Explore the comprehensive NCERT Solutions for Class 12 Science Maths Chapter 4 on Determinants, featuring easy-to-follow step-by-step explanations. These solutions have gained immense popularity among Class 12 Science students, serving as a valuable resource for efficiently completing homework assignments and preparing for exams. All the questions and answers from Chapter 4 of the NCERT Book for Class 12 Science Maths are readily available here, providing students with free access to essential study materials.
Page No 108:
Question 1:
Evaluate the determinants in Exercises 1 and 2.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6411/Chapter%204_html_54bbaa6a.gif)
ANSWER:
= 2(−1) − 4(−5) = − 2 + 20 = 18
Page No 108:
Question 2:
Evaluate the determinants in Exercises 1 and 2.
(i) (ii)
ANSWER:
(i) = (cos θ)(cos θ) − (−sin θ)(sin θ) = cos2θ+ sin2θ = 1
(ii)
= (x2 − x + 1)(x + 1) − (x − 1)(x + 1)
= x3 − x2 + x + x2 − x + 1 − (x2 − 1)
= x3 + 1 − x2 + 1
= x3 − x2 + 2
Page No 108:
Question 3:
If, then show that
ANSWER:
The given matrix is.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6413/Chapter%204_html_5a46c228.gif)
Page No 108:
Question 4:
If, then show that
ANSWER:
The given matrix is.
It can be observed that in the first column, two entries are zero. Thus, we expand along the first column (C1) for easier calculation.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6416/Chapter%204_html_275ad411.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6416/Chapter%204_html_6b5dffe6.gif)
From equations (i) and (ii), we have:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6416/Chapter%204_html_171d0b0c.gif)
Hence, the given result is proved.
Page No 108:
Question 5:
Evaluate the determinants
(i) (iii)
(ii) (iv)
ANSWER:
(i) Let.
It can be observed that in the second row, two entries are zero. Thus, we expand along the second row for easier calculation.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6418/Chapter%204_html_m1094e20c.gif)
(ii) Let.
By expanding along the first row, we have:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6418/Chapter%204_html_m380503db.gif)
(iii) Let
By expanding along the first row, we have:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6418/Chapter%204_html_162bb8c6.gif)
(iv) Let
By expanding along the first column, we have:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6418/Chapter%204_html_45cc20ca.gif)
Page No 109:
Question 6:
If, find
.
ANSWER:
Let
By expanding along the first row, we have:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6419/Chapter%204_html_7a717b7.gif)
Page No 109:
Question 7:
Find values of x, if
![](https://ddtarget.com/wp-content/uploads/2024/01/image-29.png)
ANSWER:
(i)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6427/Chapter%204_html_m4c59b137.gif)
(ii)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6427/Chapter%204_html_m3820bd31.gif)
Page No 109:
Question 8:
If, then x is equal to
(A) 6 (B) ±6 (C) −6 (D) 0
ANSWER:
Answer: B
![](https://ddtarget.com/wp-content/uploads/2024/01/image-28.png)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6432/Chapter%204_html_245292a.gif)
Hence, the correct answer is B.
Page No 119:
Question 1:
Using the property of determinants and without expanding, prove that:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6433/Chapter%204_html_m34ceacad.gif)
ANSWER:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6433/Chapter%204_html_m7e16c551.gif)
Page No 119:
Question 2:
Using the property of determinants and without expanding, prove that:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6434/Chapter%204_html_m54f310dc.gif)
ANSWER:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6434/Chapter%204_html_63901485.gif)
Here, the two rows R1 and R3 are identical.
Δ = 0.
Page No 119:
Question 3:
Using the property of determinants and without expanding, prove that:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6435/Chapter%204_html_7ba49357.gif)
ANSWER:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6435/Chapter%204_html_m7dd04180.gif)
Page No 119:
Question 4:
Using the property of determinants and without expanding, prove that:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6436/Chapter%204_html_393dc1bf.gif)
ANSWER:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6436/Chapter%204_html_7c73293c.gif)
By applying C3 → C3 + C2, we have:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6436/Chapter%204_html_3998e583.gif)
Here, two columns C1 and C3 are proportional.
Δ = 0.
Page No 119:
Question 5:
Using the property of determinants and without expanding, prove that:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6437/Chapter%204_html_21fba1a3.gif)
ANSWER:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6437/Chapter%204_html_m76aec2fc.gif)
Applying R2 → R2 − R3, we have:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6437/Chapter%204_html_m1d615a3f.gif)
Applying R1 ↔R3 and R2 ↔R3, we have:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6437/Chapter%204_html_m1cf7fb93.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6437/Chapter%204_html_m5f94847d.gif)
Applying R1 → R1 − R3, we have:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6437/Chapter%204_html_m224a7b62.gif)
Applying R1 ↔R2 and R2 ↔R3, we have:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6437/Chapter%204_html_5299b3f6.gif)
From (1), (2), and (3), we have:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6437/Chapter%204_html_m1a942179.gif)
Hence, the given result is proved.
Page No 120:
Question 6:
By using properties of determinants, show that:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6438/Chapter%204_html_79eca739.gif)
ANSWER:
We have,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6438/Chapter%204_html_m29d1e3cc.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6438/Chapter%204_html_3a09dee6.gif)
Here, the two rows R1 and R3 are identical.
∴Δ = 0.
Page No 120:
Question 7:
By using properties of determinants, show that:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6439/Chapter%204_html_53198e9d.gif)
ANSWER:
![](https://ddtarget.com/wp-content/uploads/2024/01/image-30.png)
Applying R2 → R2 + R1 and R3 → R3 + R1, we have:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6439/Chapter%204_html_6d7f53fb.gif)
Page No 120:
Question 8:
By using properties of determinants, show that:
(i)
(ii)
ANSWER:
(i)
Applying R1 → R1 − R3 and R2 → R2 − R3, we have:
![](https://ddtarget.com/wp-content/uploads/2024/01/image-31.png)
Applying R1 → R1 + R2, we have:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6440/Chapter%204_html_7cb13fc9.gif)
Expanding along C1, we have:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6440/Chapter%204_html_m27875559.gif)
Hence, the given result is proved.
(ii) Let.
Applying C1 → C1 − C3 and C2 → C2 − C3, we have:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6440/Chapter%204_html_m246c534f.gif)
Applying C1 → C1 + C2, we have:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6440/Chapter%204_html_m3415f1ee.gif)
Expanding along C1, we have:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6440/Chapter%204_html_m4f62d366.gif)
Hence, the given result is proved.
Page No 120:
Question 9:
By using properties of determinants, show that:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6442/Chapter%204_html_355485c4.gif)
ANSWER:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6442/Chapter%204_html_2c7a9498.gif)
Applying R2 → R2 − R1 and R3 → R3 − R1, we have:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6442/Chapter%204_html_3b60a4ce.gif)
Applying R3 → R3 + R2, we have:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6442/Chapter%204_html_m333415ed.gif)
Expanding along R3, we have:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6442/Chapter%204_html_2e7e34fa.gif)
Hence, the given result is proved.
Page No 120:
Question 10:
By using properties of determinants, show that:
(i)
(ii)
ANSWER:
(i)
Applying R1 → R1 + R2 + R3, we have:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6448/Chapter%204_html_m521a4384.gif)
Applying C2 → C2 − C1, C3 → C3 − C1, we have:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6448/Chapter%204_html_7159bae1.gif)
Expanding along C3, we have:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6448/Chapter%204_html_m47d09c0.gif)
Hence, the given result is proved.
(ii)
Applying R1 → R1 + R2 + R3, we have:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6448/Chapter%204_html_7dd23915.gif)
Applying C2 → C2 − C1 and C3 → C3 − C1, we have:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6448/Chapter%204_html_m18416f2f.gif)
Expanding along C3, we have:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6448/Chapter%204_html_7b7c9f87.gif)
Hence, the given result is proved.
Page No 120:
Question 11:
By using properties of determinants, show that:
(i)
(ii)
ANSWER:
(i)
Applying R1 → R1 + R2 + R3, we have:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6451/Chapter%204_html_100f2562.gif)
Applying C2 → C2 − C1, C3 → C3 − C1, we have:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6451/Chapter%204_html_5b75f335.gif)
Expanding along C3, we have:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6451/Chapter%204_html_m70739894.gif)
Hence, the given result is proved.
(ii)
Applying C1 → C1 + C2 + C3, we have:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6451/Chapter%204_html_m84f5197.gif)
Applying R2 → R2 − R1 and R3 → R3 − R1, we have:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6451/Chapter%204_html_10d6ee16.gif)
Expanding along R3, we have:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6451/Chapter%204_html_6c9dd75f.gif)
Hence, the given result is proved.
Page No 121:
Question 12:
By using properties of determinants, show that:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6455/Chapter%204_html_m31a8ec4f.gif)
ANSWER:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6455/Chapter%204_html_m5af9956a.gif)
Applying R1 → R1 + R2 + R3, we have:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6455/Chapter%204_html_m6e2e046a.gif)
Applying C2 → C2 − C1 and C3 → C3 − C1, we have:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6455/Chapter%204_html_m3ef95028.gif)
Expanding along R1, we have:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6455/Chapter%204_html_m31e7ad67.gif)
Hence, the given result is proved.
Page No 121:
Question 13:
By using properties of determinants, show that:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6458/Chapter%204_html_317e675b.gif)
ANSWER:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6458/Chapter%204_html_m38800b43.gif)
Applying R1 → R1 + bR3 and R2 → R2 − aR3, we have:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6458/Chapter%204_html_m12ea1c0b.gif)
Expanding along R1, we have:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6458/Chapter%204_html_m744fa485.gif)
Page No 121:
Question 14:
By using properties of determinants, show that:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6463/Chapter%204_html_3a589695.gif)
ANSWER:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6463/Chapter%204_html_m281f621f.gif)
Taking out common factors a, b, and c from R1, R2, and R3 respectively, we have:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6463/Chapter%204_html_m4723eaeb.gif)
Applying R2 → R2 − R1 and R3 → R3 − R1, we have:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6463/Chapter%204_html_5197aeba.gif)
Applying C1 → aC1, C2 → bC2, and C3 → cC3, we have:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6463/Chapter%204_html_349958b4.gif)
Expanding along R3, we have:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6463/Chapter%204_html_m3c8bec5f.gif)
Hence, the given result is proved.
Page No 121:
Question 15:
Choose the correct answer.
Let A be a square matrix of order 3 × 3, then is equal to
A. B.
C.
D.
ANSWER:
Answer: C
A is a square matrix of order 3 × 3.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6474/Chapter%204_html_m50c1e7b5.gif)
Hence, the correct answer is C.
Page No 121:
Question 16:
Which of the following is correct?
A. Determinant is a square matrix.
B. Determinant is a number associated to a matrix.
C. Determinant is a number associated to a square matrix.
D. None of these
ANSWER:
Answer: C
We know that to every square matrix, of order n. We can associate a number called the determinant of square matrix A, where
element of A.
Thus, the determinant is a number associated to a square matrix.
Hence, the correct answer is C.
Page No 122:
Question 1:
Find area of the triangle with vertices at the point given in each of the following:
(i) (1, 0), (6, 0), (4, 3) (ii) (2, 7), (1, 1), (10, 8)
(iii) (−2, −3), (3, 2), (−1, −8)
ANSWER:
(i) The area of the triangle with vertices (1, 0), (6, 0), (4, 3) is given by the relation,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6480/Chapter%204_html_27997392.gif)
(ii) The area of the triangle with vertices (2, 7), (1, 1), (10, 8) is given by the relation,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6480/Chapter%204_html_1ae5503b.gif)
(iii) The area of the triangle with vertices (−2, −3), (3, 2), (−1, −8)
is given by the relation,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6480/Chapter%204_html_m477135a4.gif)
Hence, the area of the triangle is.
Page No 123:
Question 2:
Show that points
are collinear
ANSWER:
Area of ΔABC is given by the relation,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6481/Chapter%204_html_4574753c.gif)
Thus, the area of the triangle formed by points A, B, and C is zero.
Hence, the points A, B, and C are collinear.
Page No 123:
Question 3:
Find values of k if area of triangle is 4 square units and vertices are
(i) (k, 0), (4, 0), (0, 2) (ii) (−2, 0), (0, 4), (0, k)
ANSWER:
We know that the area of a triangle whose vertices are (x1, y1), (x2, y2), and
(x3, y3) is the absolute value of the determinant (Δ), where
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6486/Chapter%204_html_m7d88ee23.gif)
It is given that the area of triangle is 4 square units.
∴Δ = ± 4.
(i) The area of the triangle with vertices (k, 0), (4, 0), (0, 2) is given by the relation,
Δ =
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6486/Chapter%204_html_5271de10.gif)
∴−k + 4 = ± 4
When −k + 4 = − 4, k = 8.
When −k + 4 = 4, k = 0.
Hence, k = 0, 8.
(ii) The area of the triangle with vertices (−2, 0), (0, 4), (0, k) is given by the relation,
Δ =
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6486/Chapter%204_html_19a19c67.gif)
∴k − 4 = ± 4
When k − 4 = − 4, k = 0.
When k − 4 = 4, k = 8.
Hence, k = 0, 8.
Page No 123:
Question 4:
(i) Find equation of line joining (1, 2) and (3, 6) using determinants
(ii) Find equation of line joining (3, 1) and (9, 3) using determinants
ANSWER:
(i) Let P (x, y) be any point on the line joining points A (1, 2) and B (3, 6). Then, the points A, B, and P are collinear. Therefore, the area of triangle ABP will be zero.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6490/Chapter%204_html_m4c94f002.gif)
Hence, the equation of the line joining the given points is y = 2x.
(ii) Let P (x, y) be any point on the line joining points A (3, 1) and
B (9, 3). Then, the points A, B, and P are collinear. Therefore, the area of triangle ABP will be zero.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6490/Chapter%204_html_2932c449.gif)
Hence, the equation of the line joining the given points is x − 3y = 0.
Page No 123:
Question 5:
If area of triangle is 35 square units with vertices (2, −6), (5, 4), and (k, 4). Then k is
A. 12 B. −2 C. −12, −2 D. 12, −2
ANSWER:
Answer: D
The area of the triangle with vertices (2, −6), (5, 4), and (k, 4) is given by the relation,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6501/Chapter%204_html_m1ffb3930.gif)
It is given that the area of the triangle is ±35.
Therefore, we have:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6501/Chapter%204_html_m20edecc0.gif)
When 5 − k = −7, k = 5 + 7 = 12.
When 5 − k = 7, k = 5 − 7 = −2.
Hence, k = 12, −2.
The correct answer is D.
Page No 126:
Question 1:
Write Minors and Cofactors of the elements of following determinants:
![](https://ddtarget.com/wp-content/uploads/2024/01/image-32.png)
ANSWER:
(i) The given determinant is.
Minor of element aij is Mij.
∴M11 = minor of element a11 = 3
M12 = minor of element a12 = 0
M21 = minor of element a21 = −4
M22 = minor of element a22 = 2
Cofactor of aij is Aij = (−1)i + j Mij.
∴A11 = (−1)1+1 M11 = (−1)2 (3) = 3
A12 = (−1)1+2 M12 = (−1)3 (0) = 0
A21 = (−1)2+1 M21 = (−1)3 (−4) = 4
A22 = (−1)2+2 M22 = (−1)4 (2) = 2
(ii) The given determinant is.
Minor of element aij is Mij.
∴M11 = minor of element a11 = d
M12 = minor of element a12 = b
M21 = minor of element a21 = c
M22 = minor of element a22 = a
Cofactor of aij is Aij = (−1)i + j Mij.
∴A11 = (−1)1+1 M11 = (−1)2 (d) = d
A12 = (−1)1+2 M12 = (−1)3 (b) = −b
A21 = (−1)2+1 M21 = (−1)3 (c) = −c
A22 = (−1)2+2 M22 = (−1)4 (a) = a
Page No 126:
Question 2:
(i) (ii)
ANSWER:
(i) The given determinant is.
By the definition of minors and cofactors, we have:
M11 = minor of a11=
M12 = minor of a12=
M13 = minor of a13 =
M21 = minor of a21 =
M22 = minor of a22 =
M23 = minor of a23 =
M31 = minor of a31=
M32 = minor of a32 =
M33 = minor of a33 =
A11 = cofactor of a11= (−1)1+1 M11 = 1
A12 = cofactor of a12 = (−1)1+2 M12 = 0
A13 = cofactor of a13 = (−1)1+3 M13 = 0
A21 = cofactor of a21 = (−1)2+1 M21 = 0
A22 = cofactor of a22 = (−1)2+2 M22 = 1
A23 = cofactor of a23 = (−1)2+3 M23 = 0
A31 = cofactor of a31 = (−1)3+1 M31 = 0
A32 = cofactor of a32 = (−1)3+2 M32 = 0
A33 = cofactor of a33 = (−1)3+3 M33 = 1
(ii) The given determinant is.
By definition of minors and cofactors, we have:
M11 = minor of a11=
M12 = minor of a12=
M13 = minor of a13 =
M21 = minor of a21 =
M22 = minor of a22 =
M23 = minor of a23 =
M31 = minor of a31=
M32 = minor of a32 =
M33 = minor of a33 =
A11 = cofactor of a11= (−1)1+1 M11 = 11
A12 = cofactor of a12 = (−1)1+2 M12 = −6
A13 = cofactor of a13 = (−1)1+3 M13 = 3
A21 = cofactor of a21 = (−1)2+1 M21 = 4
A22 = cofactor of a22 = (−1)2+2 M22 = 2
A23 = cofactor of a23 = (−1)2+3 M23 = −1
A31 = cofactor of a31 = (−1)3+1 M31 = −20
A32 = cofactor of a32 = (−1)3+2 M32 = 13
A33 = cofactor of a33 = (−1)3+3 M33 = 5
Page No 126:
Question 3:
Using Cofactors of elements of second row, evaluate.
ANSWER:
The given determinant is.
We have:
M21 =
∴A21 = cofactor of a21 = (−1)2+1 M21 = 7
M22 =
∴A22 = cofactor of a22 = (−1)2+2 M22 = 7
M23 =
∴A23 = cofactor of a23 = (−1)2+3 M23 = −7
We know that Δ is equal to the sum of the product of the elements of the second row with their corresponding cofactors.
∴Δ = a21A21 + a22A22 + a23A23 = 2(7) + 0(7) + 1(−7) = 14 − 7 = 7
Page No 126:
Question 4:
Using Cofactors of elements of third column, evaluate
ANSWER:
The given determinant is.
We have:
M13 =
M23 =
M33 =
∴A13 = cofactor of a13 = (−1)1+3 M13 = (z − y)
A23 = cofactor of a23 = (−1)2+3 M23 = − (z − x) = (x − z)
A33 = cofactor of a33 = (−1)3+3 M33 = (y − x)
We know that Δ is equal to the sum of the product of the elements of the second row with their corresponding cofactors.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6511/Chapter%204_html_m7f475413.gif)
Hence,
Page No 126:
Question 5:
If and Aij is Cofactors of aij, then value of Δ is given by
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6512/Chapter%204_html_c4712c.gif)
ANSWER:
Answer: D
We know that:
Δ = Sum of the product of the elements of a column (or a row) with their corresponding cofactors
∴Δ = a11A11 + a21A21 + a31A31
Hence, the value of Δ is given by the expression given in alternative D.
The correct answer is D.
Page No 131:
Question 1:
Find adjoint of each of the matrices.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6513/Chapter%204_html_413e6eb9.gif)
ANSWER:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6513/Chapter%204_html_m3bb9a490.gif)
Page No 131:
Question 2:
Find adjoint of each of the matrices.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6514/Chapter%204_html_m31b7a52.gif)
ANSWER:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6514/Chapter%204_html_15c3d01e.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6514/Chapter%204_html_6be1bcdf.gif)
Page No 131:
Question 3:
Verify A (adj A) = (adj A) A = I .
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6515/Chapter%204_html_m448574d0.gif)
ANSWER:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6515/Chapter%204_html_5446985e.gif)
Page No 131:
Question 4:
Verify A (adj A) = (adj A) A = I .
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6516/Chapter%204_html_m3858f336.gif)
ANSWER:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6516/Chapter%204_html_m319e61e.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6516/Chapter%204_html_4a4cd990.gif)
Page No 132:
Question 5:
Find the inverse of each of the matrices (if it exists).
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6523/Chapter%204_html_m4bdb47fc.gif)
ANSWER:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6523/Chapter%204_html_77db058d.gif)
Page No 132:
Question 6:
Find the inverse of each of the matrices (if it exists).
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6525/Chapter%204_html_m72284cb9.gif)
ANSWER:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6525/Chapter%204_html_537eb3f4.gif)
Page No 132:
Question 7:
Find the inverse of each of the matrices (if it exists).
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6529/Chapter%204_html_m6265657b.gif)
ANSWER:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6529/Chapter%204_html_m47a4c645.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6529/Chapter%204_html_m6bb09ccb.gif)
Page No 132:
Question 8:
Find the inverse of each of the matrices (if it exists).
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6534/Chapter%204_html_m743fb22c.gif)
ANSWER:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6534/Chapter%204_html_4a874e22.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6534/Chapter%204_html_32a06410.gif)
Page No 132:
Question 9:
Find the inverse of each of the matrices (if it exists).
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6537/Chapter%204_html_m534c34b0.gif)
ANSWER:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6537/Chapter%204_html_66eaf191.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6537/Chapter%204_html_m74dc7de7.gif)
Page No 132:
Question 10:
Find the inverse of each of the matrices (if it exists).
.
ANSWER:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6540/Chapter%204_html_m792b92cb.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6540/Chapter%204_html_m20bcb755.gif)
Page No 132:
Question 11:
Find the inverse of each of the matrices (if it exists).
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6542/Chapter%204_html_m71c01161.gif)
ANSWER:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6542/Chapter%204_html_m27178729.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6542/Chapter%204_html_3aa10bb1.gif)
Page No 132:
Question 12:
Let and
. Verify that
ANSWER:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6545/Chapter%204_html_m12c34e07.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6545/Chapter%204_html_m48497f5a.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6545/Chapter%204_html_m625f8815.gif)
From (1) and (2), we have:
(AB)−1 = B−1A−1
Hence, the given result is proved.
Page No 132:
Question 13:
If, show that
. Hence find
.
ANSWER:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6547/Chapter%204_html_m327be03a.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6547/Chapter%204_html_9b5bee6.gif)
Page No 132:
Question 14:
For the matrix, find the numbers a and b such that A2 + aA + bI = O.
ANSWER:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6548/Chapter%204_html_22d4c56d.gif)
We have:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6548/Chapter%204_html_3ba04951.gif)
Comparing the corresponding elements of the two matrices, we have:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6548/Chapter%204_html_6694237a.gif)
Hence, −4 and 1 are the required values of a and b respectively.
Page No 132:
Question 15:
For the matrixshow that A3 − 6A2 + 5A + 11 I = O. Hence, find A−1.
ANSWER:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6550/Chapter%204_html_70231627.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6550/Chapter%204_html_283bca91.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6550/Chapter%204_html_m1f042fdd.gif)
From equation (1), we have:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6550/Chapter%204_html_m294f3c5a.gif)
Page No 132:
Question 16:
If verify that A3 − 6A2 + 9A − 4I = O and hence find A−1
ANSWER:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6556/Chapter%204_html_4446908.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6556/Chapter%204_html_4ef14992.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6556/Chapter%204_html_69fa609.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6556/Chapter%204_html_m18f477ff.gif)
From equation (1), we have:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6556/Chapter%204_html_2db3f604.gif)
Page No 132:
Question 17:
Let A be a nonsingular square matrix of order 3 × 3. Then is equal to
A. B.
C.
D.
ANSWER:
Answer: B
We know that,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6572/Chapter%204_html_m49d4e361.gif)
Hence, the correct answer is B.
Page No 132:
Question 18:
If A is an invertible matrix of order 2, then det (A−1) is equal to
A. det (A) B. C. 1 D. 0
ANSWER:
Since A is an invertible matrix,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6578/Chapter%204_html_m12be1d7d.gif)
Hence, the correct answer is B.
Page No 136:
Question 1:
Examine the consistency of the system of equations.
x + 2y = 2
2x + 3y = 3
ANSWER:
The given system of equations is:
x + 2y = 2
2x + 3y = 3
The given system of equations can be written in the form of AX = B, where
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6584/Chapter%204_html_30b0a9c2.gif)
∴ A is non-singular.
Therefore, A−1 exists.
Hence, the given system of equations is consistent.
Page No 136:
Question 2:
Examine the consistency of the system of equations.
2x − y = 5
x + y = 4
ANSWER:
The given system of equations is:
2x − y = 5
x + y = 4
The given system of equations can be written in the form of AX = B, where
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6589/Chapter%204_html_625c2816.gif)
∴ A is non-singular.
Therefore, A−1 exists.
Hence, the given system of equations is consistent.
Page No 136:
Question 3:
Examine the consistency of the system of equations.
x + 3y = 5
2x + 6y = 8
ANSWER:
The given system of equations is:
x + 3y = 5
2x + 6y = 8
The given system of equations can be written in the form of AX = B, where
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6594/Chapter%204_html_2c1ee03a.gif)
∴ A is a singular matrix.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6594/Chapter%204_html_10044d87.gif)
Thus, the solution of the given system of equations does not exist. Hence, the system of equations is inconsistent.
Page No 136:
Question 4:
Examine the consistency of the system of equations.
x + y + z = 1
2x + 3y + 2z = 2
ax + ay + 2az = 4
ANSWER:
The given system of equations is:
x + y + z = 1
2x + 3y + 2z = 2
ax + ay + 2az = 4
This system of equations can be written in the form AX = B, where
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6601/Chapter%204_html_m4b4d1993.gif)
∴ A is non-singular.
Therefore, A−1 exists.
Hence, the given system of equations is consistent.
Page No 136:
Question 5:
Examine the consistency of the system of equations.
3x − y − 2z = 2
2y − z = −1
3x − 5y = 3
ANSWER:
The given system of equations is:
3x − y − 2z = 2
2y − z = −1
3x − 5y = 3
This system of equations can be written in the form of AX = B, where
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6606/Chapter%204_html_m54650ebf.gif)
∴ A is a singular matrix.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6606/Chapter%204_html_m1e5b7c2a.gif)
Thus, the solution of the given system of equations does not exist. Hence, the system of equations is inconsistent.
Page No 136:
Question 6:
Examine the consistency of the system of equations.
5x − y + 4z = 5
2x + 3y + 5z = 2
5x − 2y + 6z = −1
ANSWER:
The given system of equations is:
5x − y + 4z = 5
2x + 3y + 5z = 2
5x − 2y + 6z = −1
This system of equations can be written in the form of AX = B, where
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6608/Chapter%204_html_m631ef4b3.gif)
∴ A is non-singular.
Therefore, A−1 exists.
Hence, the given system of equations is consistent.
Page No 136:
Question 7:
Solve system of linear equations, using matrix method.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6613/Chapter%204_html_m724cf69e.gif)
ANSWER:
The given system of equations can be written in the form of AX = B, where
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6613/Chapter%204_html_m33c86301.gif)
Thus, A is non-singular. Therefore, its inverse exists.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6613/Chapter%204_html_39a6a942.gif)
Page No 136:
Question 8:
Solve system of linear equations, using matrix method.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6621/Chapter%204_html_1d13fa43.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6621/Chapter%204_html_m2be58a95.gif)
ANSWER:
The given system of equations can be written in the form of AX = B, where
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6621/Chapter%204_html_59b8eaa8.gif)
Thus, A is non-singular. Therefore, its inverse exists.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6621/Chapter%204_html_m72cd731.gif)
Page No 136:
Question 9:
Solve system of linear equations, using matrix method.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6627/Chapter%204_html_m326506ee.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6627/Chapter%204_html_m5e3784fb.gif)
ANSWER:
The given system of equations can be written in the form of AX = B, where
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6627/Chapter%204_html_5b6948cf.gif)
Thus, A is non-singular. Therefore, its inverse exists.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6627/Chapter%204_html_324d0cda.gif)
Page No 136:
Question 10:
Solve system of linear equations, using matrix method.
5x + 2y = 3
3x + 2y = 5
ANSWER:
The given system of equations can be written in the form of AX = B, where
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6632/Chapter%204_html_m2810aa19.gif)
Thus, A is non-singular. Therefore, its inverse exists.
Page No 136:
Question 11:
Solve system of linear equations, using matrix method.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6638/Chapter%204_html_54e2ee67.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6638/Chapter%204_html_mcb0a148.gif)
ANSWER:
The given system of equations can be written in the form of AX = B, where
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6638/Chapter%204_html_m4ae283eb.gif)
Thus, A is non-singular. Therefore, its inverse exists.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6638/Chapter%204_html_m71cac2ce.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6638/Chapter%204_html_bf8a970.gif)
Page No 136:
Question 12:
Solve system of linear equations, using matrix method.
x − y + z = 4
2x + y − 3z = 0
x + y + z = 2
ANSWER:
The given system of equations can be written in the form of AX = B, where
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6644/Chapter%204_html_fffddc7.gif)
Thus, A is non-singular. Therefore, its inverse exists.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6644/Chapter%204_html_59c7d826.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6644/Chapter%204_html_2398fa92.gif)
Page No 136:
Question 13:
Solve system of linear equations, using matrix method.
2x + 3y + 3z = 5
x − 2y + z = −4
3x − y − 2z = 3
ANSWER:
The given system of equations can be written in the form AX = B, where
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6647/Chapter%204_html_m6e5aee51.gif)
Thus, A is non-singular. Therefore, its inverse exists.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6647/Chapter%204_html_2a364aa6.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6647/Chapter%204_html_m480412b3.gif)
Page No 136:
Question 14:
Solve system of linear equations, using matrix method.
x − y + 2z = 7
3x + 4y − 5z = −5
2x − y + 3z = 12
ANSWER:
The given system of equations can be written in the form of AX = B, where
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6652/Chapter%204_html_6f61bdbd.gif)
Thus, A is non-singular. Therefore, its inverse exists.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6652/Chapter%204_html_214d0473.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6652/Chapter%204_html_m76863815.gif)
Page No 137:
Question 15:
If, find A−1. Using A−1 solve the system of equations
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6661/Chapter%204_html_m4a65d85.gif)
ANSWER:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6661/Chapter%204_html_m1d26ecb7.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6661/Chapter%204_html_m4a89e4df.gif)
Now, the given system of equations can be written in the form of AX = B, where
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6661/Chapter%204_html_1236668a.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6661/Chapter%204_html_24ed4cc2.gif)
Page No 137:
Question 16:
The cost of 4 kg onion, 3 kg wheat and 2 kg rice is Rs 60. The cost of 2 kg onion, 4 kg
wheat and 6 kg rice is Rs 90. The cost of 6 kg onion 2 kg wheat and 3 kg rice is Rs 70.
Find cost of each item per kg by matrix method.
ANSWER:
Let the cost of onions, wheat, and rice per kg be Rs x, Rs y,and Rs z respectively.
Then, the given situation can be represented by a system of equations as:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6666/Chapter%204_html_25a36f4.gif)
This system of equations can be written in the form of AX = B, where
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6666/Chapter%204_html_74fc5d5f.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6666/Chapter%204_html_5c8fdc20.gif)
Now,
X = A−1B
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6666/Chapter%204_html_m5d62b9ab.gif)
Hence, the cost of onions is Rs 5 per kg, the cost of wheat is Rs 8 per kg, and the cost of rice is Rs 8 per kg.
Page No 141:
Question 1:
Prove that the determinant is independent of θ.
ANSWER:
![](https://img-nm.mnimgs.com/img/study_content/editlive_ncert/77/2012_02_15_17_54_55/mathmlequation3065325959819493208.png)
Hence, Δ is independent of θ.
Page No 141:
Question 2:
Without expanding the determinant, prove that
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6679/Chapter%204_html_m35d1ab09.gif)
ANSWER:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6679/Chapter%204_html_m3c295fc.gif)
Hence, the given result is proved.
Page No 141:
Question 3:
Evaluate
ANSWER:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6684/Chapter%204_html_m2a6ea31.gif)
Expanding along C3, we have:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6684/Chapter%204_html_78843555.gif)
Page No 141:
Question 4:
If a, b and c are real numbers, and,
Show that either a + b + c = 0 or a = b = c.
ANSWER:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6688/Chapter%204_html_m7b4eb4be.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6688/Chapter%204_html_de0ec94.gif)
Expanding along R1, we have:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6688/Chapter%204_html_6069f2c1.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6688/Chapter%204_html_20447801.gif)
Hence, if Δ = 0, then either a + b + c = 0 or a = b = c.
Page No 141:
Question 5:
Solve the equations
ANSWER:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6695/Chapter%204_html_m377141b6.gif)
Page No 141:
![](https://ddtarget.com/wp-content/uploads/2024/01/image-33.png)
ANSWER:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6701/Chapter%204_html_53bf5077.gif)
Expanding along R3, we have:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6701/Chapter%204_html_m629ccfa8.gif)
Hence, the given result is proved.
Page No 141:
Question 7:
If
ANSWER:
We know that.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6705/Chapter%204_html_7fded7ff.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6705/Chapter%204_html_62c2672f.gif)
Page No 142:
Question 8:
Let verify that
(i)
(ii)
ANSWER:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6708/Chapter%204_html_m3f7303ea.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6708/Chapter%204_html_m6678b4af.gif)
(i)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6708/Chapter%204_html_5f871e24.gif)
We have,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6708/Chapter%204_html_1bb9104b.gif)
(ii)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6708/Chapter%204_html_44784180.gif)
Page No 142:
Question 9:
Evaluate
ANSWER:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6713/Chapter%204_html_317181b7.gif)
Expanding along R1, we have:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6713/Chapter%204_html_2d28f70b.gif)
Page No 142:
Question 10:
Evaluate
ANSWER:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6716/Chapter%204_html_24808112.gif)
Expanding along C1, we have:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6716/Chapter%204_html_m7a073c6a.gif)
Page No 142:
Question 11:
Using properties of determinants, prove that:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6717/Chapter%204_html_m2053ac31.gif)
ANSWER:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6717/Chapter%204_html_m13b66365.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6717/Chapter%204_html_3b1c868d.gif)
Expanding along R3, we have:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6717/Chapter%204_html_5c654038.gif)
Hence, the given result is proved.
Page No 142:
Question 12:
Using properties of determinants, prove that:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6719/Chapter%204_html_6e13f6e9.gif)
ANSWER:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6719/Chapter%204_html_37275d0f.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6719/Chapter%204_html_m79fd7df8.gif)
Expanding along R3, we have:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6719/Chapter%204_html_77d332cc.gif)
Hence, the given result is proved.
Page No 142:
Question 13:
Using properties of determinants, prove that:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6723/Chapter%204_html_35583b6a.gif)
ANSWER:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6723/Chapter%204_html_m3a9c7a67.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6723/Chapter%204_html_m74d69704.gif)
Expanding along C1, we have:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6723/Chapter%204_html_m356b29d8.gif)
Hence, the given result is proved.
Page No 142:
Question 14:
Using properties of determinants, prove that:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6726/Chapter%204_html_m14dcca89.gif)
ANSWER:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6726/Chapter%204_html_391432e3.gif)
Expanding along C1, we have:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6726/Chapter%204_html_m7c6e807f.gif)
Hence, the given result is proved.
Page No 142:
Question 15:
Using properties of determinants, prove that:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6729/Chapter%204_html_m151f0f29.gif)
ANSWER:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6729/Chapter%204_html_m9108e5c.gif)
Hence, the given result is proved.
Page No 142:
Question 16:
Solve the system of the following equations
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6733/Chapter%204_html_30a8d8b2.gif)
ANSWER:
Let
Then the given system of equations is as follows:
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6733/Chapter%204_html_6303329e.gif)
This system can be written in the form of AX = B, where
A
Thus, A is non-singular. Therefore, its inverse exists.
Now,
A11 = 75, A12 = 110, A13 = 72
A21 = 150, A22 = −100, A23 = 0
A31 = 75, A32 = 30, A33 = − 24
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6733/Chapter%204_html_68fa224.gif)
Page No 143:
Question 17:
Choose the correct answer.
If a, b, c, are in A.P., then the determinant
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6734/Chapter%204_html_m5ab9cf8c.gif)
A. 0 B. 1 C. x D. 2x
ANSWER:
Answer: A
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6734/Chapter%204_html_m4df80470.gif)
Here, all the elements of the first row (R1) are zero.
Hence, we have Δ = 0.
The correct answer is A.
Page No 143:
Question 18:
Choose the correct answer.
If x, y, z are nonzero real numbers, then the inverse of matrix is
A. B.
C. D.
ANSWER:
Answer: A
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6736/Chapter%204_html_1a5715dc.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/12/15/233/6736/Chapter%204_html_m451a05bd.gif)
The correct answer is A.
Page No 143:
Question 19:
Choose the correct answer.
Let, where 0 ≤ θ≤ 2π, then
A. Det (A) = 0
B. Det (A) ∈ (2, ∞)
C. Det (A) ∈ (2, 4)
D. Det (A)∈ [2, 4]
ANSWER:
![NCERT Solutions for Class 12 - DD Target PMT](https://ddtarget.com/wp-content/uploads/2024/01/image-34.png)