Mathematics Multiple Choice Questions on “Applications of Determinants and Matrices”.
1. Which of the following conditions holds true for a system of equations to be consistent?
a) It should have one or more solutions
b) It should have no solutions
c) It should have exactly one solution
d) It should have exactly two solutions
Answer: a
Clarification: If a given system of equations has one or more solutions then the system is said to be consistent.
2. A given systems of equations is said to be inconsistent if _________________
a) it has one or more solutions
b) it has infinitely many solutions
c) it has no solutions
d) it has exactly one solution
Answer: c
Clarification: If a given system of equations has no solutions, then the system is said to be inconsistent.
3. Find the value of x and y for the given system of equations.
3x+2y=6 5x+y=2
a) x=(frac{2}{7}), y=(frac{24}{7})
b) x=-(frac{2}{7}), y=-(frac{24}{7})
c) x=-(frac{2}{7}), y=(frac{24}{7})
d) x=(frac{2}{7}), y=-(frac{24}{7})
Answer: c
Clarification: By using the matrix method, the given equations can be expressed in the form of the equation AX=B, where
A=(begin{bmatrix}3&2\5&1end{bmatrix}), B=(begin{bmatrix}6\2end{bmatrix}), X=(begin{bmatrix}x\yend{bmatrix})
To find the value of x and y, we need to solve the matrix X
X=A-1 B
A-1=(frac{1}{|A|}) adj A=(frac{1}{-7}begin{bmatrix}1&-2\-5&3end{bmatrix})=(begin{bmatrix}-frac{1}{7}&frac{2}{7}\frac{5}{7}&-frac{3}{7}end{bmatrix})
X=A-1 B=(begin{bmatrix}-frac{1}{7}&frac{2}{7}\frac{5}{7}&-frac{3}{7}end{bmatrix}begin{bmatrix}6\2end{bmatrix})
=(begin{bmatrix}-frac{6}{7}+frac{4}{7}\frac{30}{7}-frac{6}{7}end{bmatrix})=(begin{bmatrix}-frac{2}{7}\frac{24}{7}end{bmatrix})
4. The given system of equation is inconsistent.
4x+2y=7 6x+3y=9
a) True
b) False
Answer: b
Clarification: The given statement is true. The given system of equations can be expressed in the form AX=B, where
A=(begin{bmatrix}4&2\6&3end{bmatrix}), B=(begin{bmatrix}6\9end{bmatrix}),X=(begin{bmatrix}x\yend{bmatrix})
|A|=4(3)-6(2)=12-12=0
∴A is a singular matrix.
If A is a singular matrix, then we calculate (adj A)B.
(adj A)B=(begin{bmatrix}3&-2\-6&4end{bmatrix}begin{bmatrix}7\9end{bmatrix})=(begin{bmatrix}3×7+(-2)×9\(-6)×7+4×9end{bmatrix})≠0
Thus, the solution does not exist for the given system of equations. Hence, it is inconsistent.
5. The given system of equations is consistent.
2x+3y=5 5x+y=3
a) True
b) False
Answer: a
Clarification: The given statement is true. The given system of equations can be expressed in the form AX=B, where
A=(begin{bmatrix}2&3\5&1end{bmatrix}), B=(begin{bmatrix}5\3end{bmatrix}), X=(begin{bmatrix}x\yend{bmatrix})
|A|=2(1)-5(3)=2-15=-13≠0
∴A is non-singular matrix.
⇒A-1 exists. Hence, the given system of equations is consistent.
6. Find the value of x, y, z for the given system of equations.
2x+3y+2z=50 x+4y+3z=40 3x+3y+5z=60
a) x=(frac{125}{8}), y=(frac{15}{2}), z=(frac{15}{8})
b) x=(frac{125}{8}), y=(frac{15}{2}), z=-(frac{15}{8})
c) x=(frac{125}{8}), y=-(frac{15}{2}), z=-(frac{15}{8})
d) x=-(frac{125}{8}), y=(frac{15}{2}), z=-(frac{15}{8})
Answer: b
Clarification: The given system of equations can be expressed in the form of AX=B, where
A=(begin{bmatrix}2&3&2\1&4&3\3&3&5end{bmatrix}), X=(begin{bmatrix}x\y\zend{bmatrix}), B=(begin{bmatrix}50\40\60end{bmatrix})
X=A-1 B
∴A-1=(frac{1}{|A|}) adj A
A-1=(frac{1}{16}begin{bmatrix}11&-9&1\4&4&-4\-9&3&5end{bmatrix})=(begin{bmatrix}frac{11}{16}&-frac{9}{16}&frac{1}{16}\frac{1}{4}&frac{1}{4}&-frac{1}{4}\-frac{9}{16}&frac{3}{16}&frac{5}{16}end{bmatrix})
X=A-1 B
X=(begin{bmatrix}frac{11}{16}&-frac{9}{16}&frac{1}{16}\frac{1}{4}&frac{1}{4}&-frac{1}{4}\-frac{9}{16}&frac{3}{16}&frac{5}{16}end{bmatrix}begin{bmatrix}50\40\60end{bmatrix})=(begin{bmatrix}frac{125}{8}\frac{15}{2}\-frac{15}{8}end{bmatrix})
7. Find the values of x and y for the given system of equations.
3x-2y=3 2x+2y=4
a) x=(frac{7}{5}), y=(frac{3}{5})
b) x=-(frac{7}{5}), y=-(frac{3}{5})
c) x=-(frac{7}{5}), y=(frac{3}{5})
d) x=(frac{7}{5}), y=-(frac{3}{5})
Answer: a
Clarification: The given system of equations can be expressed in the form of AX=B,
⇒X=A-1 B
A=(begin{bmatrix}3&-2\2&2end{bmatrix}), X=(begin{bmatrix}x\yend{bmatrix}), B=(begin{bmatrix}3\4end{bmatrix})
We know that, A-1=(frac{1}{|A|}) adj A
A-1=(frac{1}{10} begin{bmatrix}2&2\-2&3end{bmatrix})=(begin{bmatrix}frac{1}{5}&frac{1}{5}\-frac{1}{5}&frac{3}{10}end{bmatrix})
∴X=A-1 B=(begin{bmatrix}frac{1}{5}&frac{1}{5}\-frac{1}{5}&frac{3}{10}end{bmatrix}begin{bmatrix}3\4end{bmatrix})=(begin{bmatrix}frac{1}{5}×3+frac{1}{5}×4\-frac{1}{5}×3+frac{3}{10}×4end{bmatrix})=(begin{bmatrix}frac{7}{5}\frac{3}{5}end{bmatrix})
8. The cost of 8kg apple and 3kg is Rs 70. The cost of 10kg apple and 6kg orange is 90. Find the cost of each item if x is the cost of apples per kg and y is the cost of oranges per kg.
a) x=2, y=3
b) x=3, y=2
c) x=2, y=2
d) x=3, y=3
Answer: b
Clarification: The given situation can be represented by a system of equations as:
8x+3y=70
10x+5y=90
Consider, A=(begin{bmatrix}8&3\10&5end{bmatrix}), X=(begin{bmatrix}x\yend{bmatrix}), B=(begin{bmatrix}70\90end{bmatrix})
It can be expressed in the form of AX=B
⇒X=A-1 B
We know that, A-1=(frac{1}{|A|}) adj A=(frac{1}{10} begin{bmatrix}5&-3\-10&8end{bmatrix})
∴X=(frac{1}{10} begin{bmatrix}5&-3\-10&8end{bmatrix}begin{bmatrix}70\90end{bmatrix})=(frac{1}{10} begin{bmatrix}5×70-3×90\-10×70+8×90end{bmatrix})=(frac{1}{10} begin{bmatrix}300-270\-700+720end{bmatrix})=(frac{1}{10} begin{bmatrix}30\20end{bmatrix})
∴x=3, y=2.
i.e. The cost of apples is Rs 3 per kg and the cost of oranges is Rs 2 per kg.
9. For a given system of equations if |A|=0 and (adj A)B≠O(zero matrix), then which of the following is correct regarding the solutions of the given equations?
a) there will be exactly two solutions
b) there will be exactly one solution
c) the solution does not exist
d) there are one or more solutions
Answer: c
Clarification: If A is a singular matrix, then |A|=0
In this case, if (adj A)B≠O, then solution does not exist and the system of equations is called inconsistent.
10. Find the value of x and y for the given system of equations.
3x+4y=6 5x-4y=4
a) x=-(frac{5}{4}), y=-(frac{16}{9})
b) x=(frac{5}{4}), y=-(frac{9}{16})
c) x=-(frac{5}{4}), y=(frac{16}{9})
d) x=(frac{5}{4}), y=(frac{9}{16})
Answer: d
Clarification: The given system of equations can be expressed in the form of AX=B,
⇒X=A-1 B
We know that, A-1=(frac{1}{|A|}) adj A
A-1=(frac{1}{-32} begin{bmatrix}-4&-4\-5&3end{bmatrix})=(begin{bmatrix}frac{1}{8}&frac{1}{8}\frac{5}{32}&-frac{3}{32}end{bmatrix})
∴X=(begin{bmatrix}frac{1}{8}&frac{1}{8}\frac{5}{32}&-frac{3}{32}end{bmatrix}begin{bmatrix}6\4end{bmatrix})=(begin{bmatrix}frac{1}{8}×6+frac{1}{8}×4\frac{5}{32}×6-frac{3}{32}×4end{bmatrix})=(begin{bmatrix}frac{6}{8}+frac{4}{8}\frac{30}{32}-frac{12}{32}end{bmatrix})=(begin{bmatrix}frac{10}{8}\frac{18}{32}end{bmatrix})=(begin{bmatrix}frac{5}{4}\frac{9}{16}end{bmatrix})