This is a matrix question.
R is the matrix (3r 1)
(s 2s)
1. State, in terms of s and r the determinant of R
2. If r=1/3 and s=4 determine the inverse of R
3.State the pair of r and s not including 0, which would make the matrix R a singular matrix
det = 6rs-s
R is singular if s(6r-1) = 0, so r=1/6 and s is anything
R^-1 =
1/4 *
(8,-1)
(-4,1)
1. The determinant of a 2x2 matrix is calculated using the formula: determinant = (a*d) - (b*c), where the matrix R can be expressed as:
R = |3r 1|
|s 2s|
The determinant of R = (3r * 2s) - (1 * s) = 6rs - s = s(6r - 1).
So, the determinant of R in terms of s and r is s(6r - 1).
2. To find the inverse of matrix R, we can use the formula for a 2x2 matrix:
R^-1 = (1/determinant) * |2s -1|
|-s 3r|
Substituting r = 1/3 and s = 4 into the formula, we have:
R^-1 = (1/(s(6r - 1))) * |8 -1|
|-4 1|
Simplifying, we get:
R^-1 = (1/(4(6/3 - 1))) * |8 -1|
|-4 1|
= (1/(4(2 - 1))) * |8 -1|
|-4 1|
= (1/(4(1))) * |8 -1|
|-4 1|
= (1/4) * |8 -1|
|-4 1|
= (1/4) * |7|
|-4|
= |7/4|
|-1|
Therefore, the inverse of R when r = 1/3 and s = 4 is:
R^-1 = |7/4|
|-1|
3. A matrix is singular if its determinant is equal to 0. So, by setting the determinant of R to 0 and solving for r and s:
s(6r - 1) = 0
For the determinant to be 0, either s = 0 or 6r - 1 = 0.
If s = 0, the matrix R becomes:
R = |3r 1|
|0 2s|
This would make the second row all zeros, and the matrix R would be singular.
If 6r - 1 = 0, then:
6r = 1
r = 1/6
Thus, the pair of r and s, not including 0, that would make the matrix R singular is (1/6, 0).
To solve this matrix question, let's go step by step.
1. Determinant of R:
The determinant of a 2x2 matrix like R can be found using the formula:
Determinant = (Upper left element * Lower right element) - (Upper right element * Lower left element)
In the given matrix R:
Upper left element = 3r
Lower right element = 2s
Upper right element = 1
Lower left element = s
So, the determinant of R is:
Determinant of R = (3r * 2s) - (1 * s)
= 6rs - s
= s(6r - 1)
So, in terms of s and r, the determinant of R is s(6r - 1).
2. Inverse of R:
To find the inverse of matrix R, we can use the following formula:
Inverse of R = (1 / Determinant of R) * (Adjugate of R)
where the adjugate of R can be found by swapping the main diagonal elements and negating the off-diagonal elements.
Using the determinant of R from the previous step, we get:
Determinant of R = s(6r - 1)
Now, substitute the given values r = 1/3 and s = 4 into the formula:
Determinant of R = 4(6 * 1/3 - 1)
= 4(2 - 1)
= 4(1)
= 4
Next, find the adjugate of R by swapping the main diagonal elements and negating the off-diagonal elements:
Adjugate of R = (2s -1 -1)
(-s 3r)
Substituting the given values r = 1/3 and s = 4 into the adjugate matrix, we get:
Adjugate of R = (2 * 4 - 1 -1)
(-4 1)
Now, multiply the adjugate matrix by 1/determinant of R:
Inverse of R = (1 / Determinant of R) * (Adjugate of R)
= (1 / 4) * ( (7 -1)
(-4 1) )
= (1/4) * ( (7/4 -1/4)
(-1 1/4) )
So, the inverse of R is:
Inverse of R = (7/16 -1/4)
(-1/4 1/4)
3. Singular matrix:
A matrix is considered singular if its determinant is zero. So, to find the pair of r and s, not including 0, that would make matrix R singular, we need to set the determinant of R to zero and solve for r and s.
Using the determinant of R from step 1, we have:
s(6r - 1) = 0
For the determinant to be zero, either s = 0 or (6r - 1) = 0.
Since the question states that r and s cannot be zero, we only consider the second case:
6r - 1 = 0
6r = 1
r = 1/6
Therefore, if r = 1/6 and s is not equal to zero, the matrix R will be singular.