Gujarat Board GSEB Textbook Solutions Class 12 Maths Chapter 4 Determinants Ex 4.6 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 12 Maths Chapter 4 Determinants Ex 4.6

Question 1.

x + 2y = 2

2x + 3y = 3

Solution:

x + 2y = 2

2x + 3y = 3

Let A = \(\left[\begin{array}{ll} 1 & 2 \\ 2 & 3 \end{array}\right]\)

âˆ´ |A| = \(\left|\begin{array}{ll} 1 & 2 \\ 2 & 3 \end{array}\right|\) = 3 – 4 = – 1

So, |A| â‰ 0.

âˆ´ System of equations has a unique solution.

Hence, the system of equations is consistent.

Question 2.

2x – y = 5

x + y = 4

Solution:

2x – y = 5

x + y = 4

Let A = \(\left[\begin{array}{ll} 2 & -1 \\ 1 & 1 \end{array}\right]\)

âˆ´ |A| = \(\left|\begin{array}{ll} 2 & -1 \\ 1 & 1 \end{array}\right|\) = 2 + 1 = 3 â‰ 0.

âˆ´ System of equations is consistent.

Question 3.

x + 3y = 5

2x + 6y = 8

Solution:

x + 3y = 5

2x + 6y = 8

Let A = \(\left[\begin{array}{ll} 1 & 3 \\ 2 & 6 \end{array}\right]\)

âˆ´ |A| = \(\left|\begin{array}{ll} 1 & 3 \\ 2 & 6 \end{array}\right|\) = 6 – 6 = 0.

Now adj A = \(\left[\begin{array}{ll} 6 & -2 \\ -3 & 1 \end{array}\right]\) = \(\left[\begin{array}{ll} 6 & -3 \\ -2 & 1 \end{array}\right]\) and B = \(\left[\begin{array}{l} 5 \\ 8 \end{array}\right]\)

(adj A).B = \(\left[\begin{array}{ll} 6 & -3 \\ -2 & 1 \end{array}\right]\)\(\left[\begin{array}{l} 5 \\ 8 \end{array}\right]\) = \(\left[\begin{array}{l} 30-24 \\ -10+8 \end{array}\right]\)

= \(\left[\begin{array}{l} 6 \\ -2 \end{array}\right]\) â‰ 0.

âˆ´ Given System of equations is inconsistent.

Question 4.

x + y + z = 1

2x + 3y + 2z = 2

ax + ay + 2az = 4

Solution:

x + y + z = 1

2x + 3y + 2z = 2

ax + ay + 2az = 4 or x + y + 2z = \(\frac { 4 }{ a }\)

Let A = \(\left[\begin{array}{lll} 1 & 1 & 1 \\ 2 & 3 & 2 \\ 1 & 1 & 2 \end{array}\right]\), X = \(\left[\begin{array}{l} x \\ y \\ z \end{array}\right]\) and B = \(\left[\begin{array}{c} 1 \\ 2 \\ 4 / a \end{array}\right]\).

Now, |A| = \(\left|\begin{array}{lll} 1 & 1 & 1 \\ 2 & 3 & 3 \\ 1 & 1 & 2 \end{array}\right|\) = 1.(6-2)-1.(4-2)+1.(2-3)

= 4 – 2 – 1 = 1

âˆ´ |A| â‰ 0.

â‡’ System of equations is consistent.

Question 5.

3x – y – 2z = 2

2y – z = – 1

3x – 5y = 3

Solution:

The System of equation is

3x – y – 2z = 2

2y – z = – 1

3x – 5y = 3

âˆ´ Given system equation is inconsistent.

Question 6.

5x – y + 4z = 5

2x + 3y + 5z = 2

5x – 2y + 6z = – 1

Solution:

The given System of equation is

5x – y + 4z = 5

2x + 3y + 5z = 2

5x – 2y + 6z = – 1

â‡’ The given System of equations is consistent.

Question 7.

5x + 2y = 4

7x + 3y = 5

Solution:

The System of equation is

5x + 2y = 4

7x + 3y = 5

Question 8.

2x – y = – 2

3x + 4y = 3

Solution:

The system of equations is

2x – y = – 2

3x + 4y = 3

This may be wirtten as AX = B.

or X = A^{-1}B,

Question 9.

4x – 3y = 3

3x – 5y = 7

Solution:

The system of equations is

4x – 3y = 3

3x – 5y = 7

This system of equations may be put as AX = B or X = A^{-1}B,

Question 10.

5x + 2y = 3

3x + 2y = 5

Solution:

The system of equations is

5x + 2y = 3

3x + 2y = 5

This system of equations may be written as AX = B or X = A^{-1}B,

Question 11.

2x + y + z = 1

x – 2y – z = \(\frac { 3 }{ 2 }\)

3y – 5z = 9

Solution:

The given system of equations is

2x + y + z = 1

x – 2y – z = \(\frac { 3 }{ 2 }\)

3y – 5z = 9

This system of equations may be written as AX = B or X = A^{-1}B,

[+ or – signs are put according as ,(-i)^{i+i} is +ve or -ve respectively. Minors of corresponding elements are written in brackets ()].

Question 12.

x – y + z = 4

2x + y – 3z = 0

x + y + z = 2

Solution:

The given system of equations is

x – y + z = 4

2x + y – 3z = 0

x + y + z = 2

which may be written as AX = B or X = A^{-1}B,

where A = \(\left[\begin{array}{ccc}

1 & -1 & 1 \\

2 & 1 & -3 \\

1 & 1 & 1

\end{array}\right]\), X = \(\left[\begin{array}{l}

x \\

y \\

z

\end{array}\right]\) and B = \(\left[\begin{array}{l}

4 \\

0 \\

2

\end{array}\right]\)

To find adj A, we calculate the cofactors of the elements of A.

Question 13.

2x + + 3z = 5

x – 2y + z = – 4

3x – y – 2z = 3

Solution:

The given system of equations can be written as

Question 14.

x – y + 2z = 7

3x + 4y – 5z = – 5

2x – y + 3z = 12

Solution:

The given system of equation can be written as

âˆ´ A^{-1} exists and hence the given equations have a unique solution.

Here,

Question 15.

If A = \(\left[\begin{array}{ccc}

2 & -3 & 5 \\

3 & 2 & -4 \\

1 & 1 & -2

\end{array}\right]\), find A^{-1}. Using A^{-1}, solve the following system of linear equations:

2x – 3y + 5z = 11, 3x + 2y – 4z = – 5, x + y – 2z = – 3.

Solution:

The given system of equations can be written as

\(\left[\begin{array}{ccc}

2 & -3 & 5 \\

3 & 2 & -4 \\

1 & 1 & -2

\end{array}\right]\)\(\left[\begin{array}{l}

x \\

y \\

z

\end{array}\right]\) = \(\left[\begin{array}{l}

11 \\

-5 \\

-3

\end{array}\right]\), i.e., AX = B,

Now, |A| = \(\left|\begin{array}{ccc}

2 & -3 & 5 \\

3 & 2 & -4 \\

1 & 1 & -2

\end{array}\right|\)

= 2(- 4 + 4) + 3(- 6 + 6) + 5(3 – 2)

= 0 – 6 + 5 = – 1 â‰ 0.

âˆ´ A^{-1} exists and hence the given equations have a unique solution.

Question 16.

The cost of 4 kg onions, 3 kg wheat and 2kg rice is â‚¹ 60. The cost of 2 kg onions, 4 kg wheat and 6 kg rice â‚¹ 90. The cost of 6 kg onions, 2 kg wheat and 3 kg rice is 70. Find the cost of each item per kg by matrix method.

Solution:

Let the cost of ornons, wheat and rice per kg be â‚¹ x, â‚¹ y and â‚¹ z respectively.

The cost of 4 kg onions, 3 kg wheat and 2 kg rice = â‚¹ 60.

âˆ´ 4x + 3y + 2z = 60. … (1)

The cost of 2 kg onions, 4 kg wheat and 6 kg rice = â‚¹ 90

âˆ´ 2x + 4y + 6z = 90 … (2)

The cost of 6 kg onions, 2 kg wheat and 3 kg rice = â‚¹ 70

âˆ´ 6x + 2y + 3z = 70 … (3)

The system of equations is

4x + 3y + 2z = 60

x + 2y + 3z = 45 [Dividing equation (2) by 2]

6x + 2y + 3z = 70,

which may be written as

AX = B or X = A^{-1}B,

â‡’ The cost of onions, wheat and rice per kg are â‚¹ 5, â‚¹ 8, â‚¹ 8 respectively.

Also, we have given equations as

x – 2y = 10

2x – y – z = 8

and – 2y + z = 7

or \(\left[\begin{array}{ccc}

1 & -2 & 0 \\

2 & -1 & -1 \\

0 & -2 & 1

\end{array}\right]\)\(\left[\begin{array}{l}

x \\

y \\

z

\end{array}\right]\) = \(\left[\begin{array}{l}

10 \\

8 \\

7

\end{array}\right]\)

or CX = D,

where C = \(\left[\begin{array}{ccc}

1 & -2 & 0 \\

2 & -1 & -1 \\

0 & -2 & 1

\end{array}\right]\), X = \(\left[\begin{array}{l}

x \\

y \\

z

\end{array}\right]\)

and D = \(\left[\begin{array}{l}

10 \\

8 \\

7

\end{array}\right]\).

We know that (A’)^{-1} = (A^{-1})’.

âˆ´ C’ = \(\left[\begin{array}{ccc}

1 & -2 & 0 \\

– 2 & -1 & -2 \\

0 & – 1 & 1

\end{array}\right]\)