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Q2(ii):
Evaluate the determinant: (ii) $\begin{vmatrix} x^2 - x + 1 & x - 1 \\ x + 1 & x + 1 \end{vmatrix}$
Solution :
Step 1: Application of Determinant Definition
For a $2 \times 2$ matrix $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$, the determinant is defined as $|A| = ad - bc$ [By the definition of a second-order determinant].
Given the determinant $\Delta = \begin{vmatrix} x^2 - x + 1 & x - 1 \\ x + 1 & x + 1 \end{vmatrix}$, we identify the components:
- $a = x^2 - x + 1$
- $b = x - 1$
- $c = x + 1$
- $d = x + 1$
Step 2: Algebraic Expansion
Applying the formula $\Delta = (x^2 - x + 1)(x + 1) - (x - 1)(x + 1)$:
$\Delta = [(x^2 - x + 1)(x + 1)] - [(x - 1)(x + 1)]$
Step 3: Simplification via Factorization
We observe that $(x + 1)$ is a common factor in both terms. Factoring $(x + 1)$ out allows for easier simplification:
$\Delta = (x + 1) \cdot [(x^2 - x + 1) - (x - 1)]$
[By the Distributive Property of multiplication over subtraction: $a \cdot c - b \cdot c = c(a - b)$]
Step 4: Final Polynomial Reduction
Simplify the expression within the square brackets:
$\Delta = (x + 1) \cdot [x^2 - x + 1 - x + 1]$
$\Delta = (x + 1) \cdot [x^2 - 2x + 2]$
Now, perform the final multiplication:
$\Delta = x(x^2 - 2x + 2) + 1(x^2 - 2x + 2)$
$\Delta = x^3 - 2x^2 + 2x + x^2 - 2x + 2$
Group the like terms:
$\Delta = x^3 + (-2x^2 + x^2) + (2x - 2x) + 2$
$\Delta = x^3 - x^2 + 2$
Visual Representation of the Determinant Structure
Final Solution: x^3 - x^2 + 2
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) $\begin{vmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{vmatrix}$
- 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) $\begin{vmatrix} 2 & 3 \\ 4 & 5 \end{vmatrix} = \begin{vmatrix} x & 3 \\ 2x & 5 \end{vmatrix}$
- 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
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