Mathematics Questions and Answers – Determinants – Minors and Cofactors

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This set of Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Determinants – Minors and Cofactors”.

1. Which of the following is the formula for cofactor of an element aij ?
a) Aij=(1)i+j Mij
b) Aij=(-2)i+j Mij
c) Aij=(-1)i+j Mij
d) Aij=(-1)i-j Mij
View Answer

Answer: c
Explanation: The cofactor of an element aij, denoted by Aij is given by
Aij=(-1)i+j Mij, where Mij is the minor of the element aij.
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2. What is the minor of the element 5 in the determinant Δ=\(\begin{vmatrix}1&5&4\\2&3&6\\7&9&4\end{vmatrix}\)?
a) -34
b) 34
c) -17
d) 21
View Answer

Answer: a
Explanation: The minor of element 5 in the determinant Δ=\(\begin{vmatrix}1&5&4\\2&3&6\\7&9&4\end{vmatrix}\) is the determinant obtained by deleting the row and column containing element 5.
∴M12=\(\begin{vmatrix}2&6\\7&4\end{vmatrix}\)=2(4)-7(6)=-34.

3. Find the minor and cofactor respectively for the element 3 in the determinant Δ=\(\begin{vmatrix}1&5\\3&6\end{vmatrix}\).
a) M21=-5, A21=-5
b) M21=5, A21=-5
c) M21=-5, A21=5
d) M21=5, A21=5
View Answer

Answer: b
Explanation: The element 3 is in the second row (i=2) and first column(j=1).
∴M21=5 (obtained by deleting R2 and C1 in Δ)
A21=(-1)1+2 M21=-1×5 =-5.
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4. Find the minor of the element 1 in the determinant Δ=\(\begin{vmatrix}1&5\\3&8\end{vmatrix}\).
a) 5
b) 1
c) 8
d) 3
View Answer

Answer: c
Explanation: The minor of the element 1 can be obtained by deleting the first row and the first column
∴M11=8.

5. Find the cofactor of element -3 in the determinant Δ=\(\begin{vmatrix}1&4&4\\-3&5&9\\2&1&2\end{vmatrix}\).
a) -4
b) 4
c) -5
d) -3
View Answer

Answer: a
Explanation: The minor of element -3 is given by
M21=\(\begin{vmatrix}4&4\\1&2\end{vmatrix}\)=4(2)-4=4 (Obtained by eliminating R2 and C1)
∴A21=(-1)2+1 M21=(-1)3 4=-4.
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6. If Δ=\(\begin{vmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33} \end{vmatrix}\), then the determinant in terms of cofactors Aij can be expressed as a11 A11+a21 A21+a31 A31.
a) True
b) False
View Answer

Answer: a
Explanation: The given statement is true.
Expanding the determinant Δ=\(\begin{vmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33} \end{vmatrix}\) along R1, we get
Δ=(-1)1+1 a11 \(\begin{vmatrix}a_{22}&a_{23}\\a_{32}&a_{33} \end{vmatrix}\)+(-1)1+2 a12 \(\begin{vmatrix}a_{21}&a_{23}\\a_{31}&a_{33} \end{vmatrix}\)+(-1)1+3 a13 \(\begin{vmatrix}a_{21}&a_{22}\\a_{31}&a_{32} \end{vmatrix}\)
Δ=a11 A11+a21 A21+a31 A31, where Aij is the cofactor of aij.

7. Find the minor of the element 2 in the determinant Δ=\(\begin{vmatrix}1&9\\2&3\end{vmatrix}\)?
a) 3
b) 9
c) 1
d) 2
View Answer

Answer: b
Explanation: The minor of the element 2 can be obtained by deleting the first row and the first column
∴M11=9.
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8. For which of the elements in the determinant Δ=\(\begin{vmatrix}1&8&-6\\2&-3&4\\-7&9&5\end{vmatrix}\) the cofactor is -37.
a) 4
b) 1
c) -6
d) -3
View Answer

Answer: d
Explanation: Consider the element -3 in Δ=\(\begin{vmatrix}1&8&-6\\2&-3&4\\-7&9&5\end{vmatrix}\)
The cofactor of the element -3 is given by
A22=(-1)2+2 M22
M22=\(\begin{vmatrix}1&-6\\-7&5\end{vmatrix}\)=1(5)-(-6)(-7)=5-42=-37
A22=(-1)2+2 (-37)=-37.

9. For which of the following elements in the determinant Δ=\(\begin{vmatrix}2&8\\4&7\end{vmatrix}\), the minor of the element is 2?
a) 2
b) 7
c) 4
d) 8
View Answer

Answer: b
Explanation: Consider the element 7 in the determinant Δ=\(\begin{vmatrix}2&8\\4&7\end{vmatrix}\)
The minor of the element 7 can be obtained by deleting R2 and C2
∴M22=2
Hence, the minor of the element 7 is 2.
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10. For which of the following element in the determinant Δ=\(\begin{vmatrix}5&-5&8\\6&2&-1\\5&-6&8\end{vmatrix}\) , the minor and the cofactor both are zero.
a) -5
b) 2
c) -6
d) 8
View Answer

Answer: b
Explanation: Consider the element 2 in the determinant Δ=\(\begin{vmatrix}5&-5&8\\6&2&-1\\5&-6&8\end{vmatrix}\)
The minor of the element 2 is given by
∴M22=\(\begin{vmatrix}5&8\\5&8\end{vmatrix}\)=40-40=0
⇒A22=(-1)2+2 (0)=0.

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

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Manish Bhojasia - Founder & CTO at Sanfoundry
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn | Youtube | Instagram | Facebook | Twitter