TS Inter 1st Year Maths 1A Matrices Important Questions Long Answer Type

Students must practice these Maths 1A Important Questions TS Inter 1st Year Maths 1A Matrices Important Questions Long Answer Type to help strengthen their preparations for exams.

TS Inter 1st Year Maths 1A Matrices Important Questions Long Answer Type

Question 1.
Without expanding the determinant show that \(\left|\begin{array}{lll}
\mathbf{b}+\mathbf{c} & \mathbf{c}+\mathbf{a} & \mathbf{a}+\mathbf{b} \\
\mathbf{c}+\mathbf{a} & \mathbf{a}+\mathbf{b} & \mathbf{b}+\mathbf{c} \\
\mathbf{a}+\mathbf{b} & \mathbf{b}+\mathbf{c} & \mathbf{c}+\mathbf{a}
\end{array}\right|\) = 2\(\left|\begin{array}{lll}
\mathbf{a} & \mathbf{b} & \mathbf{c} \\
\mathbf{b} & \mathbf{c} & \mathbf{a} \\
\mathbf{c} & \mathbf{a} & \mathbf{b}
\end{array}\right|\) [Mar. 15 (AP); May 98, 96, 91]
Answer:

Question 2.
Show that \(\left|\begin{array}{ccc}
1 & a^2 & a^3 \\
1 & b^2 & b^3 \\
1 & c^2 & c^3
\end{array}\right|\) = (a – b) (b – c) (c – a) (ab + bc + ca). [Mar. 17(AP), 09: May 15 (AP); 02]
Answer:

= (a – b) (b – c) (c – a) [0 (c³ – c² (a + b + c) – (a + b) (0 – a – b – c) + (a² + ab + b²) (0 – 1)]
= (a – b) (b – c) (c – a) [0 + a² + ab + ac + ab + b² + bc – a² – ab – b²]
= (a – b) (b – c) (c – a) (ab + bc + ca) = RHS.

Question 3.
Show that \(\left|\begin{array}{ccc}
\mathbf{a}-\mathbf{b}-\mathbf{c} & \mathbf{2 a} & \mathbf{2 a} \\
\mathbf{2 b} & \mathbf{b}-\mathbf{c}-\mathbf{a} & \mathbf{2 b} \\
\mathbf{2 c} & \mathbf{2 c} & \mathbf{c}-\mathbf{a}-\mathbf{b}
\end{array}\right|\) = (a + b + c)³. [Mar. 11; May 11]
Answer:

Question 4.
Find the value of x if \(\left|\begin{array}{ccc}
x-2 & 2 x-3 & 3 x-4 \\
x-4 & 2 x-9 & 3 x-16 \\
x-8 & 2 x-27 & 3 x-64
\end{array}\right|\) = 0 [Mar 15 (TS); Mar. 06]
Answer:

⇒ (x – 2) (30 – 24) – (2x – 3) (10 – 6) + (3x – 4) (4 – 3) = 0
⇒ (x – 2)6 – (2x – 3)4 + (3x – 4)(1)
⇒ 6x – 12 – 8x + 12 + 3x – 4 = 0
⇒ x – 4 = 0
⇒ x = 4.

Question 5.
Show that \(\left|\begin{array}{ccc}
a+b+2 c & a & b \\
c & b+c+2 a & b \\
c & a & c+a+2 b
\end{array}\right|\) = 2 (a + b + c)³ [Mar. 18, 16 (AP); Mar. 16 (TS), 10 Mar.19 (TS), May 12, 10, 08, 03, 99]
Answer:
= 2 (a + b + c)² [1{(c + a + 2b) – 0} – a (0 – 0) + b (0 – 1)]
= 2(a + b + c)² [c + a + 2b – b]
= 2(a + b + c)² (a + b + c) = 2(a + b + c)³

Question 6.
Show that \(\left|\begin{array}{lll}
a & b & c \\
b & c & a \\
c & a & b
\end{array}\right|^2=\left|\begin{array}{ccc}
2 b c-a^2 & c^2 & b^2 \\
c^2 & 2 a c-b^2 & a^2 \\
b^2 & a^2 & 2 a b-c^2
\end{array}\right|\) = (a³ + b³ + c³ – 3abc)². [Mar. 19 (AP) Mar. 18 (TS): May 14. 09; Mar. 12, 01]
Answer:

Question 7.
Show that \(\left|\begin{array}{ccc}
a^2+2 a & 2 a+1 & 1 \\
2 a+1 & a+2 & 1 \\
3 & 3 & 1
\end{array}\right|\) = (a – 1)³ [Mar. 13, 07]
Answer:

= (a – 1)² [(a + 1) (1 – 0) – 1 (2 – 0) + 0 (6 – 3)]
= (a – 1)² [a + 1 – 2] = (a – 1)² (a – 1) = (a – 1)³ = R.H.S.

Question 8.
Show that \(\left|\begin{array}{ccc}
\mathbf{a} & \mathbf{b} & \mathbf{c} \\
\mathbf{a}^2 & \mathbf{b}^2 & \mathbf{c}^2 \\
\mathbf{a}^3 & \mathbf{b}^3 & \mathbf{c}^3
\end{array}\right|\) = abc (a – b) (b – c) (c – a). [May. 06]
Answer:

= abc(a – b)(b – c) [0(c² – bc – c²) – 0(c² – ac – bc) + 1(b + c – a – b)]
= abc (a – b) (b – c) (c – a) = R.H.S

Question 9.
If A = \(\left[\begin{array}{lll}
\mathbf{a}_1 & \mathbf{b}_1 & \mathbf{c}_1 \\
\mathbf{a}_2 & \mathbf{b}_2 & \mathbf{c}_2 \\
\mathbf{a}_{\mathbf{3}} & \mathbf{b}_3 & \mathbf{c}_3
\end{array}\right]\) is a non-singular matrix, then show that A is invertiable and A^-1 = \(\frac{{Adj} \mathbf{A}}{{det} \mathbf{A}}\). [Mar. 17 (AP). May 15 (AP). 13, 10, 07, 06, 02, Mar. 07, 02, 99, 94, 82, 80]
Answer:

Question 10.
Solve the system of equations 3x + 4y + 5z = 18, 2x – y + 8z = 13, 5x – 2y + 7z = 20 by using Cramer’s rule. [Mar. 12, 03; May 09]
Answer:
Given system of linear equations are 3x + 4y + 5z = 18, 2x – y + 8z = 13, 5x – 2y + 7z = 20
Let A = \(\left[\begin{array}{rrr}
3 & 4 & 5 \\
2 & -1 & 8 \\
5 & -2 & 7
\end{array}\right]\), X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\) and D = \(\left[\begin{array}{l}
18 \\
13 \\
20
\end{array}\right]\)
Then we can write the given equations in the form of matrix equation as AX = D.
Δ = det A = \(\left|\begin{array}{rrr}
3 & 4 & 5 \\
2 & -1 & 8 \\
5 & -2 & 7
\end{array}\right|\) = 3 (- 7 + 16) – 4 (14 – 40) + 5 (- 4 + 5)
= 3(9) – 4 (- 26) + 5 (1)
= 27 + 104 + 5 = 136 ≠ 0
Hence, we can solve the given equations by using Cramer’s rule.
Δ₁ = \(\left|\begin{array}{rrr}
18 & 4 & 5 \\
13 & -1 & 8 \\
20 & -2 & 7
\end{array}\right|\) = 18(- 7 + 16) – 4(91 – 160) + 5(- 26 + 20) = 18(9) – 4(- 69) + 5(- 6) = 162 + 276 – 30 = 408
Δ₂ = \(\left|\begin{array}{lll}
3 & 18 & 5 \\
2 & 13 & 8 \\
5 & 20 & 7
\end{array}\right|\) = 3(91 – 160) – 18(14 – 40) + 5(40 – 65) = 3(- 69) – 18(- 26) + 5(- 25) = – 207 + 468 – 125 = 136
Δ₃ = \(\left|\begin{array}{rrr}
3 & 4 & 18 \\
2 & -1 & 13 \\
5 & -2 & 20
\end{array}\right|\) = 3(- 20 + 26) – 4(40 – 65) + 18(- 4 + 5) = 3(6) – 4(- 25) + 18(1)
= 18 + 100 + 18 = 136
Hence, by Cramer’s rule,
x = \(\frac{\Delta_1}{\Delta}=\frac{408}{136}\) = 3, y = \(\frac{\Delta_2}{\Delta}=\frac{136}{136}\) = 1, z = \(\frac{\Delta_3}{\Delta}=\frac{136}{136}\) = 1
∴ The solution of the giveñ system of equations is x = 3, y = 1, z = 1.

Solve the system of equations 2x – y + 3z =9, x + y + z = 6, x – y + z = 2 by using Cramer’s rule. [Mar. 17 (TS), 16 (AP), 02; May 13]
Answer:
x = 1, y = 2, z = 3

Question 11.
Solve: 3x + 4y + 5z = 18, 2x – y + 8z = 13 and 5x – 2y + 7z = 20 by using the matrix inversion method. [Mar. ‘19 (TS): Mar. ‘15 (AP) ; Mar. ‘13. ‘08, ‘01, ‘00, 96]
Answer:
Given system of linear equations are 3x + 4y + 5z = 18, 2x – y + 8z = 13, 5x – 2y + 7z = 20
Let A = \(\left[\begin{array}{rrr}
3 & 4 & 5 \\
2 & -1 & 8 \\
5 & -2 & 7
\end{array}\right]\), X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\), D = \(\left[\begin{array}{l}
18 \\
13 \\
20
\end{array}\right]\)
This can be represented as AX = B and X = A1B is a solution.
∆ = det A = \(\left|\begin{array}{rrr}
3 & 4 & 5 \\
2 & -1 & 8 \\
5 & -2 & 7
\end{array}\right|\) = 3(- 7 + 16) – 4(14 – 40) + 5(- 4 + 5) = 3(9) – 4(- 26) + 5(1) = 27 + 104 + 5 = 136
Cofactor of 3 is A₁ = +(- 7 + 16) = 9
Cofactor of 5 is A₃ = +(32 + 5) = 37
Cofactor of -1 is B₂ = +(21 – 25) = – 4
Cofactor of 5 is C₁ = +(- 4 + 5) = 1
Cofactor of 7 is C₃ = +(- 3 – 8) = – 11
Cofactor of 2 is A₂ = – (28 + 10) = – 38.
Cofactor of 4 is B₁ = – (14 – 40) = 26
Cofactor of -2 is B₃ = – (24 – 10) = – 14
Cofactor of 8 is C₂ = – (- 6 – 20) = 26

Solve 2x – y + 3z = 8, – x + 2y + z = 4, 3x + y – 4z = 0 by using matrix Inversion method. [May 15 (AP); May. 12]
Answer:
x = 2, y = 2, z = 2

Solve x + y + z = 1, 2x + 2y + 3z = 6, x + 4y + 9z = 3 by using matrix Inversion method. [May 03, 93]
Answer:
x = 7, y = – 10, z = 4

Question 12.
Solve the equations 3x + 4y + 5z = 18, 2x – y + 8z = 13 and 5x – 2y + 7z = 20 by Gauss-Jordan method. [May 15(TS): May 06, 01: Mar. 01]
Answer:
Given system of Linear equations are 3x + 4y + 5z = 18, 2x – y + 8z = 13 and 5x – 2y + 7z = 20

In this case, system of equations have unique solution.
i.e., x = 3, y = 1, z = 1.
∴ The solution of the given system of equations is x = 3, y = 1, z = 1.

Question 13.
Solve the equation 2x – y + 3z = 9, x + y + z = 6, x – y + z = 2 by Gauss-Jordan method. [Mar. 18(AP): Mar. 11, 10; May 11]
Answer:
The given system of linear equations are 2x – y + 3z = 9, x + y + z = 6, x – y + z = 2

In this case, system of equations has unique solution i.e., x = 1; y = 2; z = 3.
∴ The solution of given system of equations is x = 1; y = 2; z = 3.

Question 14.
Solve the following system of equations by Gauss – Jordan method : x + y + z = 9, 2x + 5y + 7z = 52, 2x + y – z = 0. [May 10, 07; Mar. 09, 1, 99]
Answer:
Given system of linear equations are x + y + z = 9, 2x + 5y + 7z = 52; 2x + y – z = 0. The matrix form is AX = D

In this case, the system of equations has unique solution i.e., x = 1; y = 3; z 5.
∴ The solution of given system of equations is x = 1; y = 3; z = 5.

Solve the equations 2x – y + 3z = 8, – x + 2y + z = 4, 3x + y – 4z = 0 by Gauss-Jordan method. [Mar. 07]
Answer:
x = 2, y = 2, z = 2

Solve the equations 2x – y + 8z = 13, 3x + 4y + 5z = 18, 5x – 2y + 7z = 20 by Gauss-Jordan method. [May 09; Mar. 03]
Answer:
x = 3, y = 1, z = 1

Question 15.
Examine whether the system of equations x + y + z = 9, 2x + 5y + 7z = 52, 2x + y – z = 0 are consistent or Inconsistent and If consistent, find the complete solution. [May 15(TS); May 11]
Answer:
The given system of linear equations are x + y + z = 9, 2x + 5y + 7z = 52, 2x + y – z = 0.

∴ Rank [A] = 3
Now, Rank [AD] = 3
Since, the 3 × 3 sub matrix is \(\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\) whose det is 1 ≠ 0.
∴ Rank [A] = Rank [AD] = 3.
In this case, system of equations has unique solution. i.e., x = 1, y = 3, z = 5.
Hence, the given system is consistent.

Question 16.
Examine whether the system of equations x + y + z = 6, x – y + z = 2, 2x – y + 3z = 9 are consistent or inconsistent and if consistent, find the complete solution. [Mar.11, 05]
Answer:
Given system of linear equations are x + y + z = 6, x – y + z = 2, 2x – y + 3z = 9
The given system of equations can be written as AX = D

∴ Rank of AD = 3.
Rank [A] = Rank [AD] = 3.
In this case, system of equations has unique solution. i.e., x = 1, y = 2, z = 3.
Hence, given system is consistent.

Question 17.
Examine whether the system of equations x + y + z = 1, 2x + y + z = 2, x + 2y + 2z = 1 are consistent or Inconsistent and if consistent, find the complete solution. [Mar. 15 (TS); May 05]
Answer:
Given system of linear equations are x + y + z = 1; 2x + y + z = 2; x + 2y + 2z = 1
The system of equations can be written as AX = D

∴ Rank [A] = Rank [AD]
In this case, system of equations has infinitely many solutions. x = 1; y + z = 0
∴ The solution of the given system of equations is x = 1, y + z = 0.
Hence, the given system is consistent.

Question 18.
Examine whether the following system of equations x + y + z = 6, x + 2y + 3z = 10, x + 2y + 4z = 1 are consistent or Inconsistent and if consistent, find the complete solution. [May. 02]
Answer:
The given system of linear equations are x + y + z = 6; x – 2y + 3z = 10; x + 2y – 4z = 1
The system of equations can be written as AX = D

∴ Rank [AD] = 3
∴ Rank [A] = Rank [AD] = 3
∴ In this case, system of equations has unique solution.
i.e., x = – 7; y = 22; z = – 9
Hence, the given system is consistent.

Question 19.
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:

Question 20.
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 of x, y, z and a.
Answer:
Given \(\left[\begin{array}{ccc}
x-1 & 2 & 5-y \\
0 & z-1 & 7 \\
1 & 0 & a-5
\end{array}\right]\) = \(\left[\begin{array}{lll}
1 & 2 & 3 \\
0 & 4 & 7 \\
1 & 0 & 0
\end{array}\right]\)
From equality of matrices
x – 1 = 1 ⇒ x = 2
5 – y = 3 ⇒ y = 2
z – 1 = 4 ⇒ z = 5
a – 5 = 0 ⇒ a = 5
∴ x = 2, y = 2, z = 5, a = 5

Question 21.
If \(\left[\begin{array}{ccc}
x-1 & 2 & y-5 \\
z & 0 & 2 \\
1 & -1 & 1+a
\end{array}\right]=\left[\begin{array}{ccc}
1-x & 2 & -y \\
2 & 0 & 2 \\
1 & -1 & 1
\end{array}\right]\) then find the values of x, y, z and a.
Answer:
Given
\(\left[\begin{array}{ccc}
x-1 & 2 & y-5 \\
z & 0 & 2 \\
1 & -1 & 1+a
\end{array}\right]=\left[\begin{array}{ccc}
1-x & 2 & -y \\
2 & 0 & 2 \\
1 & -1 & 1
\end{array}\right]\)
From the equality of matrices,
x – 1 = 1 – x ⇒ 2x = 2 ⇒ x = 1
y – 5 = – y ⇒ 2y = 5 ⇒ y = 5/2
z = 2
1 + a = 1 ⇒ a = 0
∴ x = 1, y = 5/2, z = 2, a = 0

Question 22.
find the trace of A
if A = \(\left[\begin{array}{ccc}
1 & 2 & -1 / 2 \\
0 & -1 & 2 \\
-1 / 2 & 2 & 1
\end{array}\right]\).
Answer:
Given A = \(\left[\begin{array}{ccc}
1 & 2 & -1 / 2 \\
0 & -1 & 2 \\
-1 / 2 & 2 & 1
\end{array}\right]\)
The elements of the principal diagonal of ‘A’ are 1, – 1, 1
Hence, the trace of A = 1 + (- 1) + 1
= 1 – 1 + 1 = 1

Question 23.
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 B – A and 4A – 5B.
Answer:

Question 24.
If A = \(\left[\begin{array}{lll}
0 & 1 & 2 \\
2 & 3 & 4 \\
4 & 5 & 6
\end{array}\right]\) and B = \(\left[\begin{array}{ccc}
1 & -2 & 0 \\
0 & 1 & -1 \\
-1 & 0 & 3
\end{array}\right]\) find A – B and 4B – 3A.
Answer:

Question 25.
If A = \(\left[\begin{array}{rrr}
1 & -2 & 3 \\
-4 & 2 & 5
\end{array}\right]\) and B = \(\left[\begin{array}{ll}
2 & 3 \\
4 & 5 \\
2 & 1
\end{array}\right]\) do AB and BA exist? If they exist find them. Do A and B commute with respect to multiplication?
Answer:
Given A = \(\left[\begin{array}{rrr}
1 & -2 & 3 \\
-4 & 2 & 5
\end{array}\right]\), B = \(\left[\begin{array}{ll}
2 & 3 \\
4 & 5 \\
2 & 1
\end{array}\right]\)
The order of matrix A is 2 × 3
The order of matrix B is 3 × 2
The no.of columns in A The no.of rows in B.
∴ AB is defined

The no.of columns in B = The no.of rows in A.
∴ BA is defined.

∴ AB ≠ BA
∴ A and B is not commute with respect to multiplication.

Question 26.
Find A² where A = \(\left[\begin{array}{cc}
4 & 2 \\
-1 & 1
\end{array}\right]\).
Answer:

Question 27.
If A = \(\left[\begin{array}{ll}
\mathrm{i} & 0 \\
0 & \mathrm{i}
\end{array}\right]\) find A².
Answer:

Question 28.
If A = \(\left[\begin{array}{ccc}
1 & 1 & 3 \\
5 & 2 & 6 \\
-2 & -1 & -3
\end{array}\right]\) then find A³.
Answer:

Question 29.
If A = \(\left[\begin{array}{cc}
-1 & 2 \\
0 & 1
\end{array}\right]\) then find AA’. Do A and A’ commute with respect to multiplication of matrices?
Answer:

∴ A and A’ do not commute with respect to multiplication of matrices.

Question 30.
Find the determinant of the matrix \(\left[\begin{array}{lll}
a & h & g \\
h & b & f \\
g & f & c
\end{array}\right]\).
Answer:
Let A = \(\left[\begin{array}{lll}
a & h & g \\
h & b & f \\
g & f & c
\end{array}\right]\)
det A = a(bc – f²) – h(ch – gf) + g(hf – bg)
= abc – af² – ch² + fgh + fgh – bg²
= abc + 2fgh – af² – bg² – ch²

Question 31.
Find the determinant of the matrix \(\left[\begin{array}{lll}
a & b & c \\
b & c & a \\
c & a & b
\end{array}\right]\).
Answer:
Let A = \(\left[\begin{array}{lll}
a & b & c \\
b & c & a \\
c & a & b
\end{array}\right]\)
det A = a(bc – a²) – b(b² – ac) + c(ab – c²)
= abc – a³ – b³ + abc + abc – c³
= 3abc – a³ – b³ – c³

Question 32.
Find the adjoint and the inverse of the matrix A = \(\).
Answer:
Given A = \(\left[\begin{array}{cc}
1 & 2 \\
3 & -5
\end{array}\right]\)
Cofactor of 1 is A₁ = + (- 5) = – 5
Cofactor of 2 is B₁ = – (3) = – 3
Cofactor of 3 is A₂ = – (2) = – 2
Cofactor of – 5 is B₂ = +(1) = 1
∴ The cofactor matrix of A is

Now det A = ad – bc = 1(- 5) – 2(3)
= – 5 – 6 = – 11 ≠ 0
Hence A is invertiable.

Question 33.
If A = \(\left[\begin{array}{lll}
1 & 2 & 2 \\
2 & 1 & 2 \\
2 & 2 & 1
\end{array}\right]\) then show that A² – 4A – 5I = 0. [Mar. 16(AP)]
Answer:
Given A = \(\left[\begin{array}{lll}
1 & 2 & 2 \\
2 & 1 & 2 \\
2 & 2 & 1
\end{array}\right]\)

Question 34.
Find the adjoint and the inverse of the matrix \(\left[\begin{array}{lll}
1 & 0 & 2 \\
2 & 1 & 0 \\
3 & 2 & 1
\end{array}\right]\).
Answer:
Let A = \(\left[\begin{array}{lll}
1 & 0 & 2 \\
2 & 1 & 0 \\
3 & 2 & 1
\end{array}\right]\)
Cofactor of 1, A₁ =+(1 – 0) = 1
Cofactor of 0, B₁ = – (2 – 0) = – 2
Cofactor of 2, C₁ = + (4 – 3) = 1
Cofactor of 2, A₂ = – (0 – 4) = 4
Cofactor of 1, B₂ = + (1 – 6) = – 5
Cofactor of 0, C₂ = – (2 – 0) = – 2
Cofactor of 3, A₃ = + (0 – 2) = – 2
Cofactor of 2, B₃ = – (0 – 4) = 4
Cofactor of 1, C₃ = + (1 – 0) = 1
∴ Cofactor matrix of A = B

Question 35.
Find the adjoint and the inverse of the matrix \(\left[\begin{array}{lll}
2 & 1 & 2 \\
1 & 0 & 1 \\
2 & 2 & 1
\end{array}\right]\).
Answer:
Let A = \(\left[\begin{array}{lll}
2 & 1 & 2 \\
1 & 0 & 1 \\
2 & 2 & 1
\end{array}\right]\)
Cofactor of 2, A₁ = + (0 – 2) = – 2
Cofactor of 1, B₁ = – (1 – 2) = 1
Cofactor of 2, C₁ = + (2 – 0) = 2
Cofactor of 1, A₂ = – (1 – 4) = 3
Cofactor of 0, B₂ = + (2 – 4) = – 2
Cofactor of 1, C₂ = – (4 – 2) = —2
Cofactor of 2, A₃ = + (1 – 0) = 1
Cofactor of 2, B₃ = – (2 – 2) = 0
Cofactor of 1, C₃ = + (0 – 1) = – 1
∴ Cofactor matrix of A = B

Question 36.
Solve the system of equations
2x – y + 3z = 9, x + y + z = 6, x – y + z = 2 by using Cramer’s rule.
Answer:

Question 37.
Solve 2x – y + 3z = 8, – x + 2y + z = 4, 3x + y – 4z = 0 by using matrix inversion method.
Answer:
The given system of linear equations are
2x – y + 3z = 8, – x + 2y + z = 4, 3x + y – 4z = 0
Let A = \(\left[\begin{array}{crr}
2 & -1 & 3 \\
-1 & 2 & 1 \\
3 & 1 & -4
\end{array}\right]\), X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\), D = \(\left[\begin{array}{l}
8 \\
4 \\
0
\end{array}\right]\)
Then we can write the given equations in the form of AX = D.
detA = 2(- 8 – 1) + 1(4 – 3) + 3(- 1 – 6)
= 2(- 9) + 1 (1) + 3(- 7)
= – 18 + 1 – 21 = – 38 ≠ 0
Hence, we can solve the given equations ¡n matrix inversion method.
Cofactor of 2 is A₁ = + (- 8 – 1) = – 9
Cofactor of – 1 is B₁ = – (4 – 3) = – 1
Cofactor of 3 is C₁ = + (- 1 – 6) = – 7
Cofactor of – 1 is A₂ = – (4 – 3) = – 1
Cofactor of 2 is B₂ = + (- 8 – 9) = – 17
Cofactor of 1 is C₂ = – (2 + 3) = – 5
Cot actor of 3 is A₃ = + (- 1 – 6) = – 7
Cofactor of 1 is B₃ = – (2 + 3) = – 5
Cofactor of – 4 is C₃ = + (4 – 1) = 3

Question 38.
Solve x + y + z = 1, 2x + 2y + 3z = 6, x + 4y + 9z = 3 by using matrix inversion method.
Answer:
Given system of linear equations are
x + y + z = 1, 2x + 2y + 3z = 6, x + 4y + 9z = 3
Let A = \(\left[\begin{array}{lll}
1 & 1 & 1 \\
2 & 2 & 3 \\
1 & 4 & 9
\end{array}\right]\), X = \(\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]\), D = \(\left[\begin{array}{l}
1 \\
6 \\
3
\end{array}\right]\)
Then we can write the given equations in the form of AX = D.
det A = \(\left|\begin{array}{lll}
1 & 1 & 1 \\
2 & 2 & 3 \\
1 & 4 & 9
\end{array}\right|\) = 1 (18 – 12) – 1 (18 – 3) + 1 (8 – 2) = 1(6) – 1(15) + 1(6)
= 6 – 15 + 6 = – 3 ≠ 0
Hence we can solve the given equations using matrix inversion method.
Cofactor of 1 is A₁ = + (18 – 12) = 6
Cofactor of 1 is B₁ = – (18 – 3) = – 15
Col actor of 1 is C₁ = + (8 – 2) = 6
Cofactor of 2 is A₂ = – (9 – 4) = – 5
Cofactor of 2 is B₂ = + (9 – 1) = 8
Cofactor of 3 is C₂ = – (4 – 1) = – 3
Cofactor of 1 is A₃ = + (3 – 2) = 1
Cofactor of 4 is B₃ = – (3 – 2) = – 1
Cofactor of 9 is C₃ = + (2 – 2) = O
∴ Cofactor matrix of A = B

∴ The solution of given system of equations is x = 7, y = – 10, z = 4.

Question 39.
Solve the equations 2x – y + 3z = 8, – x + 2y + z = 4, 3x + y – 4z = 0 by Gauss – Jordan method.
Answer:
Given system of linear equations are 2x – y + 3z = 8; – x + 2y + z = 4; 3x + y – 4z = 0.
Matrix equation form is AX = D

In this case, the system of equations has unique solution. i.e., x = y = z = 2
∴ The solution of given system of equations is x = 2; y = 2; z = 2.

Question 40.
Solve the equations 2x – y + 8z = 13, 3x + 4y + 5z = 18, 5x – 2y + 7z = 20 by Gauss – Jordan method.
Answer:
Given system of linear equations are 2x – y + 8z = 13, 3x + 4y + 5z = 18, 5x – 2y + 7z = 20.
Matrix equation form is AX = D

In this case, the system of equations has unique solution. i.e., x = 3, y = 1, z = 1.
∴ The solution of given system of equations is x = 3; y = 1; z = 1.

Some More Maths 1A Matrices Important Questions

Question 1.
If A = \(\left[\begin{array}{ccc}
2 & 3 & 1 \\
6 & -1 & 5
\end{array}\right]\) and B = \(\left[\begin{array}{ccc}
1 & 2 & -1 \\
0 & -1 & 3
\end{array}\right]\) then find the matrix X such that A + B – X = 0. What is the order of the matrix X?
Answer:
Given A = \(\left[\begin{array}{ccc}
2 & 3 & 1 \\
6 & -1 & 5
\end{array}\right]\), B = \(\left[\begin{array}{ccc}
1 & 2 & -1 \\
0 & -1 & 3
\end{array}\right]\)
A and B are matrices of same order 2 × 3.
If A + B – X is to be defined the order of X also must also be 2 × 3.
Given A + B – X = 0 ⇒ X = A + B

∴ Order of X is 2 × 3.

Question 2.
Construct a 3 × 2 matrix whose elements are defined by a_ij = \(\frac{1}{2}\) |i – 3j|.
Answer:
In general a 3 × 2 matrix is given by

Question 3.
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:

Question 4.
A certain book shop has 10 dozen chemistry books, 8 dozen physics books, 10 dozen economics books. Their selling prices are Rs. 80, Rs. 60 and Rs. 40 each respectively. Using matrix algebra, find the total value of the books in the shop.
Answer:
Number of 3 types of books is expressed by the row matrix A

= [120 96 120]
Selling price of 3 types of books is expressed by the column matrix B

Total value of the books in the shop is given by AB
AB = [120 96 120] \(\left[\begin{array}{l}
80 \\
60 \\
40
\end{array}\right]\)
= [120 × 80 + 96 × 60 + 120 × 40]
= [9600 + 5760 + 4800]
= [20160]
∴ Total value of the books = Rs. 20160

Question 5.
If A = \(\left[\begin{array}{ll}
2 & 1 \\
1 & 3
\end{array}\right]\) and B = \(\left[\begin{array}{lll}
3 & 2 & 0 \\
1 & 0 & 4
\end{array}\right]\), find AB and BA, if it exists.
Answer:
Given A = \(\left[\begin{array}{ll}
2 & 1 \\
1 & 3
\end{array}\right]\), B = \(\left[\begin{array}{lll}
3 & 2 & 0 \\
1 & 0 & 4
\end{array}\right]\)
The order of matrix A is 2 × 2
The order of matrix 13 is 2 × 3
The no.of columns in A = The no.of rows in B
∴ AB is defined

The no.of columns in B ≠ The no.of rows in A
∴ BA is not defined.

Question 6.
Give examples of two square matrices A and B of the same order for which AB = 0. But BA ≠ 0.
Answer:

Question 7.
If A = \(\left[\begin{array}{rr}
7 & -2 \\
-1 & 2 \\
5 & 3
\end{array}\right]\) and B = \(\left[\begin{array}{cc}
-2 & -1 \\
4 & 2 \\
-1 & 0
\end{array}\right]\) then find AB’ and BA’. [Mar. 18 (AP)]
Answer:

Question 8.
Find the minors of – 1 and 3 in the matrix \(\left[\begin{array}{ccc}
2 & -1 & 4 \\
0 & -2 & 5 \\
-3 & 1 & 3
\end{array}\right]\).
Answer:

Question 9.
Find the cofactors of the elements 2, – 5 in the matrix \(\left[\begin{array}{ccc}
-1 & 0 & 5 \\
1 & 2 & -2 \\
-4 & -5 & 3
\end{array}\right]\).
Answer:

Question 10.
Show that the determinant of skew – symmetric matrix of order three is always zero.
Answer:
Let A = \(\left[\begin{array}{ccc}
0 & -c & -b \\
c & 0 & -a \\
b & a & 0
\end{array}\right]\) is a skew – symmetric matrix of order ‘3’.
det A = 0(0 + a²) + c(0 + ab) – b(ac – 0)
= 0 + abc – abc = 0 + 0 = 0
∴ The determinant of skew symmetric matrix of order 3 is always zero.

Question 11.
Show that \(\left[\begin{array}{ccc}
\mathbf{y}+\mathbf{z} & \mathbf{x} & \mathbf{x} \\
\mathbf{y} & \mathbf{z}+\mathbf{x} & \mathbf{y} \\
\mathbf{z} & z & \mathrm{x}+\mathbf{y}
\end{array}\right]\) = 4xyz.
Answer:
LHS = \(\left[\begin{array}{ccc}
\mathbf{y}+\mathbf{z} & \mathbf{x} & \mathbf{x} \\
\mathbf{y} & \mathbf{z}+\mathbf{x} & \mathbf{y} \\
\mathbf{z} & z & \mathrm{x}+\mathbf{y}
\end{array}\right]\)
= (y + z) [(z + x) (x + y) – yz] – x[y(x + y) – yz] + x[yz – z(z + x)]
= (y + z) [zx + xy + zy + x² – yz] – x[xy + y² – yz] + x[yz – z² – zx]
= xyz + xy² + zy² + x²y – y²z + z² x + xyz + z²y + x²z – yz² – x²y – xy² + xyz + xyz – xz² – zx²
= 4xyz
= RHS.

Question 12.
If Δ₁ = \(\left|\begin{array}{ccc}
1 & \cos \alpha & \cos \beta \\
\cos \alpha & 1 & \cos \gamma \\
\cos \beta & \cos \gamma & 1
\end{array}\right|\), Δ₂ = \(\left|\begin{array}{ccc}
0 & \cos \alpha & \cos \beta \\
\cos \alpha & 0 & \cos \gamma \\
\cos \beta & \cos \gamma & 0
\end{array}\right|\) and Δ₁ = Δ₂, then show that cos²α + cos²β + cos²γ = 1.
Answer:
Δ₁ = \(\left|\begin{array}{ccc}
1 & \cos \alpha & \cos \beta \\
\cos \alpha & 1 & \cos \gamma \\
\cos \beta & \cos \gamma & 1
\end{array}\right|\)
= 1(1 – cos²γ) – cos α(cos α – cos β cos γ) + cos β (cos α cos γ – cos β)
= 1 – cos² γ – cos² α + cos α cos β cos γ + cos α cos β cos γ – cos²β
= 1 – cos² γ – cos²α – cos²β + 2 cos α cos β cos γ

Δ₂ = \(\left|\begin{array}{ccc}
0 & \cos \alpha & \cos \beta \\
\cos \alpha & 0 & \cos \gamma \\
\cos \beta & \cos \gamma & 0
\end{array}\right|\)
= 0(0 – cos²γ) – cos α (0 – cos γ cos β) + cos β (cos α cos γ – 0)
= cos α cos β cos γ + cos α cos β cos γ
= 2 cos α cos β cos γ

Given Δ₁ = Δ₂
1 – cos² α – cos²β – cos²γ + 2 cos α cos β cos γ = 2 cos α cos β cos γ
1 – cos²α – cos²β – cos²γ = 0
cos²α + cos² β + cos² γ = 1.

Question 13.
Show that \(\left|\begin{array}{ccc}
1 & a & a^2-b c \\
1 & b & b^2-c a \\
1 & c & c^2-a b
\end{array}\right|\) = 0.
Answer:
LHS = \(\left|\begin{array}{ccc}
1 & a & a^2-b c \\
1 & b & b^2-c a \\
1 & c & c^2-a b
\end{array}\right|\)
= 1(bc² – ab² – b²c + c²a) – a(c² – ab – b² + ac) + (a² – bc) (c – b)
= bc² – ab² – b²c + c²a – ac² + a²b + ab² – a²c + a²c – a²b – bc² . cb² = 0
= RHS.

Question 14.
Solve the following system of equations by using Cramer’s rule. [Mar. 15 (TS)]
x – y + 3z = 5, 4x + 2y – z = 0, – x + 3y + z = 5
Answer:
Given system of equations can be written as:

Question 15.
If A = \(\left[\begin{array}{lll}
0 & 1 & 1 \\
1 & 0 & 1 \\
1 & 1 & 0
\end{array}\right]\) and B = \(\frac{1}{2}\left[\begin{array}{lll}
\mathbf{b}+\mathbf{c} & \mathbf{c}-\mathbf{a} & \mathbf{b}-\mathbf{a} \\
\mathbf{c}-\mathbf{b} & \mathbf{c}+\mathbf{a} & \mathbf{a}-\mathbf{b} \\
\mathbf{b}-\mathbf{c} & \mathbf{a}-\mathbf{c} & \mathbf{a}+\mathbf{b}
\end{array}\right]\) then show that ABA^-1 is a diagonal matrxi.
Answer:
Given A = \(\left[\begin{array}{lll}
0 & 1 & 1 \\
1 & 0 & 1 \\
1 & 1 & 0
\end{array}\right]\),
B = \(\frac{1}{2}\left[\begin{array}{lll}
\mathbf{b}+\mathbf{c} & \mathbf{c}-\mathbf{a} & \mathbf{b}-\mathbf{a} \\
\mathbf{c}-\mathbf{b} & \mathbf{c}+\mathbf{a} & \mathbf{a}-\mathbf{b} \\
\mathbf{b}-\mathbf{c} & \mathbf{a}-\mathbf{c} & \mathbf{a}+\mathbf{b}
\end{array}\right]\)
Cot actor of 0 is A₁ = + (0 – 1) = – 1
Cofactor of ‘1’ is B₁ = – (0 – 1) = 1
Cofactor of 1 is C₁ = + (1 – 0) = 1
Cofactor of 1 is A₂ = – (0 – 1) = 1
Cofactor of 0 is B₂ = + (0 – 1) = – 1
Cofactor of 1 is C₂ = – (0 – 1) = 1
Cofactor of 1 is A₃ = + (1 – 0) = 1
Cofactor of 1 is B₃ = – (0 – 1) = 1
Cofactor of 0 is C₃ = +(0 – 1) = – 1
∴ Cofactor matrix of

det A = 0(0 – 1) – 1 (0 – 1) + 1 (1 – 0)
= 0 + 1 + 1 = 2 ≠ 0
∴ A is invertiable.

Question 16.
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:
Given A = \(\left[\begin{array}{lll}
3 & -3 & 4 \\
2 & -3 & 4 \\
0 & -1 & 1
\end{array}\right]\)

Cofactor of 3 is A₁ =+(- 3 + 4) = 1
Cofactor of – 3 is B₁ = – (2 – 0) = – 2
Cofactor of 4 is C₁ = (- 2 + 0) = – 2
Cofactor of 2 is A₂ = – (- 3 + 4) = – 1
Cofactor of – 3 is B₂ = + (3 – 0) = 3
Cofactor of 4 is C₂ = – (- 3 + 0) =3
Cofactor of 0 is A₃ = + (- 12 + 12) = 0
Cofactor of – 1 is B₃ = – (12 – 8) = – 4
Cofactor of 1 is C₃ = + (- 9 + 6) = – 3

Question 17.
For any square matrix A, show symmetric. [Mar. 15 (AP)]
Answer:
Let ‘A” be a square matrix
(AA’)’ = (A’)’ A’ = AA’
∴ (AA’)’ = AA’
⇒ AA’ is a symmetric matrix.

Question 18.
Find the rank of the matrix \(\left[\begin{array}{ccc}
1 & 4 & -1 \\
2 & 3 & 0 \\
0 & 1 & 2
\end{array}\right]\).
Answer:
Let A = \(\left[\begin{array}{ccc}
1 & 4 & -1 \\
2 & 3 & 0 \\
0 & 1 & 2
\end{array}\right]\)
det A = 1(6 – 0) – 4(4 – 0) – 1(2 – 0)
= 6 – 16 – 2 – 12 ≠ 0
∴ A is a non – singular.
Hence Rank (A) = 3.

Question 19.
Find the rank of the matrix \(\left[\begin{array}{lll}
1 & 2 & 3 \\
2 & 3 & 4 \\
0 & 1 & 2
\end{array}\right]\). [Mar. 19 (AP), Mar. 15 (TS)]
Answer:
Let A = \(\left[\begin{array}{lll}
1 & 2 & 3 \\
2 & 3 & 4 \\
0 & 1 & 2
\end{array}\right]\)
det A = 1 (6 – 4) – 2 (4 – 0) + 3 (2 – 0) = 2 – 8 + 6 = 0
Since det A = 0, Rank (A) ≠ 3.
Now, \(\left[\begin{array}{ll}
1 & 2 \\
2 & 3
\end{array}\right]\) is a sub matrix of ‘A’ whose determinant is 3 – 4 = – 1 ≠ 0.
Hence Rank (A) = 2.

Question 20.
Solve the following system of homogeneous equations x – y + z = 0, x + 2y – z = 0, 2x + y + 3z = 0. [Mar.16 (TS)]
Answer:
The coefficient matrix is \(\left[\begin{array}{ccc}
1 & -1 & 1 \\
1 & 2 & -1 \\
2 & 1 & 3
\end{array}\right]\)
Its determinant is 1(6 + 1) + 1(3 + 2) + 1(1 – 4) = 1(7) + 1(5) + 1(- 3) = 7 + 5 – 3 = 9
Hence the system has the trivial solution x = y = z = 0 only.

Question 21.
Solve the following system of equations by using Matrix inversion method.
2x – y + 3z = 9, x + y + z = 6, x – y + z = 2. [Mar. 16 (TS)]
Answer:

Question 22.
Solve x + y + z = 9, 2x + 5y + 7z = 52 and 2x + y – z = 0 by using matrix inversion method. [Mar. 17 (AP)]
Answer:

Question 23.
Solve the following system of equations by Cramer’s rule: 2x – y + k = 8, – x + 2y + z = 4, 3x + y – 4z = 0 [Mar. 18 (TS)]
Answer:
Given equations are
2x – y + 3z = 8,
– x + 2y + z = 4,
3x + y – 4z = 0