Important Questions for Class 12 Math Chapter 4 Determinants

Very Short Answer Questions [1 Mark Questions]

Ques: Evaluate the determinant \(\begin{vmatrix} 2 & 4 \\ -1 & -5 \\ \end{vmatrix}\)

Ans: Given, 

\(\begin{vmatrix} 2 & 4 \\ -1 & -5 \\ \end{vmatrix}\)

By solving, we get 

= -2+20   

= 18

Ques: If a, b, c, are in A.P, find the value of \(\begin{vmatrix} 2y+4 & 5y+7 & 8y+a \\ 3y+5 & 6y+8 & 9y+b \\ 4y+4 & 7y+9 & 10y+c \\ \end{vmatrix}\)

Ans: Applying R₁ → R₁ + R₃ – 2R₂ to the given determinant, we obtain

(Since  2b = a + c) \(\begin{vmatrix} 0 & 0 & 0 \\ 3y+5 & 6y+8 & 9y+b \\ 4y+4 & 7y+9 & 10y+c \\ \end{vmatrix}\)

Ques: If \(\begin{vmatrix} x & 2 \\ 18 & x \\ \end{vmatrix} = \begin{vmatrix} 16 & 2 \\ 18 & 6 \\ \end{vmatrix}\), then x is equal to

1. 6
2. ± 6
3. -6
4. 0

Ans: B. ± 6

Explanation: 

Given, 

\(\begin{vmatrix} x & 2 \\ 18 & x \\ \end{vmatrix} = \begin{vmatrix} 16 & 2 \\ 18 & 6 \\ \end{vmatrix}\)

Solving it, we have:

⇒ x² – 36 = 36 – 36

⇒ x² – 36 = 0

⇒ x² = 36

Applying square root on both sides, we obtain:

\(\implies X = \pm6\)

Hence, \(B. \pm6\) is the correct answer.

Ques: Find values of x for which

Ans: We have, 

\(\begin{vmatrix} 3 & x \\ x & 1 \\ \end{vmatrix} = \begin{vmatrix} 3 & 2 \\ 4 & 1 \\ \end{vmatrix}\)

i.e. 3 – x² = 3 – 8

i.e. x² = 8 

Hence, \(x = \pm 2 \sqrt2\)

Ques: Evaluate the determinant Δ = \(\begin{vmatrix} 1 & 2 & 4 \\ -1 & 3 & 0 \\ 4 & 1 & 0 \\ \end{vmatrix}\)

Ans: Note that in the third column, two entries are zero. So expanding along third column (C₃), we get

\(\triangle = 4\begin{vmatrix} -1 & 3 \\ 4 & 1 \\ \end{vmatrix} -0 \begin{vmatrix} 1 & 2 \\ 4 & 1 \\ \end{vmatrix} +0 \begin{vmatrix} 1 & 2 \\ -1 & 3 \\ \end{vmatrix}\)

= 4(– 1 – 12)  – 0 + 1 = – 52

Ques: Without expanding, prove that \(\triangle = \begin{vmatrix} x+y & y+z & z+x \\ z & x & y \\ 1 & 1 & 1 \\ \end{vmatrix} = 0\)

Ans: Applying R₁ → R₁ + R₂ to \(\triangle\), we get

\(\triangle = \begin{vmatrix} x+y+z & x+y+z & x+y+z \\ z & x & y \\ 1 & 1 & 1 \\ \end{vmatrix}\)

Since the elements of R₁ and R₃ are proportional, Δ = 0

Ques: If Δ = \(\begin{vmatrix} 5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3 \\ \end{vmatrix}\)

Then find:

1. the minor of the element a₂₃ (CBSE. 2012)
2. the co-factor of the element a₃₂ (CBSE. 2012)

Ans:

1. a₂₃ = \(\begin{vmatrix} 5 & 3 \\ 1 & 2 \\ \end{vmatrix}\)

= (5)(2) - (1)(3) 

= 10 - 3 = 7

2. a₃₂ = (-1)³⁺²  \(\begin{vmatrix} 5 & 8 \\ 2 & 1 \\ \end{vmatrix}\)

= (-1)⁵ [(5)(1) – (2)(8)]

= (-1)⁵ (5-16)

=(-1)(-11)

=11

Also read:

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Short Answer Questions [2 Marks Questions]

Ques: Find the area of the triangle whose vertices are (3, 8), (– 4, 2) and (5, 1). 

Ans: The area of a triangle is given by

\(\triangle = \frac{1}{2}\begin{vmatrix} 3 & 8 & 1 \\ -4 & 2 & 1 \\ 5 & 1 & 1 \\ \end{vmatrix}\)

\(= \frac{1}{2}[3(2-1)-8(-4-5)+1-(4-10)]\)

\(= \frac{1}{2}(3+72-14)\)

\(= \frac{61}{2}\)

Ques: Evaluate \(\triangle = \begin{vmatrix} 0 & sin\alpha & -cos\alpha \\ -sin\alpha & 0 & sin\beta \\ cos\alpha & -sin\beta & 0 \\ \end{vmatrix}\) 

Ans: Expanding along R₁, we get

\(\triangle = 0\begin{vmatrix} 0 & sin\beta \\ -sin\beta & 0 \\ \end{vmatrix}-sin\alpha\begin{vmatrix} -sin\alpha & sin\beta \\ cos\alpha & 0 \\ \end{vmatrix}-cos\alpha\begin{vmatrix} -sin\alpha & 0 \\ cos\alpha & -sin\beta \\ \end{vmatrix}\)

\(= 0-sin\alpha ( 0-sin\beta cos\alpha)-cos\alpha(sin\alpha sin\beta-0)\)

\(= sin\alpha .sin\beta .cos\alpha-cos\alpha.sin\alpha .sin\beta=0\)

Ques: Prove that the given determinant is independent of \(\theta \begin{vmatrix} x & sin\theta & cos\theta \\ -sin\theta & -x & 1 \\ cos\theta & 1 & x \\ \end{vmatrix}\)

Ans: Let \(\triangle= \begin{vmatrix} x & sin\theta & cos\theta \\ -sin\theta & -x & 1 \\ cos\theta & 1 & x \\ \end{vmatrix}\)

Solving it, we have

\(\to \triangle = x(x^{2}-1)-sin\theta (-x sin\theta -cos\theta)+(-sin\theta +x cos\theta )\)

\(\to \triangle= x^{3}-x + xsin^{2}\theta+ sin\theta cos\theta-sin\theta cos\theta+xcos^{2}\theta\)

\(\to \triangle = x^{3}-x+x (sin^{2}\theta+cos^{2}\theta)\)

\(\to \triangle= x^{3}-x+x \)

\(\to \triangle= x^{3}\)

Hence, \(\triangle\) is dependent of \(\theta\)

Ques: Evaluate \(\begin{vmatrix} cos\alpha cos\beta & cos\alpha sin\beta & -sin\alpha \\ -sin\beta & cos\beta & 0 \\ sin\alpha cos\beta & sin\alpha sin\beta & cos\alpha \\ \end{vmatrix}\) 

Ans: Let \(\triangle = \begin{vmatrix} cos\alpha cos\beta & cos\alpha sin\beta & -sin\alpha \\ -sin\beta & cos\beta & 0 \\ sin\alpha cos\beta & sin\alpha sin\beta & cos\alpha \\ \end{vmatrix}\)

Expanding along column C₃

\(\to \triangle= -sin\alpha( -sin\alpha sin^{2}\beta +cos^{2}\beta sin\alpha +cos\alpha (cos\alpha cos^{2}\beta+cos\alpha sin^{2}\beta) \)

\(\to \triangle= sin^{2}\alpha( sin^{2}\beta +cos^{2}\beta) +cos^{2}\alpha ( cos^{2}\beta+sin^{2}\beta) \)

\(\to \triangle= sin^{2}\alpha(1)+cos^{2}\alpha(1)\)

\(\therefore \triangle=1\)

Hence, =1

Ques: Evaluate the determinant: \(\begin{vmatrix} x^{2}-x+1 & x-1 \\ x+1 & x+1 \\ \end{vmatrix}\)

Ans: Solving the determinant, 

\(\begin{vmatrix} x^{2}-x+1 & x-1 \\ x+1 & x+1 \\ \end{vmatrix}\)
We have:

\(\implies \begin{vmatrix} x^{2}-x+1 & x-1 \\ x+1 & x+1 \\ \end{vmatrix}=(x^{2}-x+1)(x+1)-(x-1)(x+1)\)

\(\implies \begin{vmatrix} x^{2}-x+1 & x-1 \\ x+1 & x+1 \\ \end{vmatrix}=x^{3}-x^{2}+x+x^{2}-x+1-(x^{2}-1)\)

So,

\( \begin{vmatrix} x^{2}-x+1 & x-1 \\ x+1 & x+1 \\ \end{vmatrix}=x^{3}+1-x^{2}+1\)

\(\therefore\begin{vmatrix} x^{2}-x+1 & x-1 \\ x+1 & x+1 \\ \end{vmatrix}=x^{3}-x^{2}+2\)

Ques: For what value of ‘x’, the matrix \(\begin{bmatrix} 5-x & x+1 \\ 2 & 4 \\ \end{bmatrix}\) is singular? (CBSE 2011)

Ans: The matrix \(\begin{bmatrix} 5-x & x+1 \\ 2 & 4 \\ \end{bmatrix}\) is singular i.e., \(\begin{bmatrix} 5-x & x+1 \\ 2 & 4 \\ \end{bmatrix}\) =0 

⇒4(5 -x) – 2 (x + 1) = 0

⇒ 20 – 4x – 2x – 2 = 0

⇒ 18 – 6x = 0

⇒ 6x = 18.

Hence, x = 3.

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Long Answer Questions [3 Marks Questions]

Ques: Prove that \(\begin{vmatrix} b+c & a & a \\ b & c+a & b \\c & c & a+b \\ \end{vmatrix}=4abc\)

Ans: Let \(\triangle=\begin{vmatrix} b+c & a & a \\ b & c+a & b \\c & c & a+b \\ \end{vmatrix}\)

Applying R₁ → R₁ – R₂ – R₃ to \(\triangle\), we get 

\(\triangle=\begin{vmatrix} 0 & -2c & -2b \\ b & c+a & b \\c & c & a+b \\ \end{vmatrix}\)

Expanding along R₁, we get 

\(\triangle=0\begin{vmatrix} c+a & b \\ c & a+b \\ \end{vmatrix}-(-2c)\begin{vmatrix} b & b \\ c & a+b \\ \end{vmatrix}+(-2b)\begin{vmatrix} b & c+a \\ c & c \\ \end{vmatrix}\)

= 2c(ab + b² – bc) – 2b(bc – c² – ac)

= 2abc +2cb² - 2bc² - 2bc² + 2bc² + 2abc

= 4abc

Ques: Using the property of determinants and without expanding, prove that:

\(\begin{vmatrix} a-b & b-c & c-a \\ b-c & c-a & a-b \\c-a & a-b & b-c \\ \end{vmatrix}=0\)

Ans:

Let \(\triangle=\begin{vmatrix} a-b & b-c & c-a \\ b-c & c-a & a-b \\c-a & a-b & b-c \\ \end{vmatrix}\)

Applying row operation, 

R₁ → R₁ + R₂

\(\implies \triangle=\begin{vmatrix} a-b+b-c & b-c+c-a & c-a+a-b \\ b-c & c-a & a-b \\c-a & a-b & b-c \\ \end{vmatrix}\)

\(\implies \triangle=\begin{vmatrix} a-c & b-a & c-b \\ b-c & c-a & a-b \\c-a & a-b & b-c \\ \end{vmatrix}\)

\(\implies \triangle=\begin{vmatrix} a-c & b-a & c-b \\ b-c & c-a & a-b \\-(a-c) & -(b-a) & -(c-b) \\ \end{vmatrix}\)

Multiplying the third row by (−1), we get:

\(\implies \triangle=\begin{vmatrix} a-c & b-a & c-b \\ b-c & c-a & a-b \\a-c & b-a & c-b \\ \end{vmatrix}\)

We know, if two rows or columns of a determinant are identical, then the value of the determinant is zero. 

Since, the two rows R₁ and R₃ are identical.

∴ Δ = 0

Hence, 

\(\begin{vmatrix} a-b & b-c & c-a \\ b-c & c-a & a-b \\c-a & a-b & b-c \\ \end{vmatrix}=0\)

Ques: By using properties of determinants, show that: \(\begin{vmatrix} 0 & a & -b \\ -a & 0 & -c \\ b & c & 0 \\ \end{vmatrix}=0\)

Ans: Given, 

\(\triangle=\begin{vmatrix} 0 & a & -b \\ -a & 0 & -c \\ b & c & 0 \\ \end{vmatrix}\)

We know, if we multiply the elements of a matrix by a scalar c, then we multiply the matrix by the scalar, 1c

Applying R₁ → cR₁:

\(\triangle=\frac{1}{c}\begin{vmatrix} 0 & ac & -bc \\ -a & 0 & -c \\ b & c & 0 \\ \end{vmatrix}\)

Applying R₁ → R₁ – bR₂

Taking a common from the first row, we have

\(\implies \triangle=\frac{a}{c}\begin{vmatrix} b & c & 0 \\ -a & 0 & -c \\ b & c & 0 \\ \end{vmatrix}\)

Since, the two rows, R₁ and R₃, are identical, 

\(\therefore \triangle =0\)

\(\text{Hence, }\begin{vmatrix} 0 & a & -b \\ -a & 0 & -c \\ b & c & 0 \\ \end{vmatrix}=0\)

Ques: If A = \(\begin{bmatrix} 1 & 1 & 1 \\1 & 0 & 2 \\3 & 1 & 1\\ \end{bmatrix}\) find A^-1

Hence, solve the system of equations :

x + y + z = 6,

x + 2z = 7,

3x + y + z = 12. (CBSE. 2019)

Ans: Here, A = \(\begin{bmatrix} 1 & 1 & 1 \\1 & 0 & 2 \\3 & 1 & 1\\ \end{bmatrix}\)

∴ |A| = 1(0–2) – (1)(1–6) + 1(1–0)

= – 2 + 5 + 1 = 4 ≠ 0

∴ A^-1 exists

Co-factor matrix of A = \(\begin{bmatrix} -2 & 5 & 1 \\0 & -2 & 2 \\2 & -1 & -1\\ \end{bmatrix}\)

Hence, A^-1 = \(\frac{1}{|A|}\)adj.A = \(\frac{1}{4}\begin{bmatrix} -2 & 0 & 2 \\5 & -2 & -1 \\1 & 2 & -1\\ \end{bmatrix}\)

(ii) For given system of equations can be written as:

\(\begin{bmatrix} 1 & 1 & 1 \\1 & 0 & 2 \\3 & 1 & 1\\ \end{bmatrix}\begin{bmatrix} x\\y\\z\\ \end{bmatrix}= \begin{bmatrix} 6\\7\\12\\ \end{bmatrix}\)

\(\implies\begin{bmatrix} x\\y\\z\\ \end{bmatrix}= \frac{1}{4}\begin{bmatrix} -2 & 0 & 2 \\5 & -2 & -1 \\1 & 2 & -1\\ \end{bmatrix}\begin{bmatrix} 6\\7\\12\\ \end{bmatrix}\)

\(=\frac{1}{4}\begin{bmatrix} -12 +0+24 \\30-14-12 \\6+14-12\\ \end{bmatrix}= \frac{1}{4}\begin{bmatrix} 12\\4\\8\\ \end{bmatrix}=\begin{bmatrix} 3\\1\\2\\ \end{bmatrix}\)

Hence x = 3, y = 1 and z = 2.

Also read:

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Very Long Answer Questions [5 Marks Questions]

Ques: By using properties of determinants, show that:

\(\begin{vmatrix} a^{2}+1 & ab & ac \\ ab & b^{2}+1 & bc \\ ca & cb & c^{2}+1 \\ \end{vmatrix}=1+a^{2}+b^{2}+c^{2}\)

Ans: Let \(\triangle = \begin{vmatrix} a^{2}+1 & ab & ac \\ ab & b^{2}+1 & bc \\ ca & cb & c^{2}+1 \\ \end{vmatrix}\)

Taking out a, b, and c from R₁, R₂, and R₃ respectively

\(\implies\triangle = abc \begin{vmatrix} a+\frac{1}{a} & b & c \\ a & b+\frac{1}{b} & c \\ a & b & c+\frac{1}{c} \\ \end{vmatrix}\)

Applying the row operations R₂ → R₂ – R₁ and R₃ → R₃ – R₁

\(\implies\triangle = abc \begin{vmatrix} a+\frac{1}{a} & b & c \\ -\frac{1}{a} & \frac{1}{b} & 0 \\ -\frac{1}{a} & 0 & \frac{1}{c} \\ \end{vmatrix}\)

Applying C₁ → aC₁, C₂ → bC₂ and C₃ → cC₃

\(\implies\triangle = abc\times\frac{1}{abc} \begin{vmatrix} a^{2}+1 & b^{2} & c^{2} \\ -1 & 1 & 0 \\ -1 & 0 & -1 \\ \end{vmatrix}\)

Expanding along C₃

\(\implies \triangle = -1\begin{vmatrix} b^{2} & c^{2} \\ 1 & 0 \\ \end{vmatrix}+1\begin{vmatrix} a^{2}+1 & b^{2} \\ -1 & 1 \\ \end{vmatrix}\)

\(\implies \triangle = 1(-c^{2})+(a^{2}+1+b^{2})\)

\(\therefore \triangle = 1+a^{2}+b^{2}+c^{2}\)

\(\text{Hence, }\begin{vmatrix} a^{2}+1 & ab & ac \\ ab & b^{2}+1 & bc \\ ca & cb & c^{2}+1 \\ \end{vmatrix}=1+a^{2}+b^{2}+c^{2}\)

Ques: The monthly incomes of Aryan and Babban are in the ratio 3 : 4 and their monthly expenditures are in the ratio of 5 : 7. If each saves ? 15,000 per month, find their monthly incomes, using the matrix method. (CBSE. 2016)

Ans:  Let Rs.3x and Rs.4x be the monthly income of Aryan and Babban respectively.

Let Rs.5y and Rs.7y be the monthly expenditure of Aryan and Babban respectively.

By the given question,

3x – 5y = 15000 …(1)

and 4x – 7y = 15000 …(2)

These equations can be written as AX = B …..(3)

Where A = \(\begin{bmatrix} 3 & -5 \\ 4 & -7 \\ \end{bmatrix}\) 

X = \(\begin{bmatrix} x \\ y \\ \end{bmatrix}\)

B = \(\begin{bmatrix} 15000 \\ 15000 \\ \end{bmatrix}\)

Now, A = \(\begin{bmatrix} 3 & -5 \\ 4 & -7 \\ \end{bmatrix}\)

= -21 + 20

\(=-1 \neq 0\)

Hence, A is a non-singular matrix and as such A^-1 exists

Now, 

adj.A = \(\begin{bmatrix} -7 & -4 \\ 5 & 3 \\ \end{bmatrix}'=\begin{bmatrix} -7 & 5 \\ -4 & 3 \\ \end{bmatrix}\)

Hence, 

 A^-1 = \(\frac{adj.A}{|A|}\)

 A^-1 = \(\frac{1}{-1}\begin{bmatrix} -7 & 5 \\ -4 & 3 \\ \end{bmatrix}=\begin{bmatrix} 7 & -5 \\ 4 & -3 \\ \end{bmatrix}\)

From (3), we get  A^-1

⇒ A^-1 (AX) = A^-1 B

⇒ (A^-1 A)X = A^-1 B

⇒ IX = A^-1 B

⇒ X = A^-1 B

x = 30,000.

Hence,monthly income of Aryan

= 3(30,000) = Rs.90,000

and monthly income of Babban

= 4(30,000) = Rs.1,20,000.

Also Read: Applications of Determinants

Ques: Using properties of determinants, prove that:

Ques: For the given matrix A,  findA^-1. 

A = 

Using A-1, solve the system of equations (CBSE. 2017):

2x+3y+10z=2

4x-6y+5z=5

6x+9y-20z=-4

Ans: The given system of equation is:

2x+3y+10z=2 …(1)

4x-6y+5z=5 …(2)

6x+9y-20z=-4 …(3)

These equations can be written as:

AX = B ….(4)

= 2(120 – 45) – 3 (- 80 – 30) + 10 (36 + 36)

= 150 + 330 + 720= 1200 ≠ 0.

∴ A is non-singular and as such A^-1 exists.

From (4), we get  A^-1

⇒ A^-1 (AX) = A^-1 B

⇒ (A^-1 A)X = A^-1 B

⇒ IX = A^-1 B

⇒ X = A^-1 B

Also read: