Exercise 1.8-Additional Problems - Chapter 1 Applications of Matrices and Determinants 12th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Additional Problems
Question 1.
$\mathrm{A}=\left[a_{i j}\right]_{m \times n}$ is a square matrix if .
(a) $\mathrm{m}<\mathrm{n}$ (b) $\mathrm{m}>\mathrm{n}$
(c) $\mathrm{m}=\mathrm{n}$
(d) None of these
Solution:

(c) $m=n$
Hint:
$\mathrm{A}=\left[\mathrm{a}_{\mathrm{ij}}\right]_{\mathrm{m} \times \mathrm{n}}$ is a square matrix if number of rows is equal to that of columns, i.e., $\mathrm{m}=\mathrm{n}$.
Question 2.
Matrices $\mathrm{A}$ and $\mathrm{B}$ will be inverse of each other only if
(a) $\mathrm{AB}=\mathrm{BA}$
(b) $\mathrm{AB}=\mathrm{BA}=\mathrm{O}$
(c) $\mathrm{AB}=\mathrm{O}, \mathrm{BA}=\mathrm{I}$
(d) $\mathrm{AB}=\mathrm{BA}=\mathrm{I}$
Solution:
(d) $\mathrm{AB}=\mathrm{BA}=\mathrm{I}$
Hint:
By the definition of invertible matrices two matrices $A$ and $B$ are inverse of each other if $A B=B A=I$.
Question 3.
If $\mathrm{A}$ is an invertible matrix of order 2 then $\operatorname{det}\left(\mathrm{A}^{-1}\right)$ is equal to
(a) $\operatorname{det}$ (A)
(b) $\frac{1}{\operatorname{det}(\mathrm{A})}$
(c) 1
(d) 0
Solution:
(b) $\frac{1}{\operatorname{det}(\mathrm{A})}$
Hint.
We know that $\mathrm{A} \mathrm{A}^{-1}=\mathrm{I}$
$
\begin{array}{lll}
\text { We know that } & \mathrm{AA}^{-1}=\mathrm{I} & \therefore\left|\mathrm{AA}^{-1}\right|=|\mathrm{I}| \\
\Rightarrow & |\mathrm{A}|\left|\mathrm{A}^{-1}\right|=1 & \Rightarrow\left|\mathrm{A}^{-1}\right|=\frac{1}{|\mathrm{~A}|}
\end{array}
$
Question 4.
Given $A=\left(\begin{array}{cc}1 & -2 \\ -5 & 7\end{array}\right)$ then $A(\operatorname{adj} A)=$
$(a)\left(\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right)$
(b) $\frac{1}{17}\left(\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right)$
(c) $\frac{1}{3}\left(\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right)$
(d) $\frac{1}{-3}\left(\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right)$
Solution:
(d) $\frac{1}{-3}\left(\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right)$

Question 5.
The inverse of $\left(\begin{array}{ll}2 & 1 \\ 4 & 2\end{array}\right)$ is
(a) $\frac{1}{2}\left(\begin{array}{cc}2 & -1 \\ -4 & 2\end{array}\right)$
(b) $\left(\begin{array}{cc}2 & -1 \\ -4 & 2\end{array}\right)$
(c) $\left(\begin{array}{ll}2 & 1 \\ 4 & 2\end{array}\right)$
(d) Inverse does not exist
Solution:
(d) Inverse does not exist
Question 6.
Given $\rho(A, B)=\rho(A)=$ number of unknowns, then the system has $\ldots . .$.
(a) unique solution
(b) no solution
(c) inconsistent
(d) infinitely many solution
Solution:
(a) unique solution
Question 7.
Given $\rho(A, B) \neq \rho(A)$ then the system has
(a) no solution
(b) unique solution
(c) infinitely many solution
(d) None
Solution:
(a) no solution
Question 8.
Given $\rho(A, B)=\rho(A)<$ number of unknowns, then the system has
(a) unique solution
(b) no solution
(c) 3 solutions
(d) infinitely many solution
Solution:
(d) infinitely many solution

Question 9.
The rank of the matrix $\left[\begin{array}{ccc}1 & 2 & 3 \\ 2 & 4 & 6 \\ 3 & 6 & 9\end{array}\right]$ is
(a) 1
(b) 2
(c) 3
(d) None of these
Solution:
(a) 1
Question 10.
If the rank of the matrix $\left(\begin{array}{ccc}1 & 0 & -1 \\ 2 & 5 & \lambda \\ 0 & 0 & 7\end{array}\right)$ is 3 , thęen the value of $\lambda$ is
(a) 1
(b) 0
(c) 4
(d) any real number
Solution:
(d) any real number
Question 11.
If $\mathrm{A}=\left[\begin{array}{lll}2 & 0 & 1\end{array}\right]$, then the rank of $\mathrm{AA}^{\mathrm{T}}$ is $\ldots \ldots \ldots$
(a) 1
(b) 2
(c) 3
(d) 0
Solution:
(a) 1
Hint.
$
\begin{gathered}
\mathrm{A}=\left[\begin{array}{lll}
2 & 0 & 1
\end{array}\right] ; \quad A^{\mathrm{T}}=\left[\begin{array}{l}
2 \\
0 \\
1
\end{array}\right] \\
\mathrm{AA}^{\mathrm{T}}=\left[\begin{array}{lll}
2 & 0 & 1
\end{array}\right]\left[\begin{array}{l}
2 \\
0 \\
1
\end{array}\right]=[4+0+1]=[5] ; \text { Rank of } \mathrm{AA}^{\mathrm{T}}=1
\end{gathered}
$

Question 12.
$
\left[\begin{array}{ccc}
\lambda & -1 & 0 \\
0 & \lambda & -1 \\
-1 & 0 & \lambda
\end{array}\right]
$
If the
(b) 2
(c) 3
(d) any real number
Solution:
(a) 1
Hint.
Since the rank of the given matrix is 2 , the value of $3^{\text {rd }}$ order determinant is zero.
$
\begin{aligned}
& \text { Now }\left|\begin{array}{ccc}
\lambda & -1 & 0 \\
0 & \lambda & -1 \\
-1 & 0 & \lambda
\end{array}\right|=\lambda\left(\lambda^2\right)+1(-1)=(\lambda)^3-1 \\
& \lambda^3-1=0 \quad \Rightarrow(\lambda-1)\left(\lambda^2+\lambda+1\right)=0 \\
& \lambda=1 \quad \text { or } \lambda^2+\lambda+1=0 \quad \therefore \lambda=1 \\
&
\end{aligned}
$
Question 13.
If $\mathrm{A}$ is a scalar matrix with scalar $\mathrm{k} \neq 0$, of order 3 , then $\mathrm{A}^{-1}$ is ......
(a) $\frac{1}{k^2} \mathrm{I}$
(b) $\frac{1}{k^3} \mathrm{I}$
(c) $\frac{1}{k} \mathrm{I}$
(d) $k \mathrm{I}$
Solution:
$
\frac{1}{k} \mathrm{I}
$

Hint:
$
\begin{aligned}
k^3 \mathrm{~A} & =\left[\begin{array}{ccc}
k & 0 & 0 \\
0 & k & 0 \\
0 & 0 & k
\end{array}\right] \quad|\mathrm{A}|=\left[\begin{array}{lll}
k & 0 & 0 \\
0 & k & 0 \\
0 & 0 & k
\end{array} \mid=k^3 ; \mathrm{A}^{\mathrm{T}}=\left[\begin{array}{ccc}
k & 0 & 0 \\
0 & k & 0 \\
0 & 0 & k
\end{array}\right]\right. \\
\operatorname{adj~} \mathrm{A} & =\left[\begin{array}{ccc}
k^2 & 0 & 0 \\
0 & k^2 & 0 \\
0 & 0 & k^2
\end{array}\right] \\
\therefore \quad \mathrm{A}^{-1} & =\frac{1}{|\mathrm{~A}|} \text { adj } \mathrm{A}=\frac{1}{k^3}\left[\begin{array}{ccc}
k^2 & 0 & 0 \\
0 & k^2 & 0 \\
0 & 0 & k^2
\end{array}\right]=\frac{1}{k}\left[\begin{array}{ccc}
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]=\frac{1}{k} \mathrm{I}
\end{aligned}
$

Question 14.
If $\mathrm{A}=\left[\begin{array}{ll}2 & 1 \\ 3 & 4\end{array}\right]$, then $(\operatorname{adj} \mathrm{A}) \mathrm{A}$
(a) $\left[\begin{array}{cc}\frac{1}{5} & 0 \\ 0 & \frac{1}{5}\end{array}\right]$
(b) $\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
(c) $\left[\begin{array}{cc}5 & 0 \\ 0 & -5\end{array}\right]$
(d) $\left[\begin{array}{ll}5 & 0 \\ 0 & 5\end{array}\right]$
Solution:
(d) $\left[\begin{array}{ll}5 & 0 \\ 0 & 5\end{array}\right]$
Hint:
$(\operatorname{adj} \mathrm{A})(\mathrm{A})=|\mathrm{A}| \mathrm{I}$
$|\mathrm{A}|=\left|\begin{array}{ll}2 & 1 \\ 3 & 4\end{array}\right|=8-3=5 \quad \therefore(\operatorname{adj~A}) \mathrm{A}=5 \mathrm{I}=5\left(\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right)=\left(\begin{array}{ll}5 & 0 \\ 0 & 5\end{array}\right)$
Question 15.
If $\mathrm{A}$ is a square matrix of order $\mathrm{n}$, then $|\operatorname{adj} \mathrm{A}|$ is ......
(a) $|\mathrm{A}|^2$

(b) $|A|^n$
(c) $|\mathrm{A}|^{\mathrm{n}-1}$
(d) $|\mathrm{A}|$
Solution:
(c) $|\mathrm{A}|^{\mathrm{n}-1}$
Hint:
When $\mathrm{A}$ is square matrix of order 3 , then $|\operatorname{adj} \mathrm{A}|=|\mathrm{A}|^2=|\mathrm{A}|^{3-1} \therefore$ When $\mathrm{A}$ is a square matrix of order $\mathrm{n}$, $|\operatorname{adj} \mathrm{A}|=|\mathrm{A}|^{\mathrm{n}-1}$
Question 16.
If $\mathrm{A}$ is a matrix of order 3 , then $\operatorname{det}(\mathrm{kA})$
(a) $\mathrm{k}^3(\operatorname{det} \mathrm{A})$
(b) $\mathrm{k}^2(\operatorname{det} \mathrm{A})$
(c) $\mathrm{k}(\operatorname{det} \mathrm{A})$
(d) $\operatorname{det}$ (A)
Solution:
(a) $\mathrm{k}^3(\operatorname{det} \mathrm{A})$
Hint:
$\mathrm{A}$ is a matrix of order $3 . \therefore \operatorname{det}(\mathrm{kA})=\mathrm{k}^3(\operatorname{det} \mathrm{A})$
Question 17.
If $\mathrm{I}$ is the unit matrix of order $\mathrm{n}$, where $\mathrm{k} \pm 0$ is a constant, then adj(kI)=
(a) $\mathrm{k}^{\mathrm{n}}$ (adj I)
(b) $k(\operatorname{adj}$ I)
(c) $\mathrm{k}^2$ (adj I)
(d) $\mathrm{k}^{\mathrm{n}-1}$ (adj I)
Solution:
(d) $\mathrm{k}^{\mathrm{n}-1}$ (adj I)
Hint:
When $I$ is a unit matrix of order 3 , then adj $(\mathrm{kI})=\mathrm{k}^2$ (adj $\left.\mathrm{I}\right)$
$\therefore$ When $\mathrm{I}$ is a unit matrix of order $\mathrm{n}$, then adj $(\mathrm{kI})=\mathrm{k}^{\mathrm{n}-1}$ (adj $\mathrm{I}$ )
Question 18.
In a system of 3 linear non-homogeneous equations with three unknowns, if $\Delta=0$ and $\Delta x=0, \Delta y \neq 0$ and $\Delta z=0$, then the system has .....
(a) unique solution
(b) two solutions
(c) infinitely many solutions
(d) no solution
Solution:
(d) no solution
Hint:
When $\Delta=0 \Delta \mathrm{x}, \Delta \mathrm{z}=0$ but $\Delta \mathrm{y} \neq 0 \Rightarrow$ that the system is inconsistent.

$\therefore \text { There is no solution. }$