250+ TOP MCQs on Invertible Matrices | Class 12 Maths

Mathematics Multiple Choice Questions on “Invertible Matrices”.

1. Which among the following is inverse of the matrix A=(begin{bmatrix}2&3\5&1end{bmatrix}) ?
a) (begin{bmatrix}frac{1}{13}&frac{3}{13}\ frac{5}{13}&frac{-2}{13}end{bmatrix})
b) (begin{bmatrix}frac{-1}{13}&frac{3}{13}\ frac{5}{13}&frac{-2}{13}end{bmatrix})
c) (begin{bmatrix}frac{-1}{13}&frac{3}{13}\1&frac{-2}{13}end{bmatrix})
d) (begin{bmatrix}frac{-1}{13}&frac{3}{13}\ frac{5}{13}&-2end{bmatrix})
Answer: b
Clarification: Consider the matrix A=(begin{bmatrix}2&3\5&1end{bmatrix})
Using elementary row operation, we write A=IA.
(begin{bmatrix}2&3\5&1end{bmatrix})=(begin{bmatrix}1&0\0&1end{bmatrix})A
(begin{bmatrix}-13&0\5&1end{bmatrix})=(begin{bmatrix}1&-3\0&1end{bmatrix})A   (Applying R₁→R₁-3R₂)
(begin{bmatrix}1&0\5&1end{bmatrix})=(begin{bmatrix}frac{-1}{13}&frac{3}{13}\0&1end{bmatrix})A   (Applying (R_1 rightarrow -frac{R_1}{13}))
(begin{bmatrix}1&0\0&1end{bmatrix})=(begin{bmatrix}frac{-1}{13}&frac{3}{13}\ frac{5}{13}&frac{-2}{13}end{bmatrix})A   (Applying R₂→R₂-5R₁)
(A^{-1}=begin{bmatrix}frac{-1}{13}&frac{3}{13}\ frac{5}{13}&frac{-2}{13}end{bmatrix}).

2. Which of the following matrices will not have an inverse?
a) (begin{bmatrix}2&4\-1&1end{bmatrix})
b) (begin{bmatrix}1&5&2\6&4&2\1&3&2end{bmatrix})
c) (begin{bmatrix}1&2\1&1end{bmatrix})
d) (begin{bmatrix}1&2&5\3&6&4end{bmatrix})
Answer: d
Clarification: The matrix A=(begin{bmatrix}1&2&5\3&6&4end{bmatrix}) will not have an inverse as it is a rectangular matrix. Rectangular matrix does not possess an inverse matrix.

3. If A and B are invertible matrices of the same order, then (AB)^-1=B^-1 A^-1.
a) True
b) False
Answer: a
Clarification: The given statement is true.
(AB) (AB)^-1=I (Using the formula AA^-1=I)
Multiplying both sides by A^-1, we get
A^-1 (AB) (AB)^-1=A^-1 I
(A^-1 A)B(AB)^-1=A^-1
IB(AB^-1)=A^-1
B(AB^-1)=A^-1
⇒B^-1 B(AB^-1)=B^-1 A^-1
(AB^-1)=B^-1 A^-1

4. A matrix A is invertible if it has all zeroes in one or more rows on L.H.S.
a) True
b) False
Answer: b
Clarification: The given statement is false. A matrix is non-invertible if it has all zeroes in one or more rows on L.H.S. This is because after applying all the elementary operations on the matrix, we should get an identity matrix on the L.H.S. to obtain an inverse of the given matrix, which is not possible if we obtain all zeroes in one or more rows.

5. The inverse of the matrix A=(begin{bmatrix}1&2&4\5&2&4\3&6&2end{bmatrix}) is
a) (begin{bmatrix}frac{-1}{4}&frac{1}{4}&0\ frac{1}{40}&frac{-1}{8}&frac{1}{5}\ frac{3}{40}&1&frac{-1}{10}end{bmatrix})
b) (begin{bmatrix}frac{-1}{4}&frac{1}{4}&1\ frac{1}{40}&frac{-1}{8}&frac{1}{5}\ frac{3}{40}&0&frac{-1}{10}end{bmatrix})
c) (begin{bmatrix}frac{-1}{4}&frac{1}{4}&0\ frac{1}{40}&frac{-1}{8}&frac{1}{5}\ frac{3}{40}&0&frac{-1}{10}end{bmatrix})
d) (begin{bmatrix}frac{-1}{4}&-frac{1}{4}&0\ frac{1}{40}&frac{1}{8}&frac{-1}{5}\ frac{3}{40}&0&frac{-1}{10}end{bmatrix})
Answer: c
Clarification: Consider the matrix A=(begin{bmatrix}1&2&4\5&2&4\3&6&2end{bmatrix})
Using the elementary row operation, we write A=IA
(begin{bmatrix}1&2&4\5&2&4\3&6&2end{bmatrix})=(begin{bmatrix}1&0&0\0&1&0\0&0&1end{bmatrix})A
Applying R₁→R₁-R₂
(R_1 rightarrow frac{R_1}{-4})
(begin{bmatrix}1&0&0\5&2&4\3&6&2end{bmatrix})=(begin{bmatrix}frac{-1}{4}&frac{1}{4}&0\0&1&0\0&0&1end{bmatrix})A
Applying R₂→R₂-5R₁ and R₃→R₃-3R₁
(begin{bmatrix}1&0&0\0&2&4\0&6&2end{bmatrix})=(begin{bmatrix}frac{-1}{4}&frac{1}{4}&0\frac{5}{4}&frac{-1}{4}&0\frac{3}{4}&frac{-3}{4}&1end{bmatrix})A
Applying R₂→R₂-2R₃ and (R_2 rightarrow frac{R_2}{-10})
(begin{bmatrix}1&0&0\0&1&0\0&6&2end{bmatrix})=(begin{bmatrix}frac{-1}{4}&frac{1}{4}&0\ frac{1}{40}&frac{-5}{40}&frac{1}{5}\frac{3}{4}&frac{-3}{4}&1end{bmatrix})A
Applying R₃→R₃-6R₂ and (R_2 rightarrow frac{R_2}{2})
(begin{bmatrix}1&0&0\0&1&0\0&0&1end{bmatrix})=(begin{bmatrix}frac{-1}{4}&frac{1}{4}&0\ frac{1}{40}&frac{-1}{8}&frac{1}{5}\ frac{3}{40}&0&frac{-1}{10}end{bmatrix})A
A^-1=(begin{bmatrix}frac{-1}{4}&frac{1}{4}&0\ frac{1}{40}&frac{-1}{8}&frac{1}{5}\ frac{3}{40}&0&frac{-1}{10}end{bmatrix}).

6. Which of the following is the inverse of the matrix A=(begin{bmatrix}8&1\1&2end{bmatrix})?
a) (begin{bmatrix}frac{2}{15}&-frac{1}{15}\frac{1}{15}&frac{8}{15}end{bmatrix})
b) (begin{bmatrix}frac{1}{15}&-frac{1}{15}\-frac{1}{15}&frac{1}{15}end{bmatrix})
c) (begin{bmatrix}frac{2}{15}&-frac{1}{15}\-frac{1}{15}&frac{8}{15}end{bmatrix})
d) (begin{bmatrix}frac{2}{15}&frac{1}{15}\frac{1}{15}&frac{4}{15}end{bmatrix})
Answer: c
Clarification: Consider the matrix A=(begin{bmatrix}8&1\1&2end{bmatrix})
Using the elementary row operation, we write A=IA
Applying R₂→8R₂-R₁ and R₂→R₂/15, we get
(begin{bmatrix}8&1\0&1end{bmatrix})=(begin{bmatrix}1&0\-frac{1}{15}&frac{8}{15}end{bmatrix})A
Applying R₁→R₁-R₂ and R₁→R₁/8, we get
(begin{bmatrix}1&0\0&1end{bmatrix})=(begin{bmatrix}frac{2}{15}&-frac{1}{15}\-frac{1}{15}&frac{8}{15}end{bmatrix})A
A^-1=(begin{bmatrix}frac{2}{15}&-frac{1}{15}\-frac{1}{15}&frac{8}{15}end{bmatrix}).

7. Which among the below matrices has the inverse A^-1=(begin{bmatrix}1&-frac{5}{8}\0&frac{1}{8}end{bmatrix})
a) (begin{bmatrix}1&5\0&8end{bmatrix})
b) (begin{bmatrix}1&5\-1&8end{bmatrix})
c) (begin{bmatrix}1&5\0&16end{bmatrix})
d) (begin{bmatrix}1&8\0&8end{bmatrix})
Answer: a
Clarification: Consider the matrix A=(begin{bmatrix}1&5\0&8end{bmatrix})
Using the elementary column operations, we write A=AI
(begin{bmatrix}1&5\0&8end{bmatrix})=A(begin{bmatrix}1&0\0&1end{bmatrix})
Applying C₂→C₂-5C₁
(begin{bmatrix}1&0\0&8end{bmatrix})=A(begin{bmatrix}1&-5\0&1end{bmatrix})
Applying C₂→C₂/8
(begin{bmatrix}1&0\0&1end{bmatrix})=A(begin{bmatrix}1&-frac{5}{8}\0&frac{1}{8}end{bmatrix})
A^-1=(begin{bmatrix}1&-frac{5}{8}\0&frac{1}{8}end{bmatrix}).

8. Find the inverse of matrix A=(begin{bmatrix}5&1&3\4&2&6\5&4&2end{bmatrix})
a) (begin{bmatrix}0&-frac{1}{6}&0\-frac{11}{30}&frac{1}{12}&frac{3}{10}\-frac{1}{10}&frac{1}{4}&-frac{1}{10}end{bmatrix})
b) (begin{bmatrix}frac{1}{3}&-frac{1}{6}&0\-frac{11}{30}&frac{1}{12}&frac{3}{10}\-frac{1}{10}&frac{1}{4}&-frac{1}{10}end{bmatrix})
c) (begin{bmatrix}frac{1}{3}&-frac{1}{6}&0\-frac{11}{30}&1&frac{3}{10}\-frac{1}{10}&frac{1}{4}&-frac{1}{10}end{bmatrix})
d) (begin{bmatrix}frac{1}{3}&frac{1}{6}&0\frac{11}{30}&frac{1}{12}&frac{3}{10}\-frac{1}{10}&frac{1}{4}&frac{1}{10}end{bmatrix})
Answer: b
Clarification: Consider the matrix A=(begin{bmatrix}5&1&3\4&2&6\5&4&2end{bmatrix})
Using the elementary row operations, we write A=IA
(begin{bmatrix}5&1&3\4&2&6\5&4&2end{bmatrix})=(begin{bmatrix}1&0&0\0&1&0\0&0&1end{bmatrix})A
Applying R₂→5R₂-4R₁ and R₃→R₃-R₁
(begin{bmatrix}5&1&3\0&6&18\0&3&-1end{bmatrix})=(begin{bmatrix}1&0&0\-4&5&0\-1&0&1end{bmatrix})A
Applying R₁→R₂-6R₁ and R₃→R₂-2R₃
(begin{bmatrix}-30&0&0\0&6&18\0&0&20end{bmatrix})=(begin{bmatrix}-10&5&0\-4&5&0\-2&5&-2end{bmatrix})A
Applying R₂→20R₂-18R₃
(begin{bmatrix}-30&0&0\0&120&0\0&0&20end{bmatrix})=(begin{bmatrix}-10&5&0\-44&10&36\-2&5&-2end{bmatrix})A
Applying R₁→R₁/(-30), R₂→R₂/120, R₃→R₃/20
(begin{bmatrix}1&0&0\0&1&0\0&0&1end{bmatrix})=(begin{bmatrix}frac{1}{3}&-frac{1}{6}&0\-frac{11}{30}&frac{1}{12}&frac{3}{10}\-frac{1}{10}&frac{1}{4}&-frac{1}{10}end{bmatrix})A
A^-1=(begin{bmatrix}frac{1}{3}&-frac{1}{6}&0\-frac{11}{30}&frac{1}{12}&frac{3}{10}\-frac{1}{10}&frac{1}{4}&-frac{1}{10}end{bmatrix}).

9. Which of the following is not a property of invertible matrices if A and B are matrices of the same order?
a) (AB)^-1=A^-1 B^-1
b) (AA^-1)=(A^-1 A)=I
c) (AB)^-1=B^-1 A^-1
d) AB=BA=I
Answer: a
Clarification: (AB)^-1=A^-1 B^-1 is incorrect. The correct formula is (AB)^-1=B^-1 A^-1. B^-1 A^-1 ≠ A^-1 B^-1 as matrix multiplication is not commutative.

10. Find the inverse of A=(begin{bmatrix}5&3\4&1end{bmatrix}).
a) (begin{bmatrix}-frac{1}{7}&frac{3}{7}\frac{4}{7}&-frac{5}{7}end{bmatrix})
b) (begin{bmatrix}-frac{1}{7}&frac{3}{7}\frac{4}{7}&frac{5}{7}end{bmatrix})
c) (begin{bmatrix}-frac{1}{7}&-frac{3}{7}\frac{4}{7}&-frac{5}{7}end{bmatrix})
d) (begin{bmatrix}0&frac{3}{7}\frac{4}{7}&frac{5}{7}end{bmatrix})
Answer: a
Clarification: Consider the matrix A=(begin{bmatrix}5&3\4&1end{bmatrix})
By using the elementary row operations, we write A=IA
(begin{bmatrix}5&3\4&1end{bmatrix})=(begin{bmatrix}1&0\0&1end{bmatrix})A
Applying R₁→R₁-3R₂ and R₁→R₁/(-7), we get
(begin{bmatrix}1&0\4&1end{bmatrix})=(begin{bmatrix}-frac{1}{7}&frac{3}{7}\0&1end{bmatrix})
Applying R₂→R₂-4R₁, we get
(begin{bmatrix}1&0\0&1end{bmatrix})=(begin{bmatrix}-frac{1}{7}&frac{3}{7}\frac{4}{7}&-frac{5}{7}end{bmatrix})A
⇒A^-1=(begin{bmatrix}-frac{1}{7}&frac{3}{7}\frac{4}{7}&-frac{5}{7}end{bmatrix}).