(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)

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 $QP_{11}$, we multiply the first row of Q by the first column of P. For $QP_{12}$, we multiply the first row of Q by the second column of P. For $QP_{13}$, 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 $QP_{21}$, $QP_{22}$, and $QP_{23}$.

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 $QP_{31}$, $QP_{32}$, and $QP_{33}$.

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 $\frac{1}{6}$, we see that the inverse of P is $\frac{1}{6}$ 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 $X = P^{-1}B$. 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 $\frac{1}{6}Q$ 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.

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