(a) If P =
1 -1 0
2 3 4
0 1 2
and Q =
2 2 -4
-4 2 -4
2 -1 5
, find (QP) and hence solve the following system of equations using matrices :
x - y=3,2x+3y+4z=17,y+2z=7
(matrices, matrix multiplication, system of equations, linear equations)
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Step-by-Step Solution
Step 1: Calculate the product QP
To find the product of two matrices, QP, we multiply the rows of the first matrix (Q) by the columns of the second matrix (P), element by element, and sum the results. This will give us each element of the resulting matrix.
Step 2: Perform matrix multiplication (1st row)
We calculate the elements of the first row of QP. For QP11, we multiply the first row of Q by the first column of P. For QP12, we multiply the first row of Q by the second column of P. For QP13, we multiply the first row of Q by the third column of P.
Step 3: Perform matrix multiplication (2nd row)
Next, we calculate the elements of the second row of QP. We repeat the process of multiplying the second row of Q by each column of P to find QP21, QP22, and QP23.
Step 4: Perform matrix multiplication (3rd row)
Finally, we calculate the elements of the third row of QP. We multiply the third row of Q by each column of P to find QP31, QP32, and QP33.
Step 5: State the product matrix QP
After performing all the multiplications and additions, we assemble the resulting elements into the matrix QP. We observe that the resulting matrix is a scalar multiple of the identity matrix, specifically 6 times the identity matrix.
Step 6: Write the system of equations in matrix form
We can represent the given system of linear equations in matrix form AX=B, where A is the coefficient matrix, X is the variable matrix, and B is the constant matrix. Notice that the coefficient matrix A is the same as matrix P given in the problem.
Step 7: Utilize the product QP to find P−1
From the previous calculation, we found that QP=6I. We can use this relationship to find the inverse of matrix P. By multiplying both sides by 61, we see that the inverse of P is 61 times matrix Q.
Step 8: Substitute P−1 into the matrix equation
Now we can solve for X by multiplying both sides of the matrix equation PX=B by P−1. This gives X=P−1B. We substitute the expression for P−1 that we found in the previous step and the constant matrix B.
Step 9: Perform the matrix multiplication
We multiply the matrix 61Q by the column matrix B. First, we multiply the elements of each row of Q by the corresponding elements of B and sum them up.
Step 10: Simplify the resulting matrix
Finally, we simplify the elements of the resulting column matrix by dividing each element by 6. This gives us the values for x, y, and z.