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Q18:
If $A$ is an invertible matrix of order 2, then det $(A^{-1})$ is equal to
Solution :
Given Variables & Initial Setup
Let $A$ be an invertible square matrix of order $2 \times 2$. By definition, for any square matrix $A$, the product of the matrix and its inverse is the identity matrix $I$:
$A \cdot A^{-1} = I$ [Definition of Matrix Inversion]
Step 1: Applying the Determinant Property
Take the determinant on both sides of the equation:
$\det(A \cdot A^{-1}) = \det(I)$
By the Multiplicative Property of Determinants, which states that $\det(AB) = \det(A) \cdot \det(B)$, we have:
$\det(A) \cdot \det(A^{-1}) = \det(I)$ [Multiplicative Property of Determinants]
Step 2: Evaluating the Determinant of Identity Matrix
For a square matrix of order $2$, the identity matrix $I$ is defined as:
$I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$
The determinant of $I$ is calculated as:
$\det(I) = (1 \times 1) - (0 \times 0) = 1$
Step 3: Algebraic Isolation of $\det(A^{-1})$
Substitute $\det(I) = 1$ into the equation derived in Step 1:
$\det(A) \cdot \det(A^{-1}) = 1$
Since $A$ is given as an invertible matrix, $\det(A) \neq 0$. Therefore, we can divide both sides by $\det(A)$:
$\det(A^{-1}) = \frac{1}{\det(A)}$
Geometric/Visual Interpretation (Linear Scaling)
Summary of Determinant Properties
| Property Name | Mathematical Formulation |
|---|---|
| Product Property | $\det(AB) = \det(A)\det(B)$ |
| Identity Property | $\det(I_n) = 1$ |
| Inverse Determinant | $\det(A^{-1}) = (\det(A))^{-1}$ |
Final Solution: The value of $\det(A^{-1})$ is equal to $\frac{1}{\det(A)}$ or $(\det(A))^{-1}$.
More Questions from Class 12 Mathematics Determinants EXERCISE 4.4
- Q1: Find adjoint of the matrix $\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$
- Q2: Find adjoint of the matrix $\begin{bmatrix} 1 & -1 & 2 \\ 2 & 3 & 5 \\ -2 & 0 & 1 \end{bmatrix}$
- Q3: Verify $A (adj A) = (adj A) A = |A| I$ for the matrix $\begin{bmatrix} 2 & 3 \\ -4 & -6 \end{bmatrix}$
- Q4: Verify $A (adj A) = (adj A) A = |A| I$ for the matrix $\begin{bmatrix} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{bmatrix}$
- Q5: Find the inverse of the matrix (if it exists): $\begin{bmatrix} 2 & -2 \\ 4 & 3 \end{bmatrix}$
- Q6: Find the inverse of the matrix (if it exists): $\begin{bmatrix} -1 & 5 \\ -3 & 2 \end{bmatrix}$
- Q7: Find the inverse of the matrix (if it exists): $\begin{bmatrix} 1 & 2 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 5 \end{bmatrix}$
- Q8: Find the inverse of the matrix (if it exists): $\begin{bmatrix} 1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & -1 \end{bmatrix}$
- Q9: Find the inverse of the matrix (if it exists): $\begin{bmatrix} 2 & 1 & 3 \\ 4 & -1 & 0 \\ -7 & 2 & 1 \end{bmatrix}$
- Q10: Find the inverse of the matrix (if it exists): $\begin{bmatrix} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{bmatrix}$
- Q11: Find the inverse of the matrix (if it exists): $\begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos\alpha & \sin\alpha \\ 0 & \sin\alpha & -\cos\alpha \end{bmatrix}$
- Q12: Let $A = \begin{bmatrix} 3 & 7 \\ 2 & 5 \end{bmatrix}$ and $B = \begin{bmatrix} 6 & 8 \\ 7 & 9 \end{bmatrix}$. Verify that $(AB)^{-1} = B^{-1} A^{-1}$.
- Q13: If $A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$, show that $A^2 – 5A + 7I = O$. Hence find $A^{-1}$.
- Q14: For the matrix $A = \begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix}$, find the numbers $a$ and $b$ such that $A^2 + aA + bI = O$.
- Q15: For the matrix $A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{bmatrix}$ Show that $A^3– 6A^2 + 5A + 11 I = O$. Hence, find $A^{-1}$.
- Q16: If $A = \begin{bmatrix} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{bmatrix}$ Verify that $A^3 – 6A^2 + 9A – 4I = O$ and hence find $A^{-1}$
- Q17: Let $A$ be a nonsingular square matrix of order $3 \times 3$. Then $|adj A|$ is equal to
CBSE Solutions for Class 12 Mathematics Determinants
Chapters in CBSE - Class 12 Mathematics
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