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Class 12 RD Sharma Solutions – Chapter 6 Determinants Exercise Ex. 6.6 | Set 2

  • Last Updated : 21 Jul, 2021

Question 19. Let A = [aij] be a square matrix of order 3 × 3 and Cij denote the cofactor of aij in A. If |A| = 5, find the value of a11 C21 + a12 C22 + a13 C23.

Solution:

As we know that, if a matrix is square matrix of order n, then the sum of the products of elements of a row or a column with the cofactors of the corresponding elements of some other row or column is zero. 

So, 

A = [aij] is a square matrix of order n.

Also we have,



\sum^n_{j = 1} a_{i j} C_{k j} = 0

And \sum^n_{i = 1} a_{i j} C_{i k} = 0

=> a11 C21 + a12 C22 + a13 C23 = 0 

Therefore, the required value is 0.

Question 20. Find the value of \begin{vmatrix}\sin 20^\circ & - \cos 20^\circ\\ \sin 70^\circ& \cos 70^\circ\end{vmatrix} .

Solution:

Given that, 

A = \begin{bmatrix} \sin 20^\circ & - \cos 20^\circ\\\sin 70^\circ & \cos 70^\circ \end{bmatrix}  

=> |A| = \begin{vmatrix} \sin 20^\circ & - \cos 20^\circ\\\sin 70^\circ & \cos 70^\circ \end{vmatrix}



= sin 20° cos 70° + cos 20° sin 70°

= sin (20° + 70°)

= sin 90

= 1

Question 21. If A is a square matrix satisfying AT A = I, write the value of |A|.

Solution:

Let us assume that A = [aij] be a square matrix of order n.

So, by using the property of determinants, we get

=> |A| = |AT|

Here, we have

=> AT A = I

=> |AT A| = 1

So, the determinants are of same order, we get

=> |AT A| = |AT| |A|

=> |AT| |A| = 1

=> \left| A \right| = \frac{1}{\left| A^T \right|}

=> \left| A \right| = \frac{1}{\left| A \right|}

=> |A|2 = 1

=> |A| = ±1

Therefore, the value of |A| is ±1.

Question 22. If A and B are square matrices of the same order such that |A| = 3 and AB = I, then write the value of |B|.

Solution:



According to the question, A and B are square matrices of the same order.

So, by using the property of determinants we get,

=> |AB| = |A| |B|

Here, |A| = 3, AB = I.

=> |AB| = 1

=> |A| |B| = 1

=> 3 |B| = 1

=> |B| = 1/3

Therefore, the value of |B| is 1/3.

Question 23. A is a skew-symmetric of order 3, write the value of |A|.

Solution:

Here, |A| = 4.

So we have,

Order of the matrix (n) = 3

Using the properties of matrices, we get

For a square matrix of order n and constant k, we know,

=> |k A| = kn |A|

=> |- A| = (-1)3 |A|

= (-1) (4)

= -4

Therefore, the value of |A| is -4.

Question 24. If A is a square matrix of order 3 with determinant 4, then write the value of |−A|.

Solution:

Given that, |A| = 4.

Order of the matrix (n) = 3

So, by using the properties of matrices, we get

=> |k A| = kn |A|

=> |- A| = (-1)3 |A|

= (-1) (4)

= -4

Therefore, the value of |A| is -4.

Question 25. If A is a square matrix such that |A| = 2, write the value of |A AT|.

Solution:



Given that, |A| = 2

As we know that in a square matrix, |A| = AT

So, they are of sane order 

Hence, |A AT| = |A| |AT|

=> |A AT| = 2 (2)

= 4

Therefore, the value of |A AT| is 4.

Question 26. Find the value of the determinant \begin{vmatrix}243 & 156 & 300 \\ 81 & 52 & 100 \\ - 3 & 0 & 4\end{vmatrix} .

Solution:

Given that, 

A = \begin{bmatrix}243 & 156 & 300 \\ 81 & 52 & 100 \\ - 3 & 0 & 4\end{bmatrix}



|A| = \begin{vmatrix}243 & 156 & 300 \\ 81 & 52 & 100 \\ - 3 & 0 & 4\end{vmatrix}

On applying R1 -> R1 – 3R2 we have,

\begin{vmatrix} 243 - \left( 81 \times 3 \right) & 156 - \left( 52 \times 3 \right) & 300 - \left( 100 \times 3 \right)\\ 81 & 52 & 100\\ - 3 & 0 & 4 \end{vmatrix}

\begin{vmatrix} 0 & 0 & 0\\ 81 & 52 & 100\\ - 3 & 0 & 4 \end{vmatrix}   

= 0

Therefore, the value of the determinant is 0.

Question 27. Find the value of the determinant \begin{vmatrix}2 & - 3 & 5 \\ 4 & - 6 & 10 \\ 6 & - 9 & 15\end{vmatrix} .

Solution:

Given that, 

A = \begin{bmatrix}2 & - 3 & 5 \\ 4 & - 6 & 10 \\ 6 & - 9 & 15\end{bmatrix}

|A| = \begin{vmatrix} 2 & - 3 & 5\\4 & - 6 & 10\\6 & - 9 & 15 \end{vmatrix}



On applying R2 -> R2 – 2R1 we get,

\begin{vmatrix} 2 & - 3 & 5\\4 - 4 & - 6 + 6 & 10 - 10\\6 & - 9 & 15 \end{vmatrix}

\begin{vmatrix} 2 & - 3 & 5\\0 & 0 & 0\\6 & - 9 & 15 \end{vmatrix}

= 0

Therefore, the value of the determinant is 0.

Question 28. If the matrix \begin{bmatrix}5x & 2 \\ - 10 & 1\end{bmatrix}  is singular, find the value of x.

Solution:

As we know that a matrix is singular only when its determinant is zero. 

According to the question, 

\begin{bmatrix}5x & 2 \\ - 10 & 1\end{bmatrix}  is a singular matrix

So, 



=> |A| = \begin{vmatrix} 5x & 2 \\- 10 & 1 \end{vmatrix}  = 0

On expanding the determinant we get,

=> 5x + 20 = 0 

=> x = -20/5

=> x = -4

Therefore, the value of x is -4.

Question 29. If A is a square matrix of order n × n such that |A| = λ, then write the value of |−A|.

Solution:

Given that,

A is a square matrix of order n × n

So, |A| = λ

=> |- A| = (-1)n A

=> |-A| = (-1)n λ

Therefore, the value of |-A| is (-1)n λ.

Question 30. Find the value of the determinant \begin{vmatrix}2^2 & 2^3 & 2^4 \\ 2^3 & 2^4 & 2^5 \\ 2^4 & 2^5 & 2^6\end{vmatrix} .

Solution:

Given that, 

A = \begin{bmatrix}2^2 & 2^3 & 2^4 \\ 2^3 & 2^4 & 2^5 \\ 2^4 & 2^5 & 2^6\end{bmatrix}

|A| = \begin{vmatrix}2^2 & 2^3 & 2^4 \\ 2^3 & 2^4 & 2^5 \\ 2^4 & 2^5 & 2^6\end{vmatrix}  

On taking out common factors from R1, R2 and R3 we get,

2^2 \times 2^3 \times 2^4 \begin{vmatrix} 1 & 2 & 2^2 \\1 & 2 & 2^2 \\1 & 2 & 2^2 \end{vmatrix}

Here, the two rows are identical, so we get



2^2 \times 2^3 \times 2^4 \times 2 \begin{vmatrix} 1 & 1 & 2^2 \\1 & 1 & 2^2 \\1 & 1 & 2^2 \end{vmatrix}

= 0 

Therefore, the value of the determinant is 0.

Question 31. If A and B are non-singular matrices of the same order, prove whether AB is singular or non-singular.

Solution:

According to the question, A and B be non-singular matrices of order n.  

Here, |A| ≠ 0 and |B| ≠ 0. 

So, the order of these matrix are same, we get

=> |AB| = |A| |B|

=> |AB| = 0 if |A| = 0 or |B| = 0

But as it is not the case here, so |AB| is non- zero matrix and AB is non-singular matrix.

Hence proved.

Question 32. A matrix of order 3 × 3 has determinant 2. What is the value of |A (3I)|, where I is the identity matrix of order 3 × 3.

Solution:

Given that a matrix of order 3 x 3 has determinant 2. 

So let us assume B be the matrix. so the order of the matrix is 3

and |B| = 2  

Let us consider I be the identity matrix, so we get

=> |I| = 1 

=> 3 |I| = 3

=> |A (3I)| = |3 A|

= 33 |A|

= 27 (2)

= 54

=> |A (3I)| = 54

Therefore, the value of |A (3I)| is 54.

Question 33. If A and B are square matrices of order 3 such that |A| = −1, |B| = 3, then find the value of |3 AB|.

Solution:

We have,

A and B are square matrices of order 3.

Also |A| = −1, |B| = 3.

We know,

As n is the order of A, we get

=> |K A| = Kn |A| 

=> |3 AB| = 33 |AB|  

If the order of A and B matrix are same and they are square matrix then |AB| = |A| |B|. 

So, we have,

=> |3 AB| = 33 |A| |B|

= 27 (-1) (3)

= -81

Therefore, the value of |3 AB| is -81.

Question 34. Write the value of \begin{vmatrix}a + ib & c + id \\ - c + id & a - ib\end{vmatrix} .

Solution:

We have,

A = \begin{bmatrix}a + ib & c + id \\ - c + id & a - ib\end{bmatrix}

|A| = \begin{vmatrix}a + ib & c + id \\ - c + id & a - ib\end{vmatrix}

= a2 – iab + iab – i2 b2 – (-c2 – icd + icd + i2 d2

= a2 – i2 b2 + c2 – i2 d2  

Here, we have, i2 = – 1.

So we get,  

|A| = a2 – (-1) b2 + c2 – (-1) d2  

= a2 + b2 + c2 + d2

Therefore, the value is a2 + b2 + c2 + d2.

Question 35. Write the cofactor of a12 in the following matrix \begin{bmatrix}2 & - 3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & - 7\end{bmatrix} .

Solution:



We have

\begin{bmatrix}2 & - 3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & - 7\end{bmatrix}

So,

=> a12 = -3

Now we find the cofactor of a12 

a12 = (-1)1+2 \begin{vmatrix}6 & 4 \\ 1 & - 7\end{vmatrix}

= – (- 42 – 4) 

= 46

Therefore, the value of the required cofactor is 46.

Question 36. If \begin{vmatrix}2x + 5 & 3 \\ 5x + 2 & 9\end{vmatrix} = 0 , find x.

Solution:

Here we have,

A = \begin{bmatrix}2x + 5 & 3 \\ 5x + 2 & 9\end{bmatrix}

=> |A| = \begin{vmatrix}2x + 5 & 3 \\ 5x + 2 & 9\end{vmatrix}  = 0

=> 9(2x + 5) – 3(5x + 2) = 0

=> 18x + 45 – 15x – 6 = 0  

=> 3x + 39 = 0

=> 3x = – 39 

=> x = -39/3 

=> x = -13

Therefore, the value of x is -13.

Question 37. Find the value of x from the following : \begin{vmatrix}x & 4 \\ 2 & 2x\end{vmatrix} = 0

Solution:

We have,

A = \begin{bmatrix}x & 4 \\ 2 & 2x\end{bmatrix}

|A| = \begin{vmatrix}x & 4 \\ 2 & 2x\end{vmatrix} = 0

=> \begin{vmatrix}x & 4 \\ 2 & 2x\end{vmatrix}  = 0 

=> 2 x2 – 8 = 0 

=> 2 x2 = 8

=> x2 = 8/2

=> x2 = 4  

=> x = √4

=> x = ±2

Therefore, the value of x is ±2.

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