TS Inter 1st Year Maths 1A Solutions Chapter 3 Matrices Ex 3(a)

Students must practice these TS Intermediate Maths 1A Solutions Chapter 3 Matrices Ex 3(a) to find a better approach to solving the problems.

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

Question 1.
Write the following as a single matrix.
(i) [2 1 3] + [0 0 0]
Answer:
[2 1 3] + [0 0 0] = [2 + 0 1 + 0 3 + 0]
= [2 1 3]

(ii) \(\left[\begin{array}{r}
0 \\
1 \\
-1
\end{array}\right]+\left[\begin{array}{r}
-1 \\
1 \\
0
\end{array}\right]\)
Answer:
\(\left[\begin{array}{r}
0 \\
1 \\
-1
\end{array}\right]+\left[\begin{array}{r}
-1 \\
1 \\
0
\end{array}\right]\) = \(\left[\begin{array}{r}
0-1 \\
1+1 \\
-1+0
\end{array}\right]\) = \(\left[\begin{array}{r}
-1 \\
2 \\
-1
\end{array}\right]\)

(iii) \(\left[\begin{array}{ccc}
3 & 9 & 0 \\
1 & 8 & -2
\end{array}\right]+\left[\begin{array}{ccc}
4 & 0 & 2 \\
7 & 1 & 4
\end{array}\right]\)
Answer:

(iv) \(\left[\begin{array}{rr}
-1 & 2 \\
1 & -2 \\
3 & -1
\end{array}\right]+\left[\begin{array}{rr}
0 & 1 \\
-1 & 0 \\
-2 & 1
\end{array}\right]\)
Answer:

Question 2.
If A = \(\left[\begin{array}{cc}
-1 & 3 \\
4 & 2
\end{array}\right]\), B = \(\left[\begin{array}{cc}
2 & 1 \\
3 & -5
\end{array}\right]\), X = \(\left[\begin{array}{ll}
\mathbf{x}_1 & \mathbf{x}_2 \\
\mathbf{x}_3 & \mathbf{x}_4
\end{array}\right]\) and A + B = X then find the values of x₁, x₂, x₃ and x₄.
Answer:
A + B = X

⇒ x₁ = 1, x₂ = 4, x₃ = 7, x₄ = – 3.

Question 3.
If A = \(\left[\begin{array}{ccc}
-1 & -2 & 3 \\
1 & 2 & 4 \\
2 & -1 & 3
\end{array}\right]\) B = \(\left[\begin{array}{ccc}
1 & -2 & 5 \\
0 & -2 & 2 \\
1 & 2 & -3
\end{array}\right]\) and C = \(\left[\begin{array}{ccc}
-2 & 1 & 2 \\
1 & 1 & 2 \\
2 & 0 & 1
\end{array}\right]\) then find A + B + C.
Answer:
A + B + C =

Question 4.
If A = \(\left[\begin{array}{ccc}
3 & 2 & -1 \\
2 & -2 & 0 \\
1 & 3 & 1
\end{array}\right]\), B = \(\left[\begin{array}{ccc}
-3 & -1 & 0 \\
2 & 1 & 3 \\
4 & -1 & 2
\end{array}\right]\) and X = A + B then find X.
Answer:
X = A + B

Question 5.
If \(\left[\begin{array}{cc}
x-3 & 2 y-8 \\
z+2 & 6
\end{array}\right]\) = \(\left[\begin{array}{cc}
5 & 2 \\
-2 & a-4
\end{array}\right]\) then find the values of x, y, z and a. [May 2006, Mar. 14]
Answer:
Given \(\left[\begin{array}{cc}
x-3 & 2 y-8 \\
z+2 & 6
\end{array}\right]\) = \(\left[\begin{array}{cc}
5 & 2 \\
-2 & a-4
\end{array}\right]\)
We have x – 3 = 5, 2y – 8 = 2, z + 2 = – 2, a – 4 = 6
⇒ x = 8, y = 5, z = – 4, a = 10

II.
Question 1.
If \(\left[\begin{array}{ccc}
x-1 & 2 & 5-y \\
0 & z-1 & 7 \\
1 & 0 & a-5
\end{array}\right]\) = \(\left[\begin{array}{lll}
1 & 2 & 3 \\
0 & 4 & 7 \\
1 & 0 & 0
\end{array}\right]\) then find the values x, y, z and a.
Answer:
Given \(\left[\begin{array}{ccc}
x-1 & 2 & 5-y \\
0 & z-1 & 7 \\
1 & 0 & a-5
\end{array}\right]\) = \(\left[\begin{array}{lll}
1 & 2 & 3 \\
0 & 4 & 7 \\
1 & 0 & 0
\end{array}\right]\)
we have x – 1 = 1, 5 – y = 3, z – 1 = 4,
a – 5 = 0
⇒ x = 2, y = 2, z = 5, a = 5

Question 2.
Find the trace of \(\left[\begin{array}{ccc}
1 & 3 & -5 \\
2 & -1 & 5 \\
1 & 0 & 1
\end{array}\right]\)
Answer:
Trace of the given matrix
= 1 – 1 + 1 = sum of the diagonal elements
= 1

Question 3.
If A = \(\left[\begin{array}{ccc}
0 & 1 & 2 \\
2 & 3 & 4 \\
4 & 5 & -6
\end{array}\right]\) and B = \(\left[\begin{array}{ccc}
-1 & 2 & 3 \\
0 & 1 & 0 \\
0 & 0 & -1
\end{array}\right]\) find A – B and 4A – 5B.
Answer:

Question 4.
If A = \(\left[\begin{array}{lll}
1 & 2 & 3 \\
3 & 2 & 1
\end{array}\right]\) and B = \(\left[\begin{array}{lll}
3 & 2 & 1 \\
1 & 2 & 3
\end{array}\right]\) find 3B – 2A.
Answer: