NCERT Solutions for Class 12 Maths Exercise 4.5 Chapter 4 Determinants – FREE PDF Download
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NCERT Solutions for Class 12 Maths Chapter 4 – Determinants
Find adjoint of each of the matrices in Exercise 1 and 2.
1. 
A11 = Cofactor of 
A12 = Cofactor of 
A21 = Cofactor of 
A22 = Cofactor of 
adj. A =∣∣∣A11A21A12A22∣∣∣′=|A11A12A21A22|′
=∣∣∣4−2−31∣∣∣′=|4−3−21|′
=∣∣∣4−3−21∣∣∣=|4−2−31|
2. 
= 
adj. A =∣∣∣∣31−11−125−1625∣∣∣∣′=|3−126152−11−15|′
=∣∣∣∣3−126152−11−15∣∣∣∣=|31−11−125−1625|
Verify A (adj. A) = (adj. A) A= in Exercise 3 and 4.
3. 
adj. A = 
A.(adj. A) =
= 
= 
…..(i)
Again (adj. A). A =
= 
= …..(ii)
And 
= 
Again 
…..(iii)
From eq. (i), (ii) and (iii) A. (adj. A) = (adj. A). A = 
4.
= 
adj. A =∣∣∣∣032−11180−13∣∣∣∣′=|0−11031−1283|′
=∣∣∣∣0−11031−1283∣∣∣∣=|032−11180−13|
A. (adj. A) = 
= 
= 
……….(i)
Again (adj. A). A =
= 
= ……….(ii)
And
= 
Also 
= ……….(iii)
From eq. (i), (ii) and (iii) A. (adj. A) = (adj. A). A =
Find the inverse of the matrix (if it exists) given in Exercise 5 to 11.
5. 
= 6-(-8) = 6+8 = 14≠014≠0
Matrix A is non-singular and hence 
exist.
Now adj. A = 
And 
6.
= 
Matrix A is non-singular and hence exist.
Now adj. A = 
And 
7.
= 
exists.
A11 = 
, A12 = 
,
A13 = 
, A21 = 
,
A22 = 
, A23 = 
,
A31 = 
, A32 = 
,
A33 = 
adj. A =∣∣∣∣10−10205−4002∣∣∣∣′=|1000−10502−42|′
=∣∣∣∣1000−10502−42∣∣∣∣=|10−10205−4002|
8.
= 
exists.
A11 = 
, A12 = 
,
A13 = 
, A21 = 
,
A22 = 
, A23 = 
,
A31 = 
, A32 = 
,
A33 = 
adj. A = 
9.
= 
exists.
A11 = 
, A12 = 
,
A13 = 
, A21 = 
,
A22 = 
, A23 = 
,
A31 = 
, A32 = 
,
A33 = 
adj. A = 
10.
= 
exists.
A11 = 
, A12 = 
,
A13 = 
, A21 = 
,
A22 = 
, A23 = 
,
A31 = 
, A32 = 
,
A33 = 
adj. A = 
11.
= 
exists.
A11 = 
,
A12 = 
, A13 = 
,
A21 = 
, A22 = 
,
A23 = 
, A31 = 
,
A32 = 
, A33 = 
adj. A = 
12. Let A = and B = verify that
verify that 
= 15 – 14 = 1 
0
= 
Matrix B = 
= 54 – 56 = 
0
Now AB = = 
= 
= 
Now L.H.S. = 
……….(i)
R.H.S. = 
= 
= 
……….(ii)
From eq. (i) and (ii), we get
L.H.S. = R.H.S.
13. If A = , show that A2 – 5A + 7I = 0. Hence find
, show that A2 – 5A + 7I = 0. Hence find 
L.H.S. = 
= 
= 
= 
= 
= 
= R.H.S.
……(i)
To find: , multiplying eq. (i) by .
= 
= 
= 
14. For the matrix A = find numbers a and b such that
find numbers a and b such that 
We have 
……….(i)
……………..(ii)
Here satisfies 
also, therefore
Putting in eq. (i), 
Here also satisfies 
, therefore
Therefore, and
15. For the matrix A = , show that Hence find
, show that 
Hence find
= 
Now 
= 
= 
L.H.S. = 
= 
= 
= 
= 
= 
= R.H.S.
Now, to find , multiplying 
by
=> 11A-1 = 
16. If A = , verify that and hence find
, verify that 
and hence find
= 
Now 
= 
= 
L.H.S. = 
= 
= 
= 
= 
= = R.H.S.
Now, to find , multiplying by
= 
17. Let A be a non-singular matrix of order 3 x 3. Then is equal to:
is equal to:
(A) 
(B) 
(C) 
(D) 
then 
Putting 
Therefore, option (B) is correct.
18. If A is an invertible matrix of order 2, then det is equal to:
is equal to:
(A) det A
(B) 
(C) 1
(D) 0
Therefore, option (B) is correct.