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Q3:
If A = $\begin{bmatrix} 1 & 2 \\ 4 & 2 \end{bmatrix}$, then show that $| 2A | = 4 | A |$
Solution :
Step 1: Definition of Given Matrix and Determinant Calculation
Given the matrix $A = \begin{bmatrix} 1 & 2 \\ 4 & 2 \end{bmatrix}$.
To calculate the determinant of $A$, denoted as $|A|$, we apply the formula for a $2 \times 2$ matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ where $|A| = ad - bc$ [Per the definition of the determinant of a square matrix].
$|A| = (1 \times 2) - (2 \times 4)$
$|A| = 2 - 8$
$|A| = -6$
Step 2: Calculating 4|A|
Multiply the result obtained in Step 1 by the scalar $4$:
$4|A| = 4 \times (-6)$
$4|A| = -24$
Step 3: Evaluating the Matrix 2A
To find $2A$, we multiply every element of matrix $A$ by the scalar $2$ [Per the definition of scalar multiplication of a matrix]:
$2A = 2 \begin{bmatrix} 1 & 2 \\ 4 & 2 \end{bmatrix} = \begin{bmatrix} 2 \times 1 & 2 \times 2 \\ 2 \times 4 & 2 \times 2 \end{bmatrix}$
$2A = \begin{bmatrix} 2 & 4 \\ 8 & 4 \end{bmatrix}$
Step 4: Calculating the Determinant |2A|
Using the determinant formula for the resultant matrix $2A$:
$|2A| = \begin{vmatrix} 2 & 4 \\ 8 & 4 \end{vmatrix}$
$|2A| = (2 \times 4) - (4 \times 8)$
$|2A| = 8 - 32$
$|2A| = -24$
Step 5: Verification and Conclusion
Comparing the results from Step 2 and Step 4:
| Expression | Result |
|---|---|
| $|2A|$ | $-24$ |
| $4|A|$ | $-24$ |
Since $|2A| = -24$ and $4|A| = -24$, it is demonstrated that $|2A| = 4|A|$. This is a specific application of the property $|kA| = k^n|A|$, where $n$ is the order of the square matrix [Per the properties of determinants for a matrix of order $n=2$].
Final Solution: $|2A| = 4|A| = -24$
More Questions from Class 12 Mathematics Determinants EXERCISE 4.1
- Q1: Evaluate the determinant: $\begin{vmatrix} 2 & 4 \\ -5 & -1 \end{vmatrix}$
- Q2(i): Evaluate the determinant: (i) $\begin{vmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{vmatrix}$
- Q2(ii): Evaluate the determinant: (ii) $\begin{vmatrix} x^2 - x + 1 & x - 1 \\ x + 1 & x + 1 \end{vmatrix}$
- Q4: If A = $\begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{bmatrix}$, then show that $| 3 A | = 27 | A |$
- Q5(i): Evaluate the determinant: (i) $\begin{vmatrix} 3 & -1 & -2 \\ 0 & 0 & -1 \\ 3 & -5 & 0 \end{vmatrix}$
- Q5(ii): Evaluate the determinant: (ii) $\begin{vmatrix} 3 & -4 & 5 \\ 1 & 1 & -2 \\ 2 & 3 & 1 \end{vmatrix}$
- Q5(iii): Evaluate the determinant: (iii) $\begin{vmatrix} 0 & 1 & 2 \\ -1 & 0 & -3 \\ -2 & 3 & 0 \end{vmatrix}$
- Q5(iv): Evaluate the determinant: (iv) $\begin{vmatrix} 2 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0 \end{vmatrix}$
- Q6: If A = $\begin{bmatrix} 1 & 1 & -2 \\ 2 & 1 & -3 \\ 5 & 4 & -9 \end{bmatrix}$, find $| A |$
- Q7(i): Find values of $x$, if (i) $\begin{vmatrix} 2 & 4 \\ 5 & 1 \end{vmatrix} = \begin{vmatrix} 2x & 4 \\ 6 & x \end{vmatrix}$
- Q7(ii): Find values of $x$, if (ii) $\begin{vmatrix} 2 & 3 \\ 4 & 5 \end{vmatrix} = \begin{vmatrix} x & 3 \\ 2x & 5 \end{vmatrix}$
- Q8: If $\begin{vmatrix} x & 2 \\ 18 & x \end{vmatrix} = \begin{vmatrix} 6 & 2 \\ 18 & 6 \end{vmatrix}$, then $x$ is equal to
CBSE Solutions for Class 12 Mathematics Determinants
Chapters in CBSE - Class 12 Mathematics
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