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Class 12 RD Sharma Solutions – Chapter 6 Determinants Exercise Ex. 6.6 | Set 1

  • Last Updated : 07 Sep, 2021

Question 1. If A is a singular matrix, then find the value of |A|.

Solution:

Given that A is a singular matrix.

So, as we know if A is a n×n matrix and it is singular, the value of its determinant is always 0.

Thus, |A| = 0.

Question 2. For what value of x, the following matrix is singular?

\begin{bmatrix}5 - x & x + 1 \\ 2 & 4\end{bmatrix}

Solution:



Given that \begin{bmatrix}5 - x & x + 1 \\ 2 & 4\end{bmatrix}

As we know if A is a n×n matrix and it is singular, so, the value of its determinant is always 0.

=> |A| = 0

=> \begin{vmatrix}5 - x & x + 1 \\ 2 & 4\end{vmatrix} = 0

=> 4(5 – x) – 2(x + 1) = 0

=> 20 – 4x – 2x – 2 = 0

=> 18 – 6x = 0

=> 18 = 6x



=> x = 3

Therefore, the value of x is 3.

Question 3. Find the value of the determinant \begin{bmatrix}2 & 3 & 4 \\ 2x & 3x & 4x \\ 5 & 6 & 8\end{bmatrix}     .

Solution:

Given that

A = \begin{bmatrix}2 & 3 & 4 \\ 2x & 3x & 4x \\ 5 & 6 & 8\end{bmatrix}

|A| = \begin{vmatrix} 2 & 3 & 4\\ 2x & 3x & 4x \\ 5 & 6  & 8 \end{vmatrix}

So, on taking out x common from R2 we get,

|A| = x\begin{vmatrix} 2 & 3 & 4\\ 2 & 3 & 4 \\ 5 & 6 & 8 \end{vmatrix}

As R1 = R2, we get

|A| = 0



Therefore, the value of the determinant is 0.

Question 4. State whether the matrix \begin{bmatrix}2 & 3 \\ 6 & 4\end{bmatrix}     is singular or non-singular.

Solution:

Given that

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

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

|A| = 2 (4) – 6 (3)

= 8 – 18 

= -10

As we know if A is a n×n matrix and it is singular, so the value of its determinant is always 0.

As |A| = -10 here, the given matrix is non-singular.



Question 5. Find the value of the determinant \begin{bmatrix}4200 & 4201 \\ 4202 & 4203\end{bmatrix}     .

Solution:

Given that

A = \begin{bmatrix}4200 & 4201 \\ 4202 & 4203\end{bmatrix}

|A| = \begin{vmatrix}4200 & 4201 \\ 4202 & 4203\end{vmatrix}

On applying C2 -> C2 – C1, we get

|A| = \begin{vmatrix}4200 & 4201-4200 \\ 4202 & 4203-4202\end{vmatrix}

|A| = \begin{vmatrix} 4200 & 1\\4202 & 1 \end{vmatrix}

|A| = 4200 – 4202 

|A| = -2

Therefore, the value of determinant is -2.



Question 6. Find the value of the determinant \begin{bmatrix}101 & 102 & 103 \\ 104 & 105 & 106 \\ 107 & 108 & 109\end{bmatrix}     .

Solution:

Given that

A = \begin{bmatrix}101 & 102 & 103 \\ 104 & 105 & 106 \\ 107 & 108 & 109\end{bmatrix}

|A| = \begin{vmatrix}101 & 102 & 103 \\ 104 & 105 & 106 \\ 107 & 108 & 109\end{vmatrix}

On applying C2 -> C2 – C1 and C3 -> C3 – C1, we get

|A| = \begin{vmatrix}101 & 102-101 & 103-101 \\ 104 & 105-104 & 106-104 \\ 107 & 108-107 & 109-107\end{vmatrix}

|A| = \begin{vmatrix} 101 & 1 & 2\\104 & 1 & 2\\107 & 1 & 2 \end{vmatrix}

On taking out 2 common from R3 we get,

|A| = 2\begin{vmatrix} 101 & 1 & 1\\104 & 1 & 1\\107 & 1 & 1 \end{vmatrix}

As R2 = R3, we get



|A| = 0

Therefore, the value of the determinant is zero.  

Question 7. Find the value of the determinant \begin{vmatrix}a & 1 & b + c \\ b & 1 & c + a \\ c & 1 & a + b\end{vmatrix}     .

Solution:

Given that

A = \begin{bmatrix}a & 1 & b + c \\ b & 1 & c + a \\ c & 1 & a + b\end{bmatrix}

|A| = \begin{vmatrix}a & 1 & b + c \\ b & 1 & c + a \\ c & 1 & a + b\end{vmatrix}

On applying C1 -> C1 + C3 we get,

\begin{vmatrix} a + b + c & 1 & b + c\\a + b + c & 1 & c + a\\a + b + c & 1 & a + b \end{vmatrix}

a + b + c \begin{vmatrix} 1 & 1& b + c\\1 & 1 & c + a\\1 & 1 & a + b \end{vmatrix}

= (a + b + c) (0)



= 0

Therefore, the value of determinant is 0.

Question 8. If A = \begin{bmatrix}0 & i \\ i & 1\end{bmatrix}     and B = \begin{bmatrix}0 & 1 \\ 1 & 0\end{bmatrix}     , find the value of |A| + |B|.

Solution:

Given that

A = \begin{bmatrix} 0 & i\\i & 1 \end{bmatrix}

|A| = \begin{vmatrix} 0 & i\\i & 1 \end{vmatrix}

= 0 – i2   

= – (-1)

= 1

Also, we have

B = \begin{bmatrix} 0 & 1\\1 & 0 \end{bmatrix}

|B| = \begin{vmatrix} 0 & 1\\1 & 0 \end{vmatrix}

= 0 – 1 

= -1

So,

|A| + |B| = 1 + (-1) 

= 1 – 1

= 0

Therefore, the value of |A| + |B| is 0.

Question 9. If A = \begin{bmatrix}1 & 2 \\ 3 & - 1\end{bmatrix}     and B = \begin{bmatrix}1 & 0 \\ - 1 & 0\end{bmatrix}     , find |AB|.

Solution:



We have,

A = \begin{bmatrix}1 & 2 \\ 3 & - 1\end{bmatrix}     and B = \begin{bmatrix}1 & 0 \\ - 1 & 0\end{bmatrix}

So, we get

AB = \begin{bmatrix}1 & 2 \\ 3 & - 1\end{bmatrix}\begin{bmatrix}1 & 0 \\ - 1 & 0\end{bmatrix}

\begin{bmatrix}1 - 2 & 0 + 0 \\ 3 + 1 & 0 + 0\end{bmatrix}

\begin{bmatrix}- 1 & 0 \\ 4 & 0\end{bmatrix}

Now we have,

|AB| = \begin{vmatrix}- 1 & 0 \\ 4 & 0\end{vmatrix}

= -1 (0) – 0 (4)

= 0 – 0 

= 0

Therefore, the value of |AB| is 0.

Question 10. Evaluate \begin{bmatrix}4785 & 4787 \\ 4789 & 4791\end{bmatrix}     .

Solution:

Given that

A = \begin{bmatrix}4785 & 4787 \\ 4789 & 4791\end{bmatrix}

|A| = \begin{vmatrix}4785 & 4787 \\ 4789 & 4791\end{vmatrix}

On applying C2 -> C2 – C1 we get,

|A| = \begin{vmatrix}4785 & 4787-4785 \\ 4789 & 4791-4789\end{vmatrix}

\begin{vmatrix}4785 & 2\\ 4789 & 2\end{vmatrix}

On taking out 2 common from R2 we get,



2 \times \begin{vmatrix}4785 & 1\\ 4789 & 1\end{vmatrix}

= 2 (4785 – 4789)

= 2 (-4)

= -8

Therefore, the value of the determinant is 0.

Question 11. If w is an imaginary cube root of unity, find the value of \begin{vmatrix}1 & w & w^2 \\ w & w^2 & 1 \\ w^2 & 1 & w\end{vmatrix}     .

Solution:

Given that,

A = \begin{bmatrix}1 & w & w^2 \\ w & w^2 & 1 \\ w^2 & 1 & w\end{bmatrix}

|A| = \begin{vmatrix}1 & w & w^2 \\ w & w^2 & 1 \\ w^2 & 1 & w\end{vmatrix}

On applying C1 -> C1 + C_2 + C_3 we get,

\begin{vmatrix} 1 + w + w^2 & w & w^2 \\ w + w^2 + 1 & w^2 & 1\\ w^2 + 1 + w & 1 & w \end{vmatrix}

\begin{vmatrix} 0 & w& 7 w^2 \\ 0 & w^2 & 1\\ 0 & 1 & 7 w \end{vmatrix}

= 0

Question 12. If A = \begin{bmatrix}1 & 2 \\ 3 & - 1\end{bmatrix}     and B = \begin{bmatrix}1 & - 4 \\ 3 & - 2\end{bmatrix}     , find |AB|.

Solution:

Given that

A = \begin{bmatrix}1 & 2 \\ 3 & - 1\end{bmatrix}

|A| = -1 – 6 

= -7

B = \begin{bmatrix}1 & - 4 \\ 3 & - 2\end{bmatrix}

|B| = – 2 + 12 



= 10  

We know if A and B are square matrices of the same order, then we have,

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

= (-7) (10) 

= -70

Therefore, the value of |AB| is -70.

Question 13. If A = [aij]  is a 3 × 3 diagonal matrix such that a11 = 1, a22 = 2 a33 = 3, then find |A|.

Solution:

Given that  a11 = 1, a22 = 2 and a33 = 3.

If A is a diagonal matrix of order n x n, then we have

=> \left| A \right| = a_{11} \times a_{22} \times a_{33} \times . . . \times a_{nn}

So, we get

|A| = 1 (2) (3)

= 6 

Therefore, the value of |A| is 6.

Question 14. If A = [aij] is a 3 × 3 scalar matrix such that a11 = 2, then find the value of |A|.

Solution:

Given that A = [aij] which is a 3 × 3 scalar matrix and a11 = 2,

As we know that a scalar matrix is a diagonal matrix, in which all the diagonal elements are equal to a given scalar number.

=> A = \begin{bmatrix} 2 & 0 & 0\\0 & 2 & 0\\0 & 0 & 2 \end{bmatrix}

\begin{vmatrix} 2 & 0 & 0\\0 & 2 & 0\\ 0 & 0 & 2 \end{vmatrix}

On expanding along C1, we get

2 \times \begin{vmatrix} 2 & 0\\ 0 & 2 \end{vmatrix}

= 2 (2) (2)

= 8

Therefore, the value of |A| is 8.

Question 15. If I3 denotes an identity matrix of order 3 × 3, find the value of its determinant.

Solution:

As we know that in an identity matrix, all the diagonal elements are 1 and the remaining elements are 0.

Here,

I3\begin{bmatrix} 1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1 \end{bmatrix}

\begin{vmatrix} 1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1 \end{vmatrix}

On expanding along C1, we get



1 \times \begin{vmatrix} 1 & 0\\0 & 1 \end{vmatrix}

= 1

Therefore, the value of the determinant is 1.

Question 16. A matrix A of order 3 × 3 has determinant 5. What is the value of |3A|?

Solution:

Given that matrix A is of order 3 x 3 and the determinant = 5.

If A is a square matrix of order n and k is a constant, then we have

=> |kA| = kn |A|

Here,  

Number of rows = n  

Also, k is a common factor from each row of k.

Hence, we get

3A = 33 |A|

= 27 (5)

= 135

Therefore, the value of |3A| is 135.

Question 17. On expanding by the first row, the value of the determinant of 3 × 3 square matrix A = [aij] is a11 C11 + a12 C12 + a13 C13, where [Cij] is the cofactor of aij in A. Write the expression for its value on expanding by the second column.

Solution:

As we know that if a square matrix(let say  A) is of order n, then the sum of the products of elements of a row or a column with their cofactors is always equal to det (A). 

So, 

\sum^n_{i = 1} a_{i j} C_{i j} = \left| A \right|

Also, \sum^n_{j = 1} a_{i j} C_{i j} = \left| A \right|

On expanding along R1 we get,

|A| = a11 C11 + a12 C12 + a13 C13

Now,  

On expanding along C2 we get,

|A| = a12 C12 + a22 C22 + a32 C32

Question 18. On expanding by the first row, the value of the determinant of 3 × 3 square matrix A = [aij ] is a11 C11 + a12 C12 + a13 C13, where [Cij] is the cofactor of aij in A. Write the expression for its value on expanding by the second column.

Solution:

As we know that if a square matrix(let say  A) is of order n, then the sum of the products of elements of a row or a column with their cofactors is always equal to det (A). 

So, 

\sum^n_{i = 1} a_{i j} C_{i j} = \left| A \right|

Also, \sum^n_{j = 1} a_{i j} C_{i j} = \left| A \right|

On expanding along R1 we get,

|A| = a11 C11 + a12 C12 + a13 C13

Now,  

On expanding along C2 we get,

|A| = a12 C12 + a22 C22 + a32 C32 = 5

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