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Q5(ii):
Evaluate the determinant: (ii) $\begin{vmatrix} 3 & -4 & 5 \\ 1 & 1 & -2 \\ 2 & 3 & 1 \end{vmatrix}$
Solution :
Step 1: Theoretical Framework for Expansion
To evaluate a $3 \times 3$ determinant, we employ the Laplace Expansion method. We expand the determinant $A = \begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{vmatrix}$ along the first row ($R_1$):
$\det(A) = a_{11}(a_{22}a_{33} - a_{23}a_{32}) - a_{12}(a_{21}a_{33} - a_{23}a_{31}) + a_{13}(a_{21}a_{32} - a_{22}a_{31})$
[Per the definition of determinant expansion by minors and cofactors].
Step 2: Assigning Given Matrix Elements
Given the matrix:
$A = \begin{vmatrix} 3 & -4 & 5 \\ 1 & 1 & -2 \\ 2 & 3 & 1 \end{vmatrix}$
- $a_{11} = 3, a_{12} = -4, a_{13} = 5$
- $a_{21} = 1, a_{22} = 1, a_{23} = -2$
- $a_{31} = 2, a_{32} = 3, a_{33} = 1$
Step 3: Calculating Minors of Row 1
We calculate the $2 \times 2$ determinants (minors) corresponding to each element of the first row:
| Element | Minor Calculation | Result |
|---|---|---|
| $a_{11} = 3$ | $(1)(1) - (-2)(3) = 1 - (-6) = 7$ | $3 \times 7 = 21$ |
| $a_{12} = -4$ | $(1)(1) - (-2)(2) = 1 - (-4) = 5$ | $-(-4) \times 5 = 20$ |
| $a_{13} = 5$ | $(1)(3) - (1)(2) = 3 - 2 = 1$ | $5 \times 1 = 5$ |
Step 4: Final Algebraic Summation
Summing the components derived from the expansion along the first row:
$\Delta = 3[(1)(1) - (-2)(3)] - (-4)[(1)(1) - (-2)(2)] + 5[(1)(3) - (1)(2)]$
$\Delta = 3[1 + 6] + 4[1 + 4] + 5[3 - 2]$
$\Delta = 3[7] + 4[5] + 5[1]$
$\Delta = 21 + 20 + 5$
$\Delta = 46$
Visual Representation of Matrix Indices
Final Solution: 46
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(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
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