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Q8:
If $\begin{vmatrix} x & 2 \\ 18 & x \end{vmatrix} = \begin{vmatrix} 6 & 2 \\ 18 & 6 \end{vmatrix}$, then $x$ is equal to
Solution :
Step 1: Evaluation of the Determinants
To solve for the unknown variable $x$, we must apply the definition of a $2 \times 2$ determinant. For any matrix $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$, the determinant is given by $|A| = ad - bc$ [Definition of Determinant for $2 \times 2$ matrices].
For the Left-Hand Side (LHS):
$\begin{vmatrix} x & 2 \\ 18 & x \end{vmatrix} = (x)(x) - (18)(2)$
$= x^2 - 36$ [Algebraic simplification]
For the Right-Hand Side (RHS):
$\begin{vmatrix} 6 & 2 \\ 18 & 6 \end{vmatrix} = (6)(6) - (18)(2)$
$= 36 - 36$
$= 0$
Step 2: Constructing the Algebraic Equation
Equating the results from Step 1 as per the problem statement:
$x^2 - 36 = 0$
Step 3: Solving for $x$
To isolate $x$, we apply the addition property of equality:
$x^2 = 36$
Taking the square root on both sides requires considering both positive and negative roots [Principle of Square Roots]:
$x = \pm \sqrt{36}$
$x = \pm 6$
Visual Representation
Step 4: Verification
| Value of $x$ | Calculation | Result |
|---|---|---|
| $6$ | $6^2 - 36$ | $0$ |
| $-6$ | $(-6)^2 - 36$ | $0$ |
Final Solution: $x = \pm 6$
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 $| 2A | = 4 | A |$
- 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) $\begin{vmatrix} 2 & 3 \\ 4 & 5 \end{vmatrix} = \begin{vmatrix} x & 3 \\ 2x & 5 \end{vmatrix}$
CBSE Solutions for Class 12 Mathematics Determinants
Chapters in CBSE - Class 12 Mathematics
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