Online Test — Determinants
10 Questions • 20 min • Chapter MCQ
20:00
Question 1 of 10
easy
If $A$ is a $3\times3$ matrix and $|A|=4$, then $|2A|$ is:
8
16
32
4
Explanation: $|kA|=k^n|A|$. Here $k=2$, $n=3$: $|2A|=2^3 \cdot 4=8\cdot4=32$.
Question 2 of 10
easy
The value of $\begin{vmatrix}1 & 2 \\ 3 & 4\end{vmatrix}$ is:
10
-2
2
-10
Explanation: $1\cdot4-2\cdot3=4-6=-2$.
Question 3 of 10
easy
If two rows of a determinant are interchanged, the value:
Doubles
Remains unchanged
Changes sign
Becomes zero
Explanation: Interchanging any two rows (or columns) of a determinant changes its sign.
Question 4 of 10
medium
The cofactor of element $a_{12}$ in $\begin{pmatrix}1&2&3\\0&5&6\\0&0&7\end{pmatrix}$ is:
$-12$
$0$
$12$
$-6$
Explanation: $A_{12}=(-1)^{1+2}M_{12}=-M_{12}$. $M_{12}=\begin{vmatrix}0&6\\0&7\end{vmatrix}=0$. So $A_{12}=0$.
Question 5 of 10
medium
If $|A|=6$ for a $3\times3$ matrix, then $|$adj$A|$ is:
6
36
216
1/6
Explanation: $|\text{adj}A|=|A|^{n-1}=6^{3-1}=6^2=36$.
Question 6 of 10
medium
For the system $AX=B$ with $|A|=0$ and (adj$A)B=O$, the system has:
Unique solution
No solution
Infinitely many solutions
Cannot be determined
Explanation: When $|A|=0$ and (adj$A)B=O$, the system is consistent with infinitely many solutions.
Question 7 of 10
easy
The area of a triangle with vertices $(0,0)$, $(1,0)$, $(0,1)$ using the determinant formula is:
$\frac{1}{4}$
$\frac{1}{2}$
$1$
$2$
Explanation: Area $=\frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|=\frac{1}{2}|0(0-1)+1(1-0)+0(0-0)|=\frac{1}{2}$.
Question 8 of 10
medium
If $A$ is a non-singular matrix, then $A(\text{adj}A)$ equals:
$|A| \cdot A$
$|A| \cdot I$
$I$
adj$A \cdot A$
Explanation: $A(\text{adj}A)=(\text{adj}A)A=|A|I$ — this is the fundamental adjoint identity.
Question 9 of 10
hard
$\begin{vmatrix}a+b & b+c & c+a \\ b+c & c+a & a+b \\ c+a & a+b & b+c\end{vmatrix}=$
$0$
$a+b+c$
$2(a+b+c)$
$a^2+b^2+c^2$
Explanation: Adding all three rows: $R_1+R_2+R_3=2(a+b+c)[1,1,1]$. Then $R_1\to R_1-R_3$, $R_2\to R_2-R_3$ creates two rows proportional — det$=0$.
Question 10 of 10
medium
If $A$ is a $2\times2$ matrix with $|A|=3$ and $|B|=4$, then $|AB^{-1}|=$
$3/4$
$4/3$
$12$
$1$
Explanation: $|AB^{-1}|=|A||B^{-1}|=|A|/|B|=3/4$.