CBSE Class 12 Mathematics Determinants Worksheet Set 07

CBSE Class 12 Mathematics Determinants Worksheet (7). 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. Order \(3 \times 3\) find \(|A^{-1}| = ?\)
Answer:
We have \(|A^{-1}| = \left|\frac{1}{|A|} \cdot \text{Adj } A\right|\)
\( = \frac{1}{|A|^3} |\text{Adj } A|\)
\( = \frac{1}{|A|^3} \cdot |A|^{3-1} = \frac{1}{|A|^3} \cdot |A|^2 \)
\( \therefore |A^{-1}| = \frac{1}{|A|} \) Ans.

Question. Order \(3 \times 3\) ; \(|A| = 3\) and \(|2 AB| = 120\) find \(|B'| = ?\)
Answer:
We have \(|2AB| = 2^3 |AB|\)
\( 120 = 2^3 |A| |B| \)
\( 120 = 8 \times 3 \times |B| \)
\( 5 = |B| \)
Since \(|B'| = |B|\)
\( \Rightarrow |B'| = 5 \) Ans.....

Question. Order \(2 \times 2\) ; \(Adj \ A = \begin{bmatrix} 2 & -1 \\ 3 & 4 \end{bmatrix}\) and Adj B = \begin{bmatrix} 1 & 2 \\ 3 & 5 \end{bmatrix} find Adj(AB) = ?
Answer:
We have \(Adj \ (AB) = (Adj \ B) (Adj \ A)\)
\( = \begin{bmatrix} 1 & 2 \\ 3 & 5 \end{bmatrix}\begin{bmatrix} 2 & -1 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 8 & 7 \\ 21 & 17 \end{bmatrix} \) Ans....

Question. Order \(2 \times 2\) ; \(A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}\) find \(A(\text{Adj } A)\) without finding Adj A.
Answer:
We have, \(A|\text{Adj } A| = |A| I\)
\( = \begin{vmatrix} 1 & 2 \\ 3 & 4 \end{vmatrix} \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)
\( = (-2) \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)
\( A(\text{Adj } A) = \begin{bmatrix} -2 & 0 \\ 0 & -2 \end{bmatrix} \) Ans....

Question. If \(n = 3 \times 3\) find \(|\text{Adj}(\text{Adj } A)|\) and \(|A| = 5\)
Answer:
We have \(|\text{Adj}(\text{Adj } A)| = |A|^{(n-1)^2}\)
\( = |\text{Adj } A|^2 \)
\( = (|A|^{3-1})^2 = |A|^4 \)
\( = (5)^4 = 625 \) Ans.

Question. If \(A = \begin{bmatrix} 1 & 2 \\ -1 & 4 \end{bmatrix}\) find \((BA)^{-1}\) and \(B^{-1} = \begin{bmatrix} 2 & 4 \\ 3 & 6 \end{bmatrix}\)
Answer:
We know \((BA)^{-1} = A^{-1} B^{-1}\)
\( |B| = 12 - 12 = 0 \Rightarrow B \) is non invertible
\( \therefore (BA)^{-1} \) not possible

Question. If \(-1 \le x < 0\) ; \(0 \le y < 1\) and \(1 \le z < 2\)
Find \(\Delta = \begin{vmatrix} [x] + 1 & [y] & [z] \\ [x] & [y] + 1 & [z] \\ [x] & [y] & [z] + 1 \end{vmatrix}\)
Answer:
Since \(-1 \le x < 0 \therefore [x] = -1\)
\( 0 \le y < 1 \therefore [y] = 0\)
\( 1 \le z < 2 \therefore [z] = 1\)
\( \therefore \Delta = \begin{vmatrix} -1 + 1 & 0 & 1 \\ -1 & 0 + 1 & 1 \\ -1 & 0 & 1 + 1 \end{vmatrix} = \begin{vmatrix} 0 & 0 & 1 \\ -1 & 1 & 1 \\ -1 & 0 & 2 \end{vmatrix} = 1 \) Ans.....

Question. \(A = \begin{bmatrix} 2 & -1 & 3 \\ 4 & 5 & 5 \\ 3 & -1 & 4 \end{bmatrix}\) find \(M_{32}\) and \(C_{23}\)
Answer:
\(M_{32} = 10 - 12 = -12\)
\(C_{23} = 8 - 9 = (-1) = +1\) ans.
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