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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

-6 6 0 Solution Set for x on the Number Line

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


CBSE Solutions for Class 12 Mathematics Determinants


Chapters in CBSE - Class 12 Mathematics


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