Inverse Matrix - Practice Questions & MCQ

Inverse of a Matrix

A non-singular square matrix A is said to be invertible if there exists a non-singular square matrix B such that

$A B=I=B A$
and the matrix B is called the inverse of matrix A. Clearly, B should also have the same order as A.
Hence, $\mathrm{A}^{-1}=\mathrm{B} \Leftrightarrow \mathrm{AB}=\mathbb{I}_{\mathrm{n}}=\mathrm{BA}$

Formula for $\mathrm{A}^{-1}$
We know
$\mathrm{A}(\operatorname{adj} \mathrm{A})=|\mathrm{A}| \mathbb{I}_{\mathrm{n}}$
Multiplying both sides by $\mathrm{A}^{-1}$

$
\begin{aligned}
& \Rightarrow \mathrm{A}^{-1} \mathrm{~A}(\operatorname{adj} \mathrm{~A})=\mathrm{A}^{-1} \mathbb{I}_{\mathrm{n}}|\mathrm{~A}| \\
& \Rightarrow \mathbb{I}_{\mathrm{n}}(\operatorname{adj} \mathrm{~A})=\mathrm{A}^{-1}|\mathrm{~A}| \mathbb{I}_{\mathrm{n}} \quad\left(\text { As } A^{-1} \cdot A=I\right) \\
& \mathrm{A}^{-1}=\frac{\operatorname{adj} \mathrm{A}}{|\mathrm{~A}|}
\end{aligned}
$
Inverse of a $2 \times 2$ matrix
Let A is a square matrix of order 2

$
\mathrm{A}=\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]
$
Then,

$
\mathrm{A}^{-1}=\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right]^{-1}=\frac{1}{\mathrm{ad}-\mathrm{bc}}\left[\begin{array}{cc}
d & -b \\
-c & a
\end{array}\right]
$

To compute the inverse of matrix A of order 3 , first, check whether the matrix is singular or non-singular.
If the matrix is singular, then its inverse does not exist.
If the matrix in non-singular, then the following ar ethe steps to find the Inverse
We use the formula $A^{-1}=\frac{1}{|A|} \cdot \operatorname{adj}(A)$
1. Calculating the Matrix of Minors,
2. then turn that into the Matrix of Cofactors,
3. then take the transpose (These 3 steps give us the adjoint of matrix A)
4. multiply that by $1 /|\mathrm{A}|$.

Properties of inverse of a matrix:

1. Inverse of a matrix is unique

Proof:

Let $A$ be a square and non-singular matrix and let $B$ and $C$ be two inverses of matrix $A$

$
\begin{aligned}
& \mathrm{AB}=\mathrm{BA}=\mathbb{I}_n \text { (since } \mathrm{B} \text { is inverse of } \mathrm{A} \text { ) } \\
& \mathrm{AC}=\mathrm{CA}=\mathbb{I}_{\mathrm{n}} \text { (since } \mathrm{C} \text { is inverse of } \mathrm{A} \text { ) } \\
& \text { now, } \mathrm{AB}=\mathbb{I}_{\mathrm{n}} \\
& \mathrm{C}(\mathrm{AB})=\mathrm{CI}_{\mathrm{n}} \quad[\text { Multiplication by } \mathrm{C}] \\
& \text { (CA) } \mathrm{B}=\mathrm{CI}_{\mathrm{n}} \quad \text { [by associativity] } \\
& \mathbb{I}_{\mathrm{n}} \mathrm{~B}=\mathrm{CI}_{\mathrm{n}} \Rightarrow B=C
\end{aligned}
$
Hence an invertible matrix has a unique inverse.
2. If $A$ and $B$ are invertible matrices of order $n$, then $A B$ will also be invertible. and $(A B)^{-1}=B^{-1} A^{-1}$.

Proof:
A and B are invertible matrices, so $|A| \neq 0$ and $|B| \neq 0$
Hence, $|A||B| \neq 0 \Rightarrow|A B| \neq 0$
now, $(A B)\left(B^{-1} A^{-1}\right)=A\left(B B^{-1}\right) A^{-1} \quad[$ by associative law]

$
\begin{array}{ll}
=A\left(I_n\right) A^{-1} & {\left[\because \mathrm{BB}^{-1}=\mathrm{I}_{\mathrm{n}}\right]} \\
& =A A^{-1}=I_n
\end{array}
$

also, $\left(B^{-1} A^{-1}\right)(A B)=B^{-1}\left(A^{-1} A\right) B$
[by associative law]

$
\begin{aligned}
& =B^{-1}\left(I_n B\right) \quad\left[\because \mathrm{A}^{-1} \mathrm{~A}=\mathrm{I}_{\mathrm{n}}\right] \\
& =B^{-1} B=I_n
\end{aligned}
$

Thus, $(A B)\left(B^{-1} A^{-1}\right)=I_n=\left(B^{-1} A^{-1}\right)(A B)$
Hence, $(A B)^{-1}=B^{-1} A^{-1}$

Properties of Inverse of a Matrix

3. If A is an invertible matrix, then

$
\left(A^{\prime}\right)^{-1}=\left(A^{-1}\right)^{\prime}
$
Proof: As $A$ is an invertible matrix, so $|A| \neq 0 \Rightarrow\left|A^{\prime}\right| \neq 0$. Hence, $A^{\prime}$ is also invertible.
Now, $\mathrm{AA}^{-1}=\mathbb{I}_{\mathrm{n}}=\mathrm{A}^{-1} \mathrm{~A}$
Taking transpose of all three sides

$
\begin{aligned}
& \Rightarrow\left(\mathrm{AA}^{-1}\right)^{\prime}=\left(\mathbb{I}_{\mathrm{n}}\right)^{\prime}=\left(\mathrm{A}^{-1} \mathrm{~A}\right)^{\prime} \\
& \Rightarrow\left(\mathrm{A}^{-1}\right)^{\prime} \mathrm{A}^{\prime}=\mathbb{I}=\mathrm{A}^{\prime}\left(\mathrm{A}^{-1}\right)^{\prime} \\
& \left(\mathrm{A}^{\prime}\right)^{-1}=\left(\mathrm{A}^{-1}\right)^{\prime}
\end{aligned}
$

4. Let $A$ be an invertible matrix, then, $\left(A^{-1}\right)^{-1}=A$

Proof:
Let A be an invertible matrix of order n .

$
\begin{aligned}
& \text { As } A \cdot A^{-1}=I=A^{-1} \cdot A \\
& \Rightarrow \quad\left(\mathrm{~A}^{-1}\right)^{-1}=\mathrm{A}
\end{aligned}
$

5. Let $A$ be an invertible matrix of order $n$ and $k$ is a natural number, then $\left(A^k\right)^{-1}=\left(A^{-1}\right)^k=A^{-k}$

Proof:

$
\begin{aligned}
\left(\mathrm{A}^{\mathrm{k}}\right)^{-1} & =(\mathrm{A} \times \mathrm{A} \times \mathrm{A} \times \ldots \times \mathrm{A})^{-1} \\
& =\left(\mathrm{A}^{-1} \times \mathrm{A}^{-1} \times \mathrm{A}^{-1} \times \ldots \times \mathrm{A}^{-1}\right) \\
& =\left(\mathrm{A}^{-1}\right)^k
\end{aligned}
$
 

Properties of Inverse of a Matrix

6. Let A be an invertible matrix of order n , then

$
\left|\mathrm{A}^{-1}\right|=\frac{1}{|\mathrm{~A}|}
$
Proof: $\because \mathrm{A}$ is invertible, then $|\mathrm{A}| \neq 0$.
now, $\mathrm{AA}^{-1}=\mathbb{I}_{\mathrm{n}}=\mathrm{A}^{-1} \mathrm{~A}$

$
\begin{aligned}
& \Rightarrow\left|\mathrm{AA}^{-1}\right|=\left|\mathbb{I}_{\mathrm{n}}\right| \\
& \Rightarrow|\mathrm{A}|\left|\mathrm{A}^{-1}\right|=1 \\
& \Rightarrow\left|\mathrm{~A}^{-1}\right|=\frac{1}{|\mathrm{~A}|}
\end{aligned}
$

7. Inverse of a non-singular diagonal matrix is a diagonal matrix
if $A=\left[\begin{array}{lll}a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c\end{array}\right]$ and $|\mathrm{A}| \neq 0$
then

$
\mathrm{A}^{-1}=\left[\begin{array}{ccc}
\frac{1}{a} & 0 & 0 \\
0 & \frac{1}{b} & 0 \\
0 & 0 & \frac{1}{c}
\end{array}\right]
$