Question 2887
[1]
-4     5 
-5     X
"The determinant of the matrix A is 1."
<br><I>
In matrix X=<table border=1><tr><td>a<td>b<tr><td>c<td>d</table>
Determinant = ab - cd
</i><br>
Determinant = (-4*X)-(-5*5)
            = -4X+25
Since we know the determinant is one,
1=-4X+25
-4X=-24
X=-24/-4
 =6
Therefore,correct answer is: <B>c.6</b>
<P>
[2]
Matrix C=
<table border=1>
<tr><td>+4<td>0<td>–7 
<tr><td>+3<td>0<td>+6
<tr><td>-5<td>+2<td>+3
</table>
In matrix X of order 3=
<table border=1>
<tr><td>x1<td>x2<td>x3 
<tr><td>y1<td>y2<td>y3
<tr><td>z1<td>z2<td>z3
</table>
Determinant of X is=
=x1(y2z3-y3z2)-x2(y1z3-y3z1)+x3(y1z2-z1y2)
Determinant of C is=
= 4[0*3-(2*6)]-0[3*3-(-5*6)]+(-7)[3*2-(-5*0)]
= 4[0-12]-0[9+30]-7[6-0]
= 4[-12]-0[39]-7[6]
= -48-0-42
= -90
<br>Therefore,correct answer is: <B>c.-90</b>
<P>
[3]
Cramer's Rule - Given the system
p1x + q1y = c1 ... [eqn1]
p2x + q2y = c2 ... [eqn2]

Determinant detA = p1q2 - p2q1
Determinant detX = c1q2 - c2q1
Determinant detY = p1c2 - p2c1

Then x=detX/detA
Then y=detY/detA
<br>
Cramer’s Rule to solve y.
x-4y=4
4x-y=1
<br>
Consider the determinant formed by the x,y and constant term of the given equations.
<table border=1>
<tr><td><b>p</b><td><b>q</b><td><b>c</b>
<tr><td>+1<td>-4<td>4
<tr><td>+4<td>-1<td>1
</table><br>

detA = p1q2 - p2q1
=(1*-1)-(4*-4)
=-1-(-16)
=-1+16
=15
detY = p1c2 - p2c1
=(1*1)-(4*4)
=1-16
=-15
<br>
By Cramer's Rule,y=detY/detA
y=-15/15
=-1
<br>Therefore,the correct answer is: <B>b. –1</b>
<P>
Hope this helps,
best of luck.