Mathematics Problems on “Determinants – Adjoint and Inverse of a Matrix”.
1. Which of the following is the adjoint of the matrix A=(begin{bmatrix}1&5\3&4end{bmatrix})?
a) (begin{bmatrix}4&-5\-3&-1end{bmatrix})
b) (begin{bmatrix}-4&5\-3&1end{bmatrix})
c) (begin{bmatrix}4&-5\-3&1end{bmatrix})
d) (begin{bmatrix}4&5\-3&1end{bmatrix})
Answer: c
Clarification: We have A11=(-1)1+1 4=4
A12=(-1)1+2 3=-3
A21=(1)2+1 5=-5
A22=(-1)2+2 1=1
∴adj A=(begin{bmatrix}A_{11}&A_{21}\A_{12}&A_{22}end{bmatrix})=(begin{bmatrix}4&-5\-3&1end{bmatrix}).
2. If A=(begin{bmatrix}5&-8\2&6end{bmatrix}), find A(adj A).
a) (begin{bmatrix}41&0\0&46end{bmatrix})
b) (begin{bmatrix}46&0\1&46end{bmatrix})
c) (begin{bmatrix}46&1\0&46end{bmatrix})
d) (begin{bmatrix}46&0\0&46end{bmatrix})
Answer: d
Clarification: Given that, A=(begin{bmatrix}5&-8\2&6end{bmatrix})
∴adj A=(begin{bmatrix}6&8\-2&5end{bmatrix})
A(adj A)=(begin{bmatrix}5&-8\2&6end{bmatrix}begin{bmatrix}6&8\-2&5end{bmatrix})
=(begin{bmatrix}5×6+(-8)×(-2)&5×8+5×(-8)\2×6+6×(-2)&2×8+6×5end{bmatrix})=(begin{bmatrix}46&0\0&46end{bmatrix}).
3. If A=(begin{bmatrix}1&0\9&4end{bmatrix}), then (adj A)A is ______________
a) (begin{bmatrix}-4&0\0&-4end{bmatrix})
b) (begin{bmatrix}4&0\1&4end{bmatrix})
c) (begin{bmatrix}4&0\0&4end{bmatrix})
d) (begin{bmatrix}4&0\0&-4end{bmatrix})
Answer: c
Clarification: Given that, A=(begin{bmatrix}1&0\9&4end{bmatrix})
We know that, A(adj A)=(adj A)A=|A|I
∴|A|=4-0=4
⇒A(adj A)=|A|I=(begin{bmatrix}4&0\0&4end{bmatrix}).
4. Which of the following is the formula for calculating the inverse of the matrix?
a) (frac{2}{|A|}) adj A
b) (frac{1}{|A|}) adj A
c) (frac{-1}{|A|}) adj A
d) (frac{1}{|2A|}) adj A
Answer: b
Clarification: The formula for calculating the inverse of the matrix is given by
A-1=(frac{1}{|A|}) adj A, where |A| is the determinant of the matrix and adj A is the adjoint of the matrix.
5. Find the inverse of the matrix A=(begin{bmatrix}8&5\4&1end{bmatrix}).
a) (begin{bmatrix}-frac{1}{12}&frac{5}{12}\frac{1}{3}&-frac{2}{3}end{bmatrix})
b) (begin{bmatrix}frac{1}{12}&frac{5}{12}\frac{1}{3}&-frac{2}{3}end{bmatrix})
c) (begin{bmatrix}-frac{1}{12}&frac{5}{12}\frac{1}{3}&frac{2}{3}end{bmatrix})
d) (begin{bmatrix}-frac{1}{12}&frac{5}{12}\-frac{1}{3}&-frac{2}{3}end{bmatrix})
Answer: a
Clarification: Give that, A=(begin{bmatrix}8&5\4&1end{bmatrix})
adj A=(begin{bmatrix}1&-5\-4&8end{bmatrix})
|A|=8×1-(-5)×(-4)=8-20=-12
A-1=(frac{1}{|A|}) adj A=(frac{1}{-12} begin{bmatrix}1&-5\-4&8end{bmatrix})=(begin{bmatrix}-frac{1}{12}&frac{5}{12}\frac{1}{3}&-frac{2}{3}end{bmatrix}).
6. Which of the below condition is incorrect for the inverse of a matrix A?
a) The matrix A must be a square matrix
b) A must be singular matrix
c) A must be a non-singular matrix
d) adj A≠0
Answer: b
Clarification: The matrix should not be a singular matrix. A square matrix is said to be singular |A|=0.
We know that, A-1=(frac{1}{|A|}) adj A,
Hence, if |A|=0 the inverse of the matrix does not exist.
7. Which of the below given matrices has the inverse (frac{1}{-6}begin{bmatrix}2&1\0&-3end{bmatrix})?
a) (begin{bmatrix}3&-1\0&2end{bmatrix})
b) (begin{bmatrix}-3&-1\0&2end{bmatrix})
c) (begin{bmatrix}-2&0\1&3end{bmatrix})
d) (begin{bmatrix}-3&-1\0&-2end{bmatrix})
Answer: b
Clarification: Consider the matrix (begin{bmatrix}-3&-1\0&2end{bmatrix})
adj A=(begin{bmatrix}2&1\0&-3end{bmatrix})
|A|=-6
∴A-1=(frac{1}{|A|}) adj A=(frac{1}{-6}begin{bmatrix}2&1\0&-3end{bmatrix}).
8. If A=(begin{bmatrix}-8&2\6&-3end{bmatrix}) and B=(begin{bmatrix}2&1\1&7end{bmatrix}). Find (AB)-1.
a) –(frac{1}{432}) (begin{bmatrix}-27&6\9&14end{bmatrix})
b) (frac{1}{432}) (begin{bmatrix}27&6\9&14end{bmatrix})
c) (frac{1}{432}) (begin{bmatrix}-27&6\9&14end{bmatrix})
d) (frac{-1}{432}) (begin{bmatrix}27&6\9&14end{bmatrix})
Answer: c
Clarification: Given that, A=(begin{bmatrix}-8&2\6&-3end{bmatrix}) and B=(begin{bmatrix}2&1\1&7end{bmatrix})
∴AB=(begin{bmatrix}-8×2+2×1&-8×1+2×7\6×2+(-3)×1&6×1+(-3)×7end{bmatrix})=(begin{bmatrix}-14&6\9&27end{bmatrix})
adj(AB)=(begin{bmatrix}27&-6\-9&-14end{bmatrix})
|AB|=27×(-14)-(-9)×(-6)=-378-54=-432
(AB)-1=(frac{1}{|AB|}) adj AB=(frac{1}{-432} begin{bmatrix}27&-6\-9&-14end{bmatrix})=(frac{1}{432} begin{bmatrix}-27&6\9&14end{bmatrix}).
9. Which of the following formula is incorrect?
a) A(adj A)=|A|I
b) |adj (A)|=|A|n-1, for an nth order matrix
c) A-1=(frac{1}{|A|}) adj A
d) A(adj A)=|A|n-1
Answer: d
Clarification: The formula A(adj A)=|A|n-1 is incorrect. The correct formula is A(adj A)=(adjA)A=|A|I.
10. A square matrix A is said to be non-singular if |A|≠0.
a) True
b) False
Answer: a
Clarification: The given statement is true. A square matrix A is said to be singular if |A|=0 and non-singular if A≠0.
Mathematics Problems,