Matrix Determinant and Adjoint Verification with AP/GP and CI

Matrices (Question 6a)

Verify that A · (\text{adj } A) = (\text{adj } A) · A = |A| · I_3 for
A = \begin{bmatrix} 2 & 3 & 4 \\ 3 & 0 & 1 \\ 2 & 1 & 5 \end{bmatrix}.

Tasks:

• Find the determinant |A|:
• Find the Adjoint (\text{adj } A): This involves finding the cofactor of each element and then transposing the resulting matrix.
• Cofactors: C₁₁ = -1, C₁₂ = -13, C₁₃ = 3, C₂₁ = -11, C₂₂ = 2, C₂₃ = 4, C₃₁ = 3, C₃₂ = 10, C₃₃ = -9
• Multiply A · (\text{adj } A)

4. Financial Arithmetic (Question 2g)

Find the compound interest on Rs. 8,000 for 1 1/2 years at 10% per annum, compounded annually.

• Amount for the first year:
• Interest for the next half year: Use simple interest on the new principal.
• Total Compound Interest:

Answers use standard AP/GP and matrix formulas.[1][2]

5(a) AP: 7th term 23, 12th term 38

Let first term be a and common difference d.

7th term: a+6d=23. 12th term: a+11d=38.

Subtracting: (a+11d)-(a+6d)=38-23 \Rightarrow 5d=15 \Rightarrow d=3.

Then a+6\cdot3=23 \Rightarrow a=5.

• First term a=5
• Common difference d=3

The AP is: 5, 8, 11, 14,  ... .[2]

5(b) Sum of GP 2, -4, 8, ..., -4096

Here a=2, common ratio r=-2.[1]

Use a_n = a r^{n-1} with last term -4096:

-4096 = 2(-2)^{n-1} \Rightarrow -2048 = (-2)^{n-1} = -2^{11} \Rightarrow n-1=11 \Rightarrow n=12.

Sum of n terms of GP: S_n = \dfrac{a(r^n-1)}{r-1}.[1]

So

S_{12} = \frac{2((-2)^{12}-1)}{-2-1} = \frac{2(4096-1)}{-3} = \frac{2\cdot4095}{-3} = -2730.

• Number of terms n=12
• Sum of the series S=-2730

6(a) Matrix verification

Given

A = \begin{bmatrix} 2 & 3 & 4 \\ 3 & 0 & 1 \\ 2 & 1 & 5 \end{bmatrix}.

Step 1: Determinant |A|

|A| = 2(0\cdot5 - 1\cdot1) - 3(3\cdot5 - 1\cdot2) + 4(3\cdot1 - 0\cdot2) \end{code}

= 2(-1) - 3(15-2) + 4(3) = -2 - 3\cdot13 + 12 = -2 - 39 + 12 = -29.

So |A| = -29.

Step 2: Adjoint \(\operatorname{adj}A\)

Cofactors:

• C_{11} = \det\begin{bmatrix}0 & 1 \\ 1 & 5\end{bmatrix} = 0\cdot5 - 1\cdot1 = -1.
• C_{12} = -\det\begin{bmatrix}3 & 1 \\ 2 & 5\end{bmatrix} = -(3\cdot5 - 1\cdot2) = -(15-2) = -13.
• C_{13} = \det\begin{bmatrix}3 & 0 \\ 2 & 1\end{bmatrix} = 3\cdot1 - 0\cdot2 = 3.
• C_{21} = -\det\begin{bmatrix}3 & 4 \\ 1 & 5\end{bmatrix} = -(3\cdot5 - 4\cdot1) = -(15-4) = -11.
• C_{22} = \det\begin{bmatrix}2 & 4 \\ 2 & 5\end{bmatrix} = 2\cdot5 - 4\cdot2 = 10 - 8 = 2.
• C_{23} = -\det\begin{bmatrix}2 & 3 \\ 2 & 1\end{bmatrix} = -(2\cdot1 - 3\cdot2) = -(2-6) = 4.
• C_{31} = \det\begin{bmatrix}3 & 4 \\ 0 & 1\end{bmatrix} = 3\cdot1 - 4\cdot0 = 3.
• C_{32} = -\det\begin{bmatrix}2 & 4 \\ 3 & 1\end{bmatrix} = -(2\cdot1 - 4\cdot3) = -(2-12) = 10.
• C_{33} = \det\begin{bmatrix}2 & 3 \\ 3 & 0\end{bmatrix} = 2\cdot0 - 3\cdot3 = -9.

Cofactor matrix:

C = \begin{bmatrix} -1 & -13 & 3 \\ -11 & 2 & 4 \\ 3 & 10 & -9 \end{bmatrix}.

Adjoint is transpose of the cofactor matrix:

\operatorname{adj}A = \begin{bmatrix} -1 & -11 & 3 \\ -13 & 2 & 10 \\ 3 & 4 & -9 \end{bmatrix}.

Step 3: Verify \(A\operatorname{adj}A = |A| I_3\)

Compute A\operatorname{adj}A by rows and columns:

• First row:
  • (1,1): 2(-1)+3(-13)+4(3) = -2 - 39 + 12 = -29
  • (1,2): 2(-11)+3(2)+4(4) = -22 + 6 + 16 = 0
  • (1,3): 2(3)+3(10)+4(-9) = 6 + 30 - 36 = 0
• Second row:
  • (2,1): 3(-1)+0(-13)+1(3) = -3 + 0 + 3 = 0
  • (2,2): 3(-11)+0(2)+1(4) = -33 + 0 + 4 = -29
  • (2,3): 3(3)+0(10)+1(-9) = 9 + 0 - 9 = 0
• Third row:
  • (3,1): 2(-1)+1(-13)+5(3) = -2 - 13 + 15 = 0
  • (3,2): 2(-11)+1(2)+5(4) = -22 + 2 + 20 = 0
  • (3,3): 2(3)+1(10)+5(-9) = 6 + 10 - 45 = -29

Thus

A\operatorname{adj}A = \begin{bmatrix} -29 & 0 & 0 \\ 0 & -29 & 0 \\ 0 & 0 & -29 \end{bmatrix} = -29 I_3 = |A| I_3.

Answers only, stepwise.

8(a)

Given: Difference between CI and SI for 2 years at 4% p.a. is Re. 1.

For 2 years:

\text{CI} - \text{SI} = P\left(\dfrac{r}{100}\right)^2

So

1 = P\left(\dfrac{4}{100}\right)^2 = P \cdot \dfrac{16}{10000} = P \cdot \dfrac{4}{2500} \Rightarrow P = \dfrac{1 \times 2500}{4} = 625.

Required sum = Rs. 625.

8(b)

Principal P = 6950.[2]

First 2 years: 6% p.a., interest payable half‑yearly.

Half‑yearly rate = 3%; number of half‑years in 2 years = 4.

Amount after 2 years:

A_2 = 6950\left(1+\dfrac{3}{100}\right)^4 = 6950(1.03)^4.

(1.03)^2 = 1.0609, \quad (1.03)^4 = (1.0609)^2 \approx 1.1255.

A_2 \approx 6950 \times 1.1255 \approx 7813.2.

Third year: 9% p.a., half‑yearly → each half‑year 4.5%, 2 half‑years.

Amount after 3rd year:

A_3 = A_2\left(1+\dfrac{4.5}{100}\right)^2 = A_2(1.045)^2.

(1.045)^2 \approx 1.0920, \quad A_3 \approx 7813.2 \times 1.0920 \approx 8539.7.

\text{CI} = A_3 - P \approx 8539.7 - 6950 \approx 1589.7 \approx 1590.

Required compound interest = Rs. 1590 (approx.).