Mathematics Questions and Answers – Determinants – Adjoint and Inverse of a Matrix

«
»

This set of Mathematics Problems focuses on “Determinants – Adjoint and Inverse of a Matrix”.

1. Which of the following is the adjoint of the matrix A=\(\begin{bmatrix}1&5\\3&4\end{bmatrix}\)?
a) \(\begin{bmatrix}4&-5\\-3&-1\end{bmatrix}\)
b) \(\begin{bmatrix}-4&5\\-3&1\end{bmatrix}\)
c) \(\begin{bmatrix}4&-5\\-3&1\end{bmatrix}\)
d) \(\begin{bmatrix}4&5\\-3&1\end{bmatrix}\)
View Answer

Answer: c
Explanation: 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=\(\begin{bmatrix}A_{11}&A_{21}\\A_{12}&A_{22}\end{bmatrix}\)=\(\begin{bmatrix}4&-5\\-3&1\end{bmatrix}\).
advertisement

2. If A=\(\begin{bmatrix}5&-8\\2&6\end{bmatrix}\), find A(adj A).
a) \(\begin{bmatrix}41&0\\0&46\end{bmatrix}\)
b) \(\begin{bmatrix}46&0\\1&46\end{bmatrix}\)
c) \(\begin{bmatrix}46&1\\0&46\end{bmatrix}\)
d) \(\begin{bmatrix}46&0\\0&46\end{bmatrix}\)
View Answer

Answer: d
Explanation: Given that, A=\(\begin{bmatrix}5&-8\\2&6\end{bmatrix}\)
∴adj A=\(\begin{bmatrix}6&8\\-2&5\end{bmatrix}\)
A(adj A)=\(\begin{bmatrix}5&-8\\2&6\end{bmatrix}\begin{bmatrix}6&8\\-2&5\end{bmatrix}\)
=\(\begin{bmatrix}5×6+(-8)×(-2)&5×8+5×(-8)\\2×6+6×(-2)&2×8+6×5\end{bmatrix}\)=\(\begin{bmatrix}46&0\\0&46\end{bmatrix}\).

3. If A=\(\begin{bmatrix}1&0\\9&4\end{bmatrix}\), then (adj A)A is ______________
a) \(\begin{bmatrix}-4&0\\0&-4\end{bmatrix}\)
b) \(\begin{bmatrix}4&0\\1&4\end{bmatrix}\)
c) \(\begin{bmatrix}4&0\\0&4\end{bmatrix}\)
d) \(\begin{bmatrix}4&0\\0&-4\end{bmatrix}\)
View Answer

Answer: c
Explanation: Given that, A=\(\begin{bmatrix}1&0\\9&4\end{bmatrix}\)
We know that, A(adj A)=(adj A)A=|A|I
∴|A|=4-0=4
⇒A(adj A)=|A|I=\(\begin{bmatrix}4&0\\0&4\end{bmatrix}\).
advertisement
advertisement

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
View Answer

Answer: b
Explanation: 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=\(\begin{bmatrix}8&5\\4&1\end{bmatrix}\).
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) \(\begin{bmatrix}-\frac{1}{12}&\frac{5}{12}\\-\frac{1}{3}&-\frac{2}{3}\end{bmatrix}\)
View Answer

Answer: a
Explanation: Give that, A=\(\begin{bmatrix}8&5\\4&1\end{bmatrix}\)
adj A=\(\begin{bmatrix}1&-5\\-4&8\end{bmatrix}\)
|A|=8×1-(-5)×(-4)=8-20=-12
A-1=\(\frac{1}{|A|}\) adj A=\(\frac{1}{-12} \begin{bmatrix}1&-5\\-4&8\end{bmatrix}\)=\(\begin{bmatrix}-\frac{1}{12}&\frac{5}{12}\\\frac{1}{3}&-\frac{2}{3}\end{bmatrix}\).
advertisement

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
View Answer

Answer: b
Explanation: 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 \(\frac{1}{-6}\begin{bmatrix}2&1\\0&-3\end{bmatrix}\)?
a) \(\begin{bmatrix}3&-1\\0&2\end{bmatrix}\)
b) \(\begin{bmatrix}-3&-1\\0&2\end{bmatrix}\)
c) \(\begin{bmatrix}-2&0\\1&3\end{bmatrix}\)
d) \(\begin{bmatrix}-3&-1\\0&-2\end{bmatrix}\)
View Answer

Answer: b
Explanation: Consider the matrix \(\begin{bmatrix}-3&-1\\0&2\end{bmatrix}\)
adj A=\(\begin{bmatrix}2&1\\0&-3\end{bmatrix}\)
|A|=-6
∴A-1=\(\frac{1}{|A|}\) adj A=\(\frac{1}{-6}\begin{bmatrix}2&1\\0&-3\end{bmatrix}\).
advertisement

8. If A=\(\begin{bmatrix}-8&2\\6&-3\end{bmatrix}\) and B=\(\begin{bmatrix}2&1\\1&7\end{bmatrix}\). Find (AB)-1.
a) –\(\frac{1}{432}\) \(\begin{bmatrix}-27&6\\9&14\end{bmatrix}\)
b) \(\frac{1}{432}\) \(\begin{bmatrix}27&6\\9&14\end{bmatrix}\)
c) \(\frac{1}{432}\) \(\begin{bmatrix}-27&6\\9&14\end{bmatrix}\)
d) \(\frac{-1}{432}\) \(\begin{bmatrix}27&6\\9&14\end{bmatrix}\)
View Answer

Answer: c
Explanation: Given that, A=\(\begin{bmatrix}-8&2\\6&-3\end{bmatrix}\) and B=\(\begin{bmatrix}2&1\\1&7\end{bmatrix}\)
∴AB=\(\begin{bmatrix}-8×2+2×1&-8×1+2×7\\6×2+(-3)×1&6×1+(-3)×7\end{bmatrix}\)=\(\begin{bmatrix}-14&6\\9&27\end{bmatrix}\)
adj(AB)=\(\begin{bmatrix}27&-6\\-9&-14\end{bmatrix}\)
|AB|=27×(-14)-(-9)×(-6)=-378-54=-432
(AB)-1=\(\frac{1}{|AB|}\) adj AB=\(\frac{1}{-432} \begin{bmatrix}27&-6\\-9&-14\end{bmatrix}\)=\(\frac{1}{432} \begin{bmatrix}-27&6\\9&14\end{bmatrix}\).

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
View Answer

Answer: d
Explanation: The formula A(adj A)=|A|n-1 is incorrect. The correct formula is A(adj A)=(adjA)A=|A|I.
advertisement

10. A square matrix A is said to be non-singular if |A|≠0.
a) True
b) False
View Answer

Answer: a
Explanation: The given statement is true. A square matrix A is said to be singular if |A|=0 and non-singular if A≠0.

Sanfoundry Global Education & Learning Series – Mathematics – Class 12.

To practice all areas of Mathematics Problems, here is complete set of 1000+ Multiple Choice Questions and Answers.

advertisement
advertisement
Subscribe to our Newsletters (Subject-wise). Participate in the Sanfoundry Certification contest to get free Certificate of Merit. Join our social networks below and stay updated with latest contests, videos, internships and jobs!

Youtube | Telegram | LinkedIn | Instagram | Facebook | Twitter | Pinterest
Manish Bhojasia - Founder & CTO at Sanfoundry
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

Subscribe to his free Masterclasses at Youtube & technical discussions at Telegram SanfoundryClasses.