CBSE Class 12 Mathematics Determinants Worksheet Set 06

CBSE Class 12 Mathematics Determinants Worksheet (6). The Determinants questions in the worksheets have been specifically designed by best mathematics teachers so that the students can practise them to clear their Determinants concepts and get better marks in class 12 mathematics tests and examinations. Students can free download these Determinants worksheets in pdf and practice them. This will help them to get better marks in examinations. Also refer to other worksheets for the Determinants chapter and other subjects too. Use them for better understanding of the subjects.

Question. Find the value of x if the area of \(\Delta\) is 35 square units with vertices (x, 4), (2, –6) and (5, 4).
Answer: Let vertices are \(A(x, 4)\) , \(B(2, – 6)\) and \(C(5, 4)\)
Area of \(\Delta\text{ABC} = \frac{1}{2} \begin{vmatrix} x & 4 & 1 \\ 2 & -6 & 1 \\ 5 & 4 & 1 \end{vmatrix}\)
\(35 = \frac{1}{2} |x(-10) - 4(-3) + 1(38)|\)
\(\Rightarrow 35 = \frac{1}{2} |– 10x + 12 + 38|\)
\(\Rightarrow 70 = |-10x + 50|\)
\(70 = – 10x + 50\)      \(-70 = – 10x + 50\)
\(10x = – 20\)      \(10x = 120\)
\(x = – 2\)            \(x = 12\)
\(\therefore x = – 2\) , \(x = 12\) ans.

Question. Find the value of x so that matrix \(A = \begin{bmatrix} (x - 1) & 1 & 1 \\ 1 & (x - 1) & 1 \\ 1 & 1 & (x - 1) \end{bmatrix}\) is singular/ Non-Invertible.
Answer: Since matrix \(A\) is singular
\(\therefore |A| = 0\)
\(\begin{vmatrix} x - 1 & 1 & 1 \\ 1 & x - 1 & 1 \\ 1 & 1 & x - 1 \end{vmatrix} = 0\)
\(\Rightarrow (x – 1) [(x – 1)^2 – 1] – 1[x – 1 – 1] + 1[1 – x + 1] = 0\)
\(\Rightarrow (x – 1) (x^2 – 2x) – 1(x – 2) + (2 – x) = 0\)
\(\Rightarrow x^3 – 2x^2 – x^2 + 2x – x + 2 + 2 – x = 0\)
\(\Rightarrow x^3 – 3x^2 + 4 = 0\)
By trial method
\((x + 1) (x – 2) (x + 1) = 0\)
\(\Rightarrow x = – 1, x = 2\) ans.

Question.
(a) Evaluate the determinant \(\Delta = \begin{vmatrix} 1 & \sin \theta & 1 \\ -\sin \theta & 1 & \sin \theta \\ -1 & -\sin \theta & 1 \end{vmatrix}\). Also prove \(2 \le \Delta \le 4\).
(b) Prove that \(\Delta = \begin{vmatrix} x & \sin \theta & \cos \theta \\ -\sin \theta & -x & 1 \\ \cos \theta & 1 & x \end{vmatrix}\) is independent of \(\theta\).
Answer:
(a) we have, \(\Delta = \begin{vmatrix} 1 & \sin \theta & 1 \\ -\sin \theta & 1 & \sin \theta \\ -1 & -\sin \theta & 1 \end{vmatrix}\)
\(\Rightarrow \Delta = 1(1 + \sin^2\theta) - \sin \theta(-\sin \theta + \sin \theta) + 1(\sin^2\theta + 1)\)
\(\Rightarrow \Delta = 1 + \sin^2\theta + 0 + \sin^2\theta + 1\)
\(\Rightarrow \Delta = 2 + 2 \sin^2\theta\)
Now, we know
\(-1 \le \sin \theta \le 1\)
\(\Rightarrow 0 \le \sin^2\theta \le 1\)
\(\Rightarrow 0 \le 2 \sin^2\theta \le 2\) (multiply by 2)
\(\Rightarrow 2 \le 2 + 2 \sin^2\theta \le 4\) (adding 2)
\(\Rightarrow 2 \le \Delta \le 4\) (proved)

(b) \(\Delta = x(-x^2 - 1) - \sin \theta(-x \sin \theta - \cos \theta) + \cos \theta(-\sin \theta + x \cos \theta)\)
\(\Delta = -x^3 - x + x\sin^2\theta + \sin \theta \cos \theta - \sin \theta \cos \theta + x \cos^2\theta\)
\(\Delta = -x^3 - x + x(\sin^2\theta + \cos^2\theta)\)
\(\Delta = -x^3 - x + x(1)\)
\(\Delta = -x^3\) which is independent of \(\theta\).

Short Questions

Question. Order \(3 \times 3\) , \(|A| = 5\). Find \(|\text{Adj } A| = ?\)
Answer: We have \(n = 3 , |A| = 5\)
and \(|\text{Adj } A| = |A|^{n-1}\)
\(= (5)^{3-1} = 25\) ans.

Question. Order \(3 \times 3\) , \(|\text{Adj } A| = 81\) find \(|A| = ?\)
Answer: We have \(n = 3 , |\text{Adj } A| = 81\)
\(\Rightarrow |\text{Adj } A| = |A|^{n-1}\)
\(\Rightarrow 81 = |A|^2\)
\(\Rightarrow |A| = \pm9\) ans.

Question. Order \(3 \times 3\) ; \(|A| = 3\) find \(|4A| = ?\)
Answer: We have \(n = 3 , |A| = 3\)
\(|4A| = 4^3 |A|\)         \(\{ \because |kA| = k^n |A| \}\)
\(= 64 \times 3\)
\(= 192\) ans.

Question. Order \(3 \times 3\) ; \(|A| = 5\) find \(|2\text{Adj } A| =?\)
Answer: \(|2\text{Adj } A| = 2^3 |\text{Adj } A| = 2^3 |A|^{3-1}\)
\(= 8 (5)^2 = 200\) ans.

Question. Order \(4 \times 4\) ; \(|3 \text{Adj } A| = 243\) Find \(|A| =?\)
Answer: We have \(|3 \text{Adj } A| = 3^4 |\text{Adj } A|\)
\(243 = 3^4 |A|^{4-1}\)
\(243 = 81 |A|^3\)
\(|A|^3 = 3\)
\(|A| = (3)^{1/3}\) ans.

Question. Order \(4 \times 4\) ; \(|A| = 5\) find \(|A'| =?\)
Answer: We know \(|A'| = |A|\)
\(\Rightarrow |A'| = 5\) ans.

Please click the link below to downloadCBSE Class 12 Mathematics Determinants Worksheet (6).