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Q6:
If A = $\begin{bmatrix} 1 & 1 & -2 \\ 2 & 1 & -3 \\ 5 & 4 & -9 \end{bmatrix}$, find $| A |$
Solution :
Step 1: Establishing the Determinant Formula
To find the determinant of a $3 \times 3$ matrix $A = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}$, we use the expansion by the first row [Per Laplace's expansion theorem]:
$|A| = a_{11}(a_{22}a_{33} - a_{23}a_{32}) - a_{12}(a_{21}a_{33} - a_{23}a_{31}) + a_{13}(a_{21}a_{32} - a_{22}a_{31})$
Step 2: Mapping Given Matrix Elements
From the given matrix $A = \begin{bmatrix} 1 & 1 & -2 \\ 2 & 1 & -3 \\ 5 & 4 & -9 \end{bmatrix}$, we identify:
- $a_{11} = 1, a_{12} = 1, a_{13} = -2$
- $a_{21} = 2, a_{22} = 1, a_{23} = -3$
- $a_{31} = 5, a_{32} = 4, a_{33} = -9$
Step 3: Calculating the Determinant
Expanding along the first row:
$|A| = 1 \begin{vmatrix} 1 & -3 \\ 4 & -9 \end{vmatrix} - 1 \begin{vmatrix} 2 & -3 \\ 5 & -9 \end{vmatrix} + (-2) \begin{vmatrix} 2 & 1 \\ 5 & 4 \end{vmatrix}$
Applying the $2 \times 2$ determinant formula $(ad - bc)$:
Term 1: $1 \times [(1)(-9) - (-3)(4)] = 1 \times [-9 + 12] = 1 \times 3 = 3$
Term 2: $-1 \times [(2)(-9) - (-3)(5)] = -1 \times [-18 + 15] = -1 \times [-3] = 3$
Term 3: $-2 \times [(2)(4) - (1)(5)] = -2 \times [8 - 5] = -2 \times 3 = -6$
Step 4: Summation of Terms
$|A| = 3 + 3 - 6$
$|A| = 6 - 6$
$|A| = 0$
Final Solution: |A| = 0
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) $\begin{vmatrix} 2 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0 \end{vmatrix}$
- 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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