TS Inter 1st Year Maths 1A Matrices Solutions Exercise 3(e)

I.
Question 1.
Find the adjoint and inverse of the following matrices. (March 2002)

i) $\left[\begin{array}{rr} 2 & -3 \\ 4 & 6 \end{array}\right]$
Answer:
If A = $\left[\begin{array}{ll} a & b \\ c & d \end{array}\right]$ then adj A = $\left[\begin{array}{rr} \mathrm{d} & -\mathrm{b} \\ -\mathrm{c} & \mathrm{a} \end{array}\right]$

ii) $\left[\begin{array}{cc} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right]$
Answer:

iii) Find the adjoint and inverse of the matrix $\left[\begin{array}{lll} 1 & 0 & 2 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{array}\right]$.
Answer:
Find cofactors of elements in the matrix as

iv) $\left|\begin{array}{lll} 2 & 1 & 2 \\ 1 & 0 & 1 \\ 2 & 2 & 1 \end{array}\right|$
Answer:

Question 2.
If A = $\left[\begin{array}{cc} \mathrm{a}+\mathrm{i b} & \mathrm{c}+\mathrm{i d} \\ -\mathrm{c}+\mathrm{i d} & \mathrm{a}-\mathrm{i b} \end{array}\right]$, a² + b² + c² + d² = 1

Answer:
det A = (a + ib) (a – ib) – (c + id) (- c + id)
= (a² – i² b²) – (- c² + i²d²)
= a² + b² + c² + d² (∵ i² = -1)
= 1
Adj A = $\left[\begin{array}{cc} a-i b & -c-i d \\ c-i d & a+i b \end{array}\right]$
A^-1 = $\frac{{Adj} A}{{det} A}=\left[\begin{array}{cc} a-i b & -c-i d \\ c-i d & a+i b \end{array}\right]$

Question 3.
If A = $\left[\begin{array}{ccc} 1 & -2 & 3 \\ 0 & -1 & 4 \\ -2 & 2 & 1 \end{array}\right]$, then find (A’)^-1. (Board Model Paper)

Answer:

Question 4.
If A = $\left[\begin{array}{rrr} -1 & -2 & -2 \\ 2 & 1 & -2 \\ 2 & -2 & 1 \end{array}\right]$, then show that the adjoint of A = 3A, find A^-1

Answer:

Question 5.
If abc ≠ 0; find the inverse of $\left[\begin{array}{lll} \mathrm{a} & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{array}\right]$ (May 2006)

Answer:

II.
Question 1.
If A = $\left[\begin{array}{lll} \mathrm{b}+\mathrm{c} & \mathrm{c}-\mathrm{a} & \mathrm{b}-\mathrm{a} \\ \mathrm{c}-\mathrm{b} & \mathrm{c}+\mathrm{a} & \mathrm{a}-\mathrm{b} \\ \mathrm{b}-\mathrm{c} & \mathrm{a}-\mathrm{c} & \mathrm{a}+\mathrm{b} \end{array}\right]$ and B = $\frac{1}{2}\left[\begin{array}{lll} \mathrm{b}+\mathrm{c} & \mathrm{c}-\mathrm{a} & \mathrm{b}-\mathrm{a} \\ \mathrm{c}-\mathrm{b} & \mathrm{c}+\mathrm{a} & \mathrm{a}-\mathrm{b} \\ \mathrm{b}-\mathrm{c} & \mathrm{a}-\mathrm{c} & \mathrm{a}+\mathrm{b} \end{array}\right]$, then show that ABA^-1 is a diagonal matrix.

Answer:

Question 2.
If 3A = $\left[\begin{array}{ccc} 1 & 2 & 2 \\ 2 & 1 & -2 \\ -2 & 2 & -1 \end{array}\right]$, then show that A^-I = A’.

Answer:

∴ A.A’ = I and by definition A’ = A^-1
similarly A’.A = I

Question 3.
If A = $\left[\begin{array}{rrr} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{array}\right]$, then show that A^-1 = A³

Answer:

So, the multiplicative inverse of A exists and it is A³.
∴ A^-1 = A³

Question 4.
If AB = I or BA = I, then prove that A is invertible and B = A^-1.
Answer:
Given AB = I
⇒ |AB| = |I|
⇒ |A| |B| = 1
⇒ |A| ≠ 0
∴ A is a non-singular matrix.
Also BA = I
⇒ |B| |A| = |I|
⇒ |A| |B| = 1
⇒ |A| *0
∴ A is a non-singular matrix.
⇒ A is invertible
⇒ A^-1 exists AB = I
⇒ A^-1 AB = A^-1I
⇒ (A^-1 A) B = A^-1I
⇒ IB = A^-1I
⇒ B = A^-1.