(b) Given that P=[[2,−1],[3,4]],Q=[[5,2],[7,4]] and R=[[2,5],[3,8]] find a matrix S such that PQ - RS is a null matrix.
Get the complete, step-by-step math solution for: "(b) Given that P = [[2, -1], [3, 4]], Q = [[5, 2], [7, 4]] and R = [[2, 5], [3, 8]] find a matrix S such that PQ - RS is a null matrix.". Powered by SolveForX AI math tutor.
Step-by-Step Solution
Step 1: Define the condition
The problem states that the difference between the product of matrices P and Q, and the product of matrices R and S, results in a null matrix. A null matrix, denoted by O, is a matrix where all its elements are zero.
Step 2: Rearrange the equation
To solve for S, we first rearrange the equation by moving RS to the right side of the equality, making it positive. This isolates the terms involving S on one side.
Step 3: Calculate PQ
We need to calculate the product of matrix P and matrix Q. For matrix multiplication, we multiply the rows of the first matrix by the columns of the second matrix and sum the products.
Step 4: Simplify PQ
Now we perform the arithmetic operations within each element of the PQ matrix to simplify it. This gives us the final matrix for PQ.
Step 5: Let S be a general matrix
Since we need to find matrix S, and we expect it to be a 2×2 matrix (as R is 2×2 and RS must also be 2×2 to match PQ), we represent S with general elements a, b, c, d.
Step 6: Calculate RS
Next, we multiply matrix R by the general matrix S. Again, we multiply rows by columns to find the elements of the product matrix RS.
Step 7: Equate PQ and RS
Now we set the calculated matrix PQ equal to the expression for RS, based on our initial rearranged equation PQ=RS. This gives us a system of linear equations.
Step 8: Formulate system of equations for 'a' and 'c'
By equating the corresponding elements of the matrices, we get two systems of linear equations. The first system involves variables a and c, corresponding to the first column of the resulting matrix.
Step 9: Solve for 'a' and 'c' using elimination (multiply Eq 1 by 3, Eq 2 by 2)
To solve for a and c, we use the elimination method. We multiply equation (1) by 3 and equation (2) by 2, so the coefficients of a become equal (6a in both equations).
Step 10: Solve for 'c'
Subtracting equation (3) from equation (4) eliminates a, allowing us to directly solve for c. The result is c=77.
Step 11: Solve for 'a'
Substitute the value of c=77 back into equation (1) (or equation (2)) and solve for a. After isolating a, we find a=−191.
Step 12: Formulate system of equations for 'b' and 'd'
Similarly, we equate the elements of the second column from PQ and RS to form a system of linear equations for b and d.
Step 13: Solve for 'b' and 'd' using elimination (multiply Eq 5 by 3, Eq 6 by 2)
Again, using the elimination method, we multiply equation (5) by 3 and equation (6) by 2 to make the coefficients of b equal (6b in both equations).
Step 14: Solve for 'd'
Subtracting equation (7) from equation (8) eliminates b, allowing us to solve for d. The result is d=44.
Step 15: Solve for 'b'
Substitute the value of d=44 back into equation (5) (or equation (6)) and solve for b. After isolating b, we find b=−110.
Step 16: Construct Matrix S
Finally, we substitute the calculated values of a, b, c, d back into the general matrix S to find the required matrix.