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Q17:
Let $A$ be a nonsingular square matrix of order $3 \times 3$. Then $|adj A|$ is equal to
Solution :
Step 1: Establishing Fundamental Definitions
Let $A$ be a square matrix of order $n \times n$. A square matrix is termed nonsingular if its determinant is non-zero, i.e., $|A| \neq 0$. The adjoint of a matrix, denoted by $\text{adj } A$, is defined as the transpose of the matrix of cofactors of $A$. [Per the definition of the Adjoint of a Matrix].
Step 2: Utilizing the Fundamental Identity
The relationship between a square matrix $A$ and its adjoint is governed by the following identity: $A(\text{adj } A) = (\text{adj } A)A = |A|I_n$ where $I_n$ is the identity matrix of order $n$. [By the Property of Adjoint Matrices].
Step 3: Applying Determinant Properties
Taking the determinant on both sides of the identity: $|A(\text{adj } A)| = | |A|I_n |$ Applying the multiplicative property of determinants ($|XY| = |X||Y|$): $|A| \cdot |\text{adj } A| = | |A|I_n |$
Step 4: Scalar Multiplication in Determinants
For a scalar $k$ and a matrix $I_n$ of order $n$, the determinant property states $|kI_n| = k^n|I_n|$. Since $|I_n| = 1$: $| |A|I_n | = |A|^n \cdot |I_n| = |A|^n$ Substituting this into the equation from Step 3: $|A| \cdot |\text{adj } A| = |A|^n$
Step 5: Solving for $|\text{adj } A|$
Since the matrix $A$ is nonsingular, $|A| \neq 0$. We can divide both sides by $|A|$: $|\text{adj } A| = \frac{|A|^n}{|A|} = |A|^{n-1}$
Step 6: Substituting Specific Parameters
The problem specifies that $A$ is a square matrix of order $3 \times 3$, implying $n = 3$. $|\text{adj } A| = |A|^{3-1}$ $|\text{adj } A| = |A|^2$
Final Solution: |adj A| = |A|^2
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 $A^{-1}$
- Q18: If $A$ is an invertible matrix of order 2, then det $(A^{-1})$ is equal to
CBSE Solutions for Class 12 Mathematics Determinants
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