Diagunabisable the matrix \( A=\left[\begin{array}{ccc}8 & -8 & -2 \\ 4 & -3 & -2 \\ 3 & -4 & 1\end{array}\right] \) and find \( A^{4} \)
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A = \(\begin{bmatrix}8&-8&-2\\4&-3&-2\\3&-4&1\end{bmatrix}\)
A2 = \(\begin{bmatrix}8&-8&-2\\4&-3&-2\\3&-4&1\end{bmatrix}\)\(\begin{bmatrix}8&-8&-2\\4&-3&-2\\3&-4&1\end{bmatrix}\) = \(\begin{bmatrix}26&-32&-2\\14&-15&-4\\11&-16&3\end{bmatrix}\)
A4 = A2. A2 = \(\begin{bmatrix}26&-32&-2\\14&-15&-4\\11&-16&3\end{bmatrix}\)\(\begin{bmatrix}26&-32&-2\\14&-15&-4\\11&-16&3\end{bmatrix}\)
= \(\begin{bmatrix}206&-320&-70\\110&-159&20\\95&-160&51\end{bmatrix}\)
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8. If \( A=\left[\begin{array}{ll}1 & 3 \\ 4 & 1\end{array}\right] \), then find \( \left|3 A^{\prime}\right| \).
Matrix A \(=\begin{bmatrix}1&3\\4&1\end{bmatrix}_{2\times2}\) |3 A'| = 2 \(A'=\begin{bmatrix}1&4\\3&1\end{bmatrix}_{2\times2}\) We know that ⇒ |K A| = (K)3 |A| ⇒ |3 A'| = 32 |A'| \(A'=\begin{bmatrix}1&4\\3&1\end{bmatrix}\) ⇒ |A'| = 1 x 1 - 4 x 3 ⇒ 1 - 12 = -11 ⇒...
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In \( III ^{ rd } \) group of salt analysis for cations \( NH _{4} Cl \) and \( NH _{4} OH \) are added as group reagents. Which of the following can be used instead of \( NH _{4} Cl \) ? (a) \( NaCl \) (b) \( \left( NH _{4}\right)_{2} SO _{4} \) (c) \( \left( NH _{4}\right)_{2} C _{2} O _{4} \) (d) \( NH _{4} NO _{3} \)
d) NH4NO3 As it provides NH+4 ions which suppresses the ionisation of NH4OH so that only the group 3rd cations are precipitated as hydroxides because of their low solubility products and NO−3 ions...
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\[ \begin{array}{l} \text { Part - II } \\ \text { Section - D } \end{array} \] Note: answer the following question. 18. From following receipts and payments A/C of Sri Charitable Trust, \[ 1 \times 20=20 \] Additional information A. Subscriptions receivable for the year 2020 R5 2,500 B. Prepaid Rent Rs. 300 C. Outstanding stationary Bills Rs. 150 D. Capitalize donations E. Half of the entrance fees capitalized Interest receivable for the year \( 2020 Rs .200 \)
receipts. rs. payment. rs to donation. 25,000 by rent and taxes. 1300 to membership 1500 by startionary. 1200 fees to entrance fees 2400. byscholar ship 200 to...
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4. The value of \( \left(\frac{1}{2}\right)^{-2}+\left(\frac{1}{3}\right)^{-2}+\left(\frac{1}{4}\right)^{-2} \) is (a) \( \frac{61}{144} \) (b) \( \frac{144}{61} \) (c) 29 (d) none of these
Correct answer is:- (c) 29 Explanation:- 1/2-2 + 1/3-2 + 1/4-2 = 22 + 32 + 42 (x-1 = 1/x) = 4 + 9 + 16 = 29
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Consider a particle of mass \( m \) having linear momentum \( P \) at position \( r \) relative to the origin \( O \). Which of the following equations correctly relates \( r , P , L \) ? a) \( \left[\left(\frac{d L}{d t}\right)+r x\left(\frac{d P}{d t}\right)\right]=0 \) b) \( \left[\left(\frac{d L}{d t}\right)-r x\left(\frac{d P}{d t}\right)\right]=0 \) c) \( \left[\left(\frac{d L}{d t}\right) \times\left(\frac{d r}{d t}\right) \times P\right]=0 \) d) \( \left[\left(\frac{d L}{d t}\right)-\left(\frac{d r}{d t}\right) \times P\right]=0 \)
Correct answer is (b) We know that \(\tau\) = r × F ∵ \(\tau\) \(= \frac{dL}{dt},\) \(F = \frac{dP}{dt}\) \(\left(\frac{dL}{dt}\right) = r \times \left(\frac{dP}{dt}\right)\) \(\left(\frac{dL}{dt}\right) - r \times \left(\frac{dP}{dt}\right) = 0\)
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Find the Inverse?\[A=\left[\begin{array}{lll}2 & 1 & 1 \\3 & 2 & 1 \\2 & 1 & 2\end{array}\right] \quad A^{-1}=?\]
We have A = IA \(\begin{bmatrix}2&1&1\\3&2&1\\2&1&2\end{bmatrix}=\) \(\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}\)A Applying R3 → R3 — R1 R2 → 2R2 — 3R1 \(\begin{bmatrix}2&1&1\\0&1&-1\\0&0&1\end{bmatrix}=\)\(\begin{bmatrix}1&0&0\\-3&2&0\\-1&0&1\end{bmatrix}\)A Applying R2 → R2 + R3 \(\begin{bmatrix}2&1&1\\0&1&0\\0&0&1\end{bmatrix}=\)\(\begin{bmatrix}1&0&0\\-4&2&1\\-1&0&1\end{bmatrix}\)A Applying R1 → R1 — R2 — R3 \(\begin{bmatrix}2&0&0\\0&1&0\\0&0&1\end{bmatrix}=\) \(\begin{bmatrix}6&-2&-2\\-4&2&1\\-1&0&1\end{bmatrix}\)A Applying R1 → \(\frac{R_1}2\) \(\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}=\)\(\begin{bmatrix}3&-1&-1\\-4&2&1\\-1&0&1\end{bmatrix}\)A Hence A-1 = \(\begin{bmatrix}3&-1&-1\\-4&2&1\\-1&0&1\end{bmatrix}\)
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(1.) If \( T={ }_{B}^{A}\left(\begin{array}{cc}0.25 & 0.75 \\ 0.3 & \alpha\end{array}\right) \) is a transition probability matrix, then the value of \( \alpha \) is a) \( 0.5 \) b) \( 0.7 \) c) 0 /d) \( 0.6 \) 2. Rank of matrix \( \left(\begin{array}{cccc}1 & -1 & 2 & 0 \\ 0 & 2 & 0 & 3\end{array}\right) \) is a) 1 b) 0 c) 2 d) \( -1 \)
Is there any answer
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