A=\left[\begin{array}{cc}1 & -2 \\ 5 & 3\end{array}\right], B=\left[\begin{array}{cc}1 & -3 \\ 4 & -7\end{array}\right]$, then find the matrix $A-2 B+6 I$, where $I$ is the unit matrix of order 2 .
Answer$
\begin{aligned}
A-2 B+6 I & =\left[\begin{array}{rr}
1 & -2 \\
5 & 3
\end{array}\right]-2\left[\begin{array}{ll}
1 & -3 \\
4 & -7
\end{array}\right]+6\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right] \\
& =\left[\begin{array}{rr}
1 & -2 \\
5 & 3
\end{array}\right]-\left[\begin{array}{rr}
2 & -6 \\
8 & -14
\end{array}\right]+\left[\begin{array}{ll}
6 & 0 \\
0 & 6
\end{array}\right] \\
& =\left[\begin{array}{rr}
1-2+6 & -2-(-6)+0 \\
5-8+0 & 3-(-14)+6
\end{array}\right] \\
& =\left[\begin{array}{rr}
5 & 4 \\
-3 & 23
\end{array}\right]
\end{aligned}
$
View full question & answer→Question 182 Marks
Construct the matrix $A=\left[a_{i j}\right]_{3 \times 3}$, where $a_{i j}=i-j$. State whether $A$ is symmetric or skew-symmetric.
Answer$
A =\left[a_{i j}\right]_{3 \times 3}=\left(\begin{array}{lll}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33}
\end{array}\right)
$
Now, $a_{i j}=i-j$ for all $i$ and $j$
$
\begin{aligned}
\therefore & a_{11}=1-1=0, a_{12}=1-2=-1 \\
& a_{13}=1-3=-2, a_{21}=2-1=1 \\
& a_{22}=2-2=0, a_{23}=2-3=-1 \\
& a_{31}=3-1=2, a_{32}=3-2=1, a_{33}=3-3=0 \\
\therefore & A =\left(\begin{array}{rrr}
0 & -1 & -2 \\
1 & 0 & -1 \\
2 & 1 & 0
\end{array}\right)
\end{aligned}
$
Since, $a_{i j}=i-j=-(j-i)=-a_{j i}$ for all $i$ and $j$,
A is skew-symmetric matrix.
View full question & answer→Question 192 Marks
Which of the following matrices are singular or non-singular : $\left[\begin{array}{ccc}a & b & c \\ p & q & r \\ 2 a-p & 2 b-q & 2 c-r\end{array}\right]$
AnswerLet $A =\left[\begin{array}{ccc}a & b & c \\ p & q & r \\ 2 a-p & 2 b-q & 2 c-r\end{array}\right]$
$
\therefore| A |=\left|\begin{array}{ccc}
a & b & c \\
p & q & r \\
2 a-p & 2 b-q & 2 c-r
\end{array}\right|
$
By $R_3+R_2$, we get,
$
\begin{aligned}
| A | & =\left|\begin{array}{ccc}
a & b & c \\
p & q & r \\
2 a & 2 b & 2 c
\end{array}\right| \\
& =2\left|\begin{array}{lll}
a & b & c \\
p & q & r \\
a & b & c
\end{array}\right| \\
& =2 \times 0 \\
& =0
\end{aligned}
$
$\therefore A$ is a singular matrix.
View full question & answer→Question 202 Marks
Construct a matrix $A =\left[ a _{ i j}\right]_{3 \times 2}$ whose elements aij isgiven by : $a _{ ij }=\frac{(i+j)^3}{5}$
Answer$
\begin{aligned}
& a _{ ij }=\frac{(i+j)^3}{5} \\
& \therefore a 11=\frac{(1+1)^3}{5}=\frac{2^3}{5}=\frac{8}{5}, a_{12}=\frac{(1+2)^3}{5}=\frac{3^3}{5}=\frac{27}{5} \\
& a _{21}=\frac{(2+1)^3}{5}=\frac{3^3}{5}=\frac{27}{5}, a_{22}=\frac{(2+2)^2}{5}=\frac{4^3}{5}=\frac{64}{5} \\
& a _{31}=\frac{(3+1)^3}{5}=\frac{4^3}{5}=\frac{64}{5}, a_{32}=\frac{(3+2)^2}{5}=\frac{5^3}{5}=\frac{125}{5} \\
& \therefore A =\left[\begin{array}{cc}
\frac{8}{5} & \frac{27}{5} \\
\frac{27}{5} & \frac{64}{5} \\
\frac{64}{5} & \frac{125}{5}
\end{array}\right] \\
& =\left[\begin{array}{cc}
8 & 27 \\
27 & 64 \\
64 & 125
\end{array}\right] .
\end{aligned}
$
View full question & answer→Question 212 Marks
Construct a matrix $A =\left[ a _{ i j}\right]_{3 \times 2}$ whose elements aij isgiven by : $a_{ij} = i – 3j$
Answer$\begin{aligned} & a _{ ij }= i -3 j \\ & \therefore a _{11}=1-3(1)=1-3=-2 \\ & a _{12}=1-3(2)=1-6=-5 \\ & a _{21}=2-3(1)=2-3=-1 \\ & a _{22}=2-3(2)=2-6=-4 \\ & a _{31}=3-3(1)=3-3=0 \\ & a _{32}=3-3(2)=3-6=-3 \\ & \therefore A =\left[\begin{array}{cc}-2 & -5 \\ -1 & -4 \\ 0 & -3\end{array}\right]\end{aligned}$
View full question & answer→Question 222 Marks
Construct a matrix $A =\left[ a _{ i j}\right]_{3 \times 2}$ whose elements aij isgiven by : $a _{ ij }=\frac{(i-j)^2}{5-i}$
Answer$
A =\left[a_{i j}\right]_{3 \times 2}=\left(\begin{array}{ll}
a_{11} & a_{12} \\
a_{21} & a_{22} \\
a_{31} & a_{32}
\end{array}\right)
$
Now, $a_{i j}=\frac{(i-j)^2}{5-i}$
$
\begin{aligned}
\therefore a_{11} & =\frac{(1-1)^2}{5-1}=\frac{0}{4}=0 \\
a_{12} & =\frac{(1-2)^2}{5-1}=\frac{1}{4}
\end{aligned}
$
$
\begin{aligned}
a_{21} & =\frac{(2-1)^2}{5-2}=\frac{1}{3} \\
a_{22} & =\frac{(2-2)^2}{5-2}=\frac{0}{3}=0 \\
a_{31} & =\frac{(3-1)^2}{5-3}=\frac{4}{2}=2 \\
a_{32} & =\frac{(3-2)^2}{5-3}=\frac{1}{2} \\
\therefore A & =\left[\begin{array}{cc}
0 & \frac{1}{4} \\
\frac{1}{3} & 0 \\
2 & \frac{1}{2}
\end{array}\right]
\end{aligned}
$
View full question & answer→