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Q4:
If A = $\begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{bmatrix}$, then show that $| 3 A | = 27 | A |$
Solution :
Step 1: Calculating the Determinant of Matrix $A$
Given the matrix $A = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{bmatrix}$. We calculate the determinant $|A|$ by expanding along the first column $C_1$ because it contains the most zeros [Per Laplace Expansion Theorem].
$|A| = 1 \begin{vmatrix} 1 & 2 \\ 0 & 4 \end{vmatrix} - 0 \begin{vmatrix} 0 & 1 \\ 0 & 4 \end{vmatrix} + 0 \begin{vmatrix} 0 & 1 \\ 1 & 2 \end{vmatrix}$
$|A| = 1((1)(4) - (2)(0)) - 0 + 0 = 4$
Thus, $27|A| = 27 \times 4 = 108$.
Step 2: Scalar Multiplication of Matrix $A$ by 3
To find $3A$, we multiply every entry of matrix $A$ by the scalar $3$ [Per Scalar Multiplication Law].
$3A = 3 \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{bmatrix} = \begin{bmatrix} 3 & 0 & 3 \\ 0 & 3 & 6 \\ 0 & 0 & 12 \end{bmatrix}$
Step 3: Calculating the Determinant of Matrix $3A$
Now, we compute $|3A|$ using the determinant of the resulting matrix from Step 2. We expand along the first column $C_1$.
$|3A| = \begin{vmatrix} 3 & 0 & 3 \\ 0 & 3 & 6 \\ 0 & 0 & 12 \end{vmatrix}$
$|3A| = 3 \begin{vmatrix} 3 & 6 \\ 0 & 12 \end{vmatrix} - 0 \begin{vmatrix} 0 & 3 \\ 0 & 12 \end{vmatrix} + 0 \begin{vmatrix} 0 & 3 \\ 3 & 6 \end{vmatrix}$
$|3A| = 3((3)(12) - (6)(0)) = 3(36) = 108$
Step 4: Theoretical Verification
The property $|kA| = k^n|A|$ holds for any $n \times n$ matrix, where $n$ is the order of the matrix [Per Determinant Property Theorem]. Here, $n=3$ and $k=3$.
$|3A| = 3^3 |A| = 27 |A|$
Substituting the values derived:
| Expression | Result |
|---|---|
| $|3A|$ | $108$ |
| $27|A|$ | $27 \times 4 = 108$ |
Final Solution: Since $|3A| = 108$ and $27|A| = 27(4) = 108$, it is proven that $|3A| = 27|A|$.
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) $\begin{vmatrix} x^2 - x + 1 & x - 1 \\ x + 1 & x + 1 \end{vmatrix}$
- Q3: If A = $\begin{bmatrix} 1 & 2 \\ 4 & 2 \end{bmatrix}$, then show that $| 2A | = 4 | A |$
- 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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Hello, my name is Sreedevi. I live in a beautiful village in Kerala. Those who are looking to learn to talk in Malayalam, I can help you. I do not hold a degree in the language. But I am a native speaker and my grandfather was a renowned poet of his times and we love the language to the core as a family. I can teach to speak in simple and fun methods that's not textbook oriented referring to everything in daily life from food to movies to asking directions :)
