Matrices And Determinants - Types, Adjoint and Inverse of a Matrix (Lecture-03)
Matrices:
A rectangular arrangement of numbers is rows & columns is called a matrix.
A matrix of order mxn contains mn elements. A matrix of order mxn is of the form
$A=\left[\begin{array}{cccc} a _{11} & a _{12} & \ldots \ldots \ldots & a _{1 n} \\ a _{21} & a _{22} & \ldots \ldots \ldots & a _{2 n} \\ \vdots & & & \\ a _{m 1} & a _{m 2} & \ldots \ldots . . & a _{m n} \end{array}\right]=\left[a _{i j}\right] _{m \times n}$
If $m=1$, the matrix is a row matrix. If $n=1$ the matrix is a column matrix.
Equality: If $A \& B$ are of same order, then $A=B$ if $a _{i j}=b _{i j} \forall i \& j$
Types of matrices:
- Null (zero) matrix
- Square matrix
- Diagonal matrix
$\hspace {0.7 cm}\left(\text { If } \mathrm{A}=\operatorname{diag}\left(\mathrm{d} _{1}, \mathrm{~d} _{2} \ldots \ldots . \mathrm{d} _{\mathrm{n}}\right) \text {, then } \mathrm{A}^{\mathrm{n}}=\left(\mathrm{d} _{1}^{\mathrm{n}}, \mathrm{d} _{2}^{\mathrm{n}} \ldots \ldots . \mathrm{d} _{\mathrm{n}}^{\mathrm{n}}\right)\right)$
- Identity matrix
- Triangular matrix
- (Determinant of upper triangular or lower triangular matrix is the product of principal diagonal elements). Also minimum number of zeros in a triangular matrix is given by $\dfrac{\mathrm{n}(\mathrm{n}-1)}{2}$ where ’ $n$ ’ is the order of matrix.
Properties of Matrix Multiplication
i. $\mathrm{AB} \neq \mathrm{BA}$
ii. $\mathrm{A}(\mathrm{BC})=(\mathrm{AB}) \mathrm{C}$
iii. $A(B+C)=A B+A C$
iv. $\mathrm{AI}=\mathrm{A}=\mathrm{IA}$
v. $\mathrm{AB}=\mathrm{AC}$ need not imply $\mathrm{B}=\mathrm{C}$
vi. $\mathrm{AB}=0$ need not imply $\mathrm{A}=0$ or $\mathrm{B}=0$
vii. $\mathrm{I}^{2}=\mathrm{I}^{3}$ $\mathrm{I}^{\mathrm{m}}=\mathrm{I}(\mathrm{m} \in \mathrm{z})$
Trace (spur) a Matrix
Sum of diagonal elements of a square matrix is called the trace of matrix A.
i.e. $\operatorname{tr} A=\sum\limits _{\mathrm{i}=1}^{\mathrm{n}} \mathrm{a} _{\mathrm{ii}}=\mathrm{a} _{11}+\mathrm{a} _{22}+\ldots \ldots . .+\mathrm{a} _{\mathrm{nn}}$
Trace of a skew symmetric matrix is zero.
Properties:
Let $A=\left[a _{i j}\right] _{\text {nxn }}, B=\left[b _{i j}\right] _{\text {nxn }} \& \lambda$ is a scalar
-
$\operatorname{tr}(\mathrm{A} \pm \mathrm{B})=\operatorname{tr}(\mathrm{A}) \pm \operatorname{tr}(\mathrm{B})$
-
$\operatorname{tr}(A B)=\operatorname{tr}(B A)(\operatorname{tr}(A B) \neq \operatorname{tr}(A) \cdot \operatorname{tr}(B))$
-
$\operatorname{tr}\left(\mathrm{A}^{\mathrm{T}}\right)=\operatorname{tr}(\mathrm{A})$
-
$\operatorname{tr}(\lambda \mathrm{A})=\lambda \operatorname{tr}(\mathrm{A})$
-
$\operatorname{tr}\left(\mathrm{I} _{\mathrm{n}}\right)=\mathrm{n}$
Transpose of a Matrix
the matrix obtained by interchanging the rows and columns of the given matrix a is called transpose of A.
If $A$ is of order mxn, then $\mathrm{A}^{\mathrm{T}}$ is of order nxm.
Properties:
- $\left(A^{T}\right)^{T}=A$
- $(A \pm B)^{T}=A^{T} \pm B^{T}$
- $(\lambda \mathrm{A})^{\mathrm{T}}=\lambda \mathrm{A}^{\mathrm{T}}$, where $\lambda$ is a scalar
- $(\mathrm{AB})^{\mathrm{T}}=\mathrm{B}^{\mathrm{T}} \mathrm{A}^{\mathrm{T}}$ (reversal law of transposes) (if $\mathrm{A} \& \mathrm{~B}$ are conformable for multiplication) Also, $(\mathrm{ABC})^{\mathrm{T}}=\mathrm{C}^{\mathrm{T}} \mathrm{B}^{\mathrm{T}} \mathrm{A}^{\mathrm{T}}$ etc.
- $\mathrm{I}^{\mathrm{T}}=\mathrm{I}$
Conjugate of a Matrix
Conjugate of a matrix A is obtained by replacing the elements of A by their corresponding complex conjugates. It is denoted by $\bar{A}$.
Properties:
-
$(\overline{\mathrm{A}})=\mathrm{A}$
-
$(\overline{\mathrm{A}+\mathrm{B}})=\overline{\mathrm{A}}+\overline{\mathrm{B}}$
-
$(\overline{\mathrm{AB}})=\overline{\mathrm{A}} \cdot \overline{\mathrm{B}}$
-
$(\overline{\mathrm{kA}})=\overline{\mathrm{k}} \cdot \overline{\mathrm{A}}$
-
$\left(\overline{\mathrm{A}^{\mathrm{n}}}\right)=(\overline{\mathrm{A}})^{\mathrm{n}}$
Tranjugate (Transposed conjugate of a Matrix)
Transposed conjugate is obtained by interchanging the rows and columns of the matrix obtained by replacing the elements of A by their corresponding complex conjugate. It is denoted by $\mathrm{A}^{*}$.
Properties:
- $\left(\mathrm{A}^{\ast}\right)^{*}=\mathrm{A}$
- $(A+B)^{\ast}=A^{\ast}+B^{\ast}$
- $(\mathrm{AB})^{\ast}=\mathrm{B}^{\ast} \mathrm{~A}^{\ast}$
- $(\mathrm{kA})^{\ast}=\overline{\mathrm{k}} \mathrm{A}^{\ast}$
- $\left(A^{n}\right)^{\ast}=\left(A^{\ast}\right)^{n}$
Symmetric and skew symmetric matrices:
A square matrix $\mathrm{A}$ is called symmetric if $\mathrm{A}^{\mathrm{T}}=\mathrm{A}$ and skew symmetric if $\mathrm{A}^{\mathrm{T}}=-\mathrm{A}$. All principal diagonal elements of a skew-sym metric matrix are zero.
Properties:
- If $\mathrm{A}$ is a square matrix, then $\mathrm{A}+\mathrm{A}^{\mathrm{T}}, \mathrm{AA}^{\mathrm{T}}, \mathrm{A}^{\mathrm{T}} \mathrm{A}$ are symmetric matrices, while $\mathrm{A}-\mathrm{A}^{\mathrm{T}}$ is skew-symmetric matrix.
- If $\mathrm{A}$ is symmetric matrix, then - $\mathrm{A}, \mathrm{KA}, \mathrm{A}^{\mathrm{T}}, \mathrm{A}^{\mathrm{n}}, \mathrm{A}^{-1}, \mathrm{~B}^{\mathrm{T}} \mathrm{AB}$ are also symmetric matrices where $B$ is a square matrix of same order that of $A$.
- If $A$ is a skew symmetric matrix, then $A^{2 n}$ is symmetric where as $A^{2 n+1}$ and $B^{T} A B$ are skew symmetric ( $n \in N \& B$ is a square matrix of same order that of $A$ )
- If $A \& B$ are two symmetric matrices, then $A \pm B, A B+B A$ and $A B-B A$ are skew symmetric.
- If $\mathrm{A} \& \mathrm{~B}$ are two skew symmetric matrices, then $\mathrm{A} \pm \mathrm{B}, \mathrm{AB}-\mathrm{BA}$ are skew symmetric and $\mathrm{AB}+\mathrm{BA}$ is symmetric.
- Every square matrix can be uniquely expressed as sum of a symmetric and a skew symmetric matrix.
$\hspace {0.7 cm}\mathrm{A}=\dfrac{1}{2}\left(\mathrm{~A}+\mathrm{A}^{\mathrm{T}}\right)+\dfrac{1}{2}\left(\mathrm{~A}-\mathrm{A}^{\mathrm{T}}\right)=\mathrm{P}+\mathrm{Q}$. Here $\mathrm{P}$ is symmetric and $\mathrm{Q}$ is skew symmetric.
-
If $A$ is a skew symmetric matrix \& $C$ is a column matrix, then $C^{\mathrm{T}} \mathrm{AC}$ is a null matrix.
-
If $A$ is a skew symmetric matrix of odd order, then $\mathrm{A}^{-1}$ does not exists $(\because|\mathrm{A}|=0)$
-
Null matrix is both symmetric and skew symmetric
-
All elements on the principal diagonal of a skew-symmetric matrix are always zero.
Hermitian Skew-Hermitian matrix
A square matrix is said to be Hermitian if $\overline{\mathrm{A}^{\mathrm{T}}}=\mathrm{A}$ and skew-hermitian if $\overline{\mathrm{A}^{\mathrm{T}}}=-\mathrm{A}$.
Properties:
The diagonal elements of a Hermitian matrix are real where that of a skew-Hermitian matrix are either purely imaginary or zero.
- Every square matrix (with complex elements) can be uniquely expressed as the sum of Hermitian and skew-Hermitian matrices.
$\hspace {1 cm}\mathrm{A}=\underbrace{\dfrac{1}{2}\left(\mathrm{~A}+\overline{\mathrm{A}^{\mathrm{T}}}\right)} _{\text {Hermitian }}+\underbrace{\dfrac{1}{2}\left(\mathrm{~A}-\overline{\mathrm{A}^{\mathrm{T}}}\right)} _{\text {Skew-hermitian }}$
Orthogonal matrix
- A square matrix $A$ is called on orthogonal matrix if $\mathrm{AA}^{\mathrm{T}}=\mathrm{A}^{\mathrm{T}} \mathrm{A}=\mathrm{I}$.
- If $\mathrm{A}$ is orthogonal, then $|\mathrm{A}|= \pm 1$. Hence it is non-singular.
- If $A$ is orthogonal, it is invertible with $\mathrm{A}^{-1}=\mathrm{A}^{\mathrm{T}}$.
- If $\mathrm{A} \& \mathrm{~B}$ area orthogonal matrices of order $\mathrm{n}$, then $\mathrm{AB}, \mathrm{BA}, \mathrm{A}^{-1}, \mathrm{~A}^{\mathrm{T}}$ are orthogonal.
- If $A$ is orthogonal with $|A|=1$, then each element of $A$ is equal to its cofactor is $|A|$.
- If $A$ is orthogonal with $|A|=1$, then each element of $A$ is equal to the negative of its cofactor is $|\mathrm{A}|$.
- If $\mathrm{A} _{3 \times 3}$ is orthogonal and $\mathrm{B} _{3 \times 3}$ is a skew symmetric matrix, then $|\mathrm{AB}|=1$.
Unitary Matrix
A square matrix $A$ is called a unitary matrix. If $A \overline{A^{\mathrm{T}}}=\overline{\mathrm{A}^{\mathrm{T}}} \mathrm{A}=\mathrm{I}$.
Properties:
- Determinant of a unitary matrix is of unit modulus.
- If $\mathrm{A}$ is a unitary matrix, then $\mathrm{A}^{\mathrm{T}}, \overline{\mathrm{A}}, \overline{\mathrm{A}^{\mathrm{T}}}$ and $\mathrm{A}^{-1}$ are unitary.
- Product of two unit matrices is unitary.
Idempotent matrix
A square matrix is called idempotent if $\mathrm{A}^{2}=\mathrm{A}$.
Properties:
- If $\mathrm{A}$ is idempotent, then I-A is also idempotent.
- If $\mathrm{A}, \mathrm{B}$ are two idempotent matrices and $\mathrm{AB}=\mathrm{BA}=0$, then $(\mathrm{A}+\mathrm{B})$ is idempotent.
- If $\mathrm{AB}=\mathrm{A} ; \mathrm{BA}=\mathrm{B}$, then $\mathrm{A} \& \mathrm{~B}$ are idempotent matrices and $\mathrm{A}^{\mathrm{n}}+\mathrm{B}^{\mathrm{n}}=\mathrm{A}+\mathrm{B}$ where $\mathrm{n} \in \mathrm{N}$.
Periodic matrix
A square matrix $A$ is called periodic if $\mathrm{A}^{k+1}=\mathrm{A} ; \mathrm{k} \in \mathrm{Z}^{+}$. The least value of $\mathrm{k}$ is called period of A.
When $\mathrm{k}=1$, we get $\mathrm{A}^{2}=\mathrm{A}$ and it becomes an idempotent matrix.
Nilpotent Matrix
A square matrix $A$ is called Nilpotent of order $k$ if $A^{k}=0$ and $A^{k-1} \neq 0, k \in Z^{+}$. Here $k$ is called the order of the nilpotent matrix $A$.
Involutory Matrix
A square matrix $\mathrm{A}$ is called involutory if $\mathrm{A}^{2}=\mathrm{I}$.
i.e. $\mathrm{A}^{-1}=\mathrm{A}(\mathrm{A}$ is the inverse of itself)
Adjoint of a square matrix
The transpose of the matrix of cofactors $\mathrm{C}$ is called the adjoint of matrix $\mathrm{A}$ and is denoted by adj A.
Properties:
For square matrices $A \& B$ of order $n$,
-
$A(\operatorname{adj} A)=(\operatorname{adj} A) A=|A| I _{n}$
-
$\operatorname{adj}(\mathrm{AB})=(\operatorname{adjB})(\operatorname{adj} A)$
-
$\left(\operatorname{adj} A^{T}\right)=(\operatorname{adj} A)^{\mathrm{T}}$
-
$\left(\operatorname{adj} A^{m}\right)=(\operatorname{adj} A)^{m} ; m \in N$
-
$\operatorname{adj}(k A)=k^{n-1}(\operatorname{adj} A) ; k \in R$
-
Adjoint of a diagonal matrix is a diagonal matrix.
-
$|\operatorname{adj} \mathrm{A}|=|\mathrm{A}|^{\mathrm{n}-1}$
-
$\operatorname{adj}(\operatorname{adj} A)=|A|^{n-2} A ;|A| \neq 0$
-
$|\operatorname{adj}(\operatorname{adj} A)|=|A|^{(\mathrm{n}-1)^{2}} ;|\mathrm{A}| \neq 0$
Inverse of a matrix
For a non singular matrix A of order $\mathrm{n}, \mathrm{A}^{-1}=\dfrac{1}{|\mathrm{~A}|}(\operatorname{adj} \mathrm{A})$
Properties:
- $\left(\mathrm{A}^{-1}\right)^{-1}=\mathrm{A}$
- $\left(A^{T}\right)^{-1}=\left(A^{-1}\right)^{T}$
- $\left(\operatorname{adj} \mathrm{A}^{-1}\right)=(\operatorname{adj} \mathrm{A})^{-1}$
- $\left|\mathrm{A}^{-1}\right|=\dfrac{1}{|\mathrm{~A}|}=|\mathrm{A}|^{-1}$
- If $\mathrm{A}=\operatorname{diag}\left(\mathrm{a} _{11}, \mathrm{a} _{22} \ldots . . \mathrm{a} _{\mathrm{nn}}\right)$
$\hspace {0.7 cm}\mathrm{A}^{-1}=\operatorname{diag}\left(\mathrm{a} _{11}^{-1}, \mathrm{a} _{22}^{-1}, \ldots \mathrm{a} _{\mathrm{nn}}^{-1}\right)$
-
$(\mathrm{AB})=\mathrm{B}^{-1} \mathrm{~A}^{-1}$ (reversal law)
-
$\mathrm{AB}=\mathrm{AC} \Rightarrow \mathrm{A}=\mathrm{C}$ if $|\mathrm{A}| \neq 0$.
Solved Examples
1. If $A=\left[\begin{array}{ll}\alpha & 2 \\ 2 & \alpha\end{array}\right]$ and $\left|A^{3}\right|=125$, then the value of $\alpha$ is
(a) $\pm 1$
(b) $\pm 2$
(c) $\pm 3$
(d) $\pm 5$
Show Answer
Solution: $|A|=\left|\begin{array}{ll}\alpha & 2 \ 2 & \alpha\end{array}\right|=\alpha^{2}-4$
Also $\left|A^{3}\right|=125 \Rightarrow|A|^{3}=125$ gives $\left(\alpha^{2}-4\right)^{3}=5^{3}$
$\Rightarrow \alpha^{2}-4=5 \Rightarrow \alpha^{2}=9 \Rightarrow \alpha^{2}= \pm 3$.
Answer: c
2. If $\mathrm{A}=\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]$, then $\mathrm{I}+\mathrm{A}+\mathrm{A}^{2}+\mathrm{A}^{3}+—-\infty$ equals to
(a) $\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
(b) $\left[\begin{array}{rr}-1 & -2 \\ -3 & -4\end{array}\right]$
(c) $\left[\begin{array}{cc}\dfrac{1}{2} & -\dfrac{1}{3} \\ -\dfrac{1}{2} & 0\end{array}\right]$
(d) none of these
Show Answer
Solution: Let $\mathrm{B}=\mathrm{I}+\mathrm{A}+\mathrm{A}^{2}+\mathrm{A}^{3}+—–\infty$
$\Rightarrow \quad \mathrm{AB}=\mathrm{A}+\mathrm{A}^{2}+\mathrm{A}^{3}+—–\infty$
$\Rightarrow \quad \mathrm{B}-\mathrm{AB}=\mathrm{I}$
$B(I-A)=I$
$\Rightarrow \quad \mathrm{B}=(\mathrm{I}-\mathrm{A})^{-1}$
$\therefore \quad B=\left[\begin{array}{cc}0 & -2 \\ -3 & -3\end{array}\right]^{-1}=-\dfrac{1}{6}\left[\begin{array}{cc}-3 & 2 \\ 3 & 0\end{array}\right]=\left[\begin{array}{cc}\dfrac{1}{2} & -\dfrac{1}{3} \\ \dfrac{-1}{2} & 0\end{array}\right]$
Answer: c
3. If $\mathrm{A}$ is non-singular and $(\mathrm{A}-2 \mathrm{I})(\mathrm{A}-4 \mathrm{I})=0$, then $\dfrac{1}{6} \mathrm{~A}+\dfrac{4}{3} \mathrm{~A}^{-1}=$
(a) $\mathrm{I}$
(b) 0
(c) $2 \mathrm{I}$
(d) $6 \mathrm{I}$
Show Answer
Solution: $(A-2 I)(A-4 I)=0$
$\begin{array}{ll} \Rightarrow & A^{2}-2 A-4 A+8 I=0 \\ \Rightarrow & A^{2}-6 A+8 I=0 \\ & A^{-1}\left(A^{2}-6 A+8 I\right)=A^{-1} 0\left(\text { Pre multiply by } A^{-1}\right) \\ & A^{-1} A^{2}-6 A^{-1}+8 A^{-1} \mathrm{I}=0 \\ \Rightarrow & \mathrm{A}-6 \mathrm{I}+8 \mathrm{~A}^{-1}=0 \\ \Rightarrow & \mathrm{A}+8 \mathrm{~A}^{-1}=6 \mathrm{I} \\ \Rightarrow & \dfrac{1}{6} \mathrm{~A}+\dfrac{4}{3} \mathrm{~A}^{-1}=\mathrm{I} \end{array}$
Answer: a
4. If $A$ and $B$ are square matrices such that $B=-A^{-1} B A$, then
(a) $A B+B A=0$
(b) $(A+B)^{2}=A^{2}+B^{2}$
(c) $(A+B)^{2}=A^{2}+2 A B+B^{2}$
(d) $(A+B)^{2}=A+B$
Show Answer
Solution: $B=-A^{-1} B A \Rightarrow A B=-\left(A A^{-1}\right)(B A) \Rightarrow A B=-I B A$
$\Rightarrow \quad \mathrm{AB}=-\mathrm{BA} \Rightarrow \mathrm{AB}+\mathrm{BA}=0$
Now $(A+B)^{2}=(A+B)(A+B)=A^{2}+A B+B A+B^{2}=A^{2}+B^{2}$ $\hspace {2 cm}(\therefore \mathrm{AB}+\mathrm{BA}=0)$
Answer: a, b
5. If $A=\dfrac{1}{3}\left[\begin{array}{ccc}1 & 2 & 2 \\ 2 & 1 & -2 \\ \mathrm{a} & 2 & \mathrm{~b}\end{array}\right]$ is an orthogonal matrix, then
(a) $\mathrm{a}=2, \mathrm{~b}=1$
(b) $\mathrm{a}=-2, \mathrm{~b}=-1$
(c) $\mathrm{a}=2, \mathrm{~b}=-1$
(d) $\mathrm{a}=-2, \mathrm{~b}=1$
Show Answer
Solution: $\because \mathrm{AA}^{\mathrm{T}}=\mathrm{I}$
$\Rightarrow \quad \dfrac{1}{3}\left[\begin{array}{ccc}1 & 2 & 2 \\ 2 & 1 & -2 \\ \mathrm{a} & 2 & \mathrm{~b}\end{array}\right] \dfrac{1}{3}\left[\begin{array}{ccc}1 & 2 & \mathrm{a} \\ 2 & 1 & 2 \\ 2 & -2 & \mathrm{~b}\end{array}\right]=\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
$\Rightarrow \quad \left[\begin{array}{ccc}9 & 0 & \mathrm{a}+4+2 \mathrm{~b} \\ 0 & 9 & 2 \mathrm{a}+2-2 \mathrm{~b} \\ \mathrm{a}+4+2 \mathrm{~b} & 2 \mathrm{a}+2-2 \mathrm{~b} & \mathrm{a}^{2}+4+\mathrm{b}^{2}\end{array}\right]=\left[\begin{array}{ccc}9 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 9\end{array}\right]$
$\Rightarrow \quad \mathrm{a}+4+2 \mathrm{~b}=0,2 \mathrm{a}+2-2 \mathrm{~b}=0 \& \mathrm{a}^{2}+4+\mathrm{b}^{2}=9$ gives $\mathrm{a}=-2, \mathrm{~b}=-1$.
Answer: b
6. If $\mathrm{P}=\left[\begin{array}{cc}\dfrac{\sqrt{3}}{2} & \dfrac{1}{2} \\ \dfrac{-1}{2} & \dfrac{\sqrt{3}}{2}\end{array}\right], \mathrm{A}=\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]$ and $\mathrm{Q}=\mathrm{PAP}^{\mathrm{T}}$ then $\mathrm{P}^{\mathrm{T}} \mathrm{Q}^{2005} \mathrm{P}=$
(a) $\left[\begin{array}{cc}1 & 2005 \\ 0 & 1\end{array}\right]$
(b) $\left[\begin{array}{cc}4+2005 \sqrt{3} & 6015 \\ 2005 & 4-2005 \sqrt{3}\end{array}\right]$
(c) $\dfrac{1}{4}\left[\begin{array}{cc}1 & 2005 \\ 0 & 1\end{array}\right]$
(d) none of these
Show Answer
Solution: Here $\mathrm{P}$ is an orthogonal matrix
$\therefore \quad \mathrm{PP}^{\mathrm{T}}=\mathrm{P}^{\mathrm{T}} \mathrm{P}=\mathrm{I}$
Now $P^{\mathrm{T}} \cdot \mathrm{Q}^{5005} \cdot \mathrm{P}=\mathrm{P}^{\mathrm{T}}$.Q.Q.Q.Q…….Q.P.
$\begin{array}{ll} = & \mathrm{P}^{\mathrm{T}}\left(\mathrm{PAP}^{\mathrm{T}}\right)\left(\mathrm{PAP}^{\mathrm{T}}\right)\left(\mathrm{PAP}^{\mathrm{T}}\right)——–\left(\mathrm{PAP}^{\mathrm{T}}\right) \mathrm{P} \\ = & \left(\mathrm{P}^{\mathrm{T}} \mathrm{P}\right) \mathrm{A}\left(\mathrm{P}^{\mathrm{T}}\right) \mathrm{A}\left(\mathrm{P}^{\mathrm{T}} \mathrm{P}\right) \mathrm{A}——-\left(\mathrm{P}^{\mathrm{T}} \mathrm{P}\right) \mathrm{A}\left(\mathrm{P}^{\mathrm{T}} \mathrm{P}\right) \\ = & \text { I.A.I.A.I.A.———– I.AI } \\ = & \text { A.A.A.——. } 2005 \text { times } \\ = & \mathrm{A}^{2005} \end{array}$
Now $A^{2}=A \cdot A=\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]=\left[\begin{array}{ll}1 & 2 \\ 0 & 1\end{array}\right]$
$\mathrm{A}^{3}=\mathrm{A}^{2} \cdot \mathrm{A}=\left[\begin{array}{ll}1 & 2 \\ 0 & 1\end{array}\right]\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]=\left[\begin{array}{ll}1 & 3 \ 0 & 1\end{array}\right]$
$\therefore \quad \mathrm{A}^{2005}=\left[\begin{array}{cc}1 & 2005 \\ 0 & 1\end{array}\right]$
Answer: a
7. If $A=\left[\begin{array}{cc}2 & 1 \\ -4 & -2\end{array}\right]$, then the value of $I+2 A+3 A^{2}+——–\infty \infty$ is
(a) $\left[\begin{array}{cc}4 & 1 \\ -4 & 0\end{array}\right]$
(b) $\left[\begin{array}{cc}3 & 1 \\ -4 & -1\end{array}\right]$
(c) $\left[\begin{array}{cc}5 & 2 \\ -8 & -3\end{array}\right]$
(d) none of these
Show Answer
Solution: $\mathrm{A}^{2}=\left[\begin{array}{cc}2 & 1 \\ -4 & -2\end{array}\right]\left[\begin{array}{cc}2 & 1 \\ -4 & -2\end{array}\right]=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]=0$
$\therefore \quad \mathrm{I}+2 \mathrm{~A}+3 \mathrm{~A}^{2}+4 \mathrm{~A}^{3}$——- $\infty=\mathrm{I}+2 \mathrm{~A}\left(\therefore \mathrm{A}^{2}=0, \mathrm{~A}^{3}=\mathrm{A}^{4}=\ldots \ldots .=0\right)$
$=\left[\begin{array}{cc}5 & 2 \\ -8 & -3\end{array}\right]$
Answer: c
Exercise
1. Consider an arbitrary $3 \times 3$ matrix $A=\left[a _{i j}\right]$, a matrix $B=\left[b _{i j}\right]$ is formed such that $b _{i j}$ is the sum of all the elements except $a _{\mathrm{ij}}$ in the $\mathrm{i} _{\text {th }}$ row of $A$. If there exists a matrix $X$ with constant elements such that $\mathrm{AX}=\mathrm{B}$, then $\mathrm{X}$ is
(a) skew symmetric
(b) null matrix
(c) diagonal matrix
(d) none of these
Show Answer
Answer: d2. Let A be a square matrix all of whose entries are integers. Then which one of the following is true?
(a) If $|\mathrm{A}|= \pm 1$, then $\mathrm{A}^{-1}$ exists but all its entries are not necessarily integers.
(b) If $|A|= \pm 1$, then $\mathrm{A}^{-1}$ exists and all its entries are non integers.
(c) If $|\mathrm{A}|= \pm 1$, then $\mathrm{A}^{-1}$ exists and all its entries are integers.
(d) If $|A|= \pm 1$, then $A^{-1}$ need not exist.
Show Answer
Answer: c3. $\mathrm{X}, \mathrm{Y} \& \mathrm{Z}$ are positive numbers greater than 10 such, that $\mathrm{Y}$ and $\mathrm{Z}$ have respectively $1 \& 0$ at their unit’s place and $\Delta$ is the determinant $\left|\begin{array}{lll}\mathrm{X} & 4 & 1 \\ \mathrm{Y} & 0 & 1 \\ \mathrm{Z} & 1 & 0\end{array}\right|$. If $(\Delta+1)$ is divisible by 10 then $\mathrm{X}$ has its unit’s place
(a) 1
(b) 0
(c) 2
(d) none of these
Show Answer
Answer: c4. If $\mathrm{P}$ is non singular matrix, then value of $\operatorname{adj}\left(\mathrm{P}^{-1}\right)$ in terms of $\mathrm{P}$ is
(a) $\dfrac{\mathrm{P}}{|\mathrm{P}|}$
(b) $\mathrm{P}|\mathrm{P}|$
(c) $\mathrm{P}$
(d) none of these
Show Answer
Answer: a5. If $\mathrm{A}=\left[\begin{array}{ll}\mathrm{a} & \mathrm{b} \\ 0 & \mathrm{a}\end{array}\right]$ is $\mathrm{n}^{\text {th }}$ root of $\mathrm{I} _{2}$ then choose the correct statements.
i. if $\mathrm{n}$ is odd, $\mathrm{a}=1, \mathrm{~b}=0$
ii. if $\mathrm{n}$ is odd, $\mathrm{a}=-1, \mathrm{~b}=0$
iii. if $\mathrm{n}$ is even, $\mathrm{a}=1, \mathrm{~b}=0$
iv. if $\mathrm{n}$ is even $\mathrm{a}=-1, \mathrm{~b}=0$
(a) i, ii, iii
(b) ii, iii, iv
(c) i,ii,iii
(d) i, iii,iv
Show Answer
Answer: d6. If $A^{2}=I$, then the value of $|A-I|$ (where $A$ has order 3$)$
(a) 1
(b) -1
(c) 0
(d) cannot say anything
Show Answer
Answer: d7. If $A=\left[\begin{array}{ll}1 & 2 \\ 2 & 1\end{array}\right] \& f(x)=\dfrac{1+x}{1-x}$ then $f(A)$ is
(a) $\left[\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right]$
(b) $\left[\begin{array}{ll}2 & 2 \\ 2 & 2\end{array}\right]$
(c) $\left[\begin{array}{ll}-1 & -1 \\ -1 & -1\end{array}\right]$
(d) none of these
Show Answer
Answer: c8. If $\mathrm{f}(\mathrm{x})$ satisfies the $\left|\begin{array}{ccc}\mathrm{f}(\mathrm{x}-3) & \mathrm{f}(\mathrm{x}+4) & \mathrm{f}\left((\mathrm{x}+1)(\mathrm{x}-2)-(\mathrm{x}-1)^{2}\right) \\ 5 & 4 & -5 \\ 5 & 6 & 15\end{array}\right|=0 \forall$ equation real $\mathrm{x}$, then
(a) $\mathrm{f}(\mathrm{x})$ is not periodic
(b) $\mathrm{f}(\mathrm{x})$ is periodic and of period 1
(c) $\mathrm{f}(\mathrm{x})$ is periodic of period 7
(d) $\mathrm{f}(\mathrm{x})$ is an odd function
Show Answer
Answer: c9. Let $\mathrm{M} \& \mathrm{~N}$ be two $3 \times 3$ non-singular skew-symmetric matrices such that $\mathrm{MN}=\mathrm{NM}$. If $\mathrm{P}^{\mathrm{T}}$ denotes the transpose of $\mathrm{P}$, then $\mathrm{M}^{2} \mathrm{~N}^{2}(\mathrm{MN})^{-1}\left(\mathrm{M}^{\mathrm{T}} \mathrm{N}^{-1}\right)^{\mathrm{T}}$ is equals
(a) $\mathrm{M}^{2}$
(b) $-\mathrm{N}^{2}$
(c) $-\mathrm{M}^{2}$
(d) $\mathrm{MN}$
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Answer: c10. Read the following and answer the questions.
Let $\mathrm{p}$ be an odd prime number and $\mathrm{T} _{\mathrm{p}}$ be the following set of $2 \times 2$ matrices $\mathrm{T} _{\mathrm{P}}=\left\{\mathrm{A}=\left[\begin{array}{ll}\mathrm{a} & \mathrm{b} \\ \mathrm{c} & \mathrm{a}\end{array}\right] ; \mathrm{a}, \mathrm{b}, \mathrm{c} \in\{0,1,2, \ldots . \mathrm{p}-1\}\right\}$.
i. The number $\mathrm{A}$ is $\mathrm{T} _{\mathrm{p}}$ such that $\mathrm{A}$ is either symmetric or skew-symmetric or both, and $\operatorname{det}(\mathrm{A})$ is divisible by $\mathrm{p}$ is
(a) $(\mathrm{p}-1)^{2}$
(b) $2(\mathrm{p}-1)$
(c) $(\mathrm{p}-1)^{2}+1$
(d) $2 \mathrm{p}-1$
ii. The number of $A$ in $T _{p}$ such that the trace of $A$ is not divisible by $p$ but det $(A)$ is divisible by $\mathrm{p}$ is
(a) $(\mathrm{p}-1)\left(\mathrm{p}^{2}-\mathrm{p}+1\right)$
(b) $\mathrm{p}^{3}-(\mathrm{p}-1)^{2}$
(c) $(\mathrm{p}-1)^{2}$
(d) $(p-1)\left(p^{2}-2\right)$
iii. The number of $A$ is $T _{p}$ such that $\operatorname{det}(\mathrm{A})$ is not divisible by $\mathrm{p}$ is
(a) $2 \mathrm{p}^{2}$
(b) $\mathrm{p}^{3}-5 \mathrm{p}$
(c) $\mathrm{p}^{3}-3 \mathrm{p}$
(d) $p^{3}-p^{2}$
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Answer: (i) d (ii) c (iii) d11.* An item of column I can be matched with more than one item of column II. All the items of column II are to be matched
Column I | Column II | ||
---|---|---|---|
(a) | If $a, b, c$ are all different from 0 such that $\dfrac{1}{\mathrm{a}}+\dfrac{1}{\mathrm{~b}}+\dfrac{1}{\mathrm{c}}=0$, then the matrix $\mathrm{A}=\left[\begin{array}{ccc}1+\mathrm{a} & 1 & 1 \\ 1 & 1+\mathrm{b} & 1 \\ 1 & 1 & 1+\mathrm{c}\end{array}\right]$ is | (p) | symmetric |
(b) | If $\alpha, \beta, \gamma$ are three real numbers, then $\text { the matrix } A=\left[\begin{array}{ccc} 1 & \cos (\alpha-\beta) & \cos (\alpha-\gamma) \\ \cos (\beta-\alpha) & 1 & \cos (\beta-\gamma) \\ \cos (\gamma-\alpha) & \cos (\gamma-\beta) & 1 \end{array}\right]$ | (q) | singular |
(c) | If $\mathrm{A}, \mathrm{B}, \mathrm{C}$ are the angles of a triangle, then the matrix $A=\left[\begin{array}{ccc} \sin 2 A & \sin C & \sin B \\ \sin C & \sin 2 B & \sin A \\ \sin B & \sin A & \sin 2 C \end{array}\right]$ is | (r) | non singular |
(s) | invertible | ||
(t) | non invertible |
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Answer: $\mathrm{a} \rightarrow \mathrm{p}, \mathrm{r}, \mathrm{s} ; \mathrm{b} \rightarrow \mathrm{p}, \mathrm{q}, \mathrm{t} ; \mathrm{c} \rightarrow \mathrm{p}, \mathrm{q}, \mathrm{t}$12. In $\triangle \mathrm{ABC}$, if $\left|\begin{array}{lll}1 & \mathrm{a} & \mathrm{b} \\ 1 & \mathrm{c} & \mathrm{a} \\ 1 & \mathrm{~b} & \mathrm{c}\end{array}\right|=0$, then the value of $\left(\sin ^{2} \mathrm{~A}+\sin ^{2} \mathrm{~B}+\sin ^{2} \mathrm{C}\right) 64$ must be
(a) 64
(b) -64
(c) 144
(d) none of these