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.

(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.

Answer: The matrix $S$ is $\begin{bmatrix} -191 & -110 \\ 77 & 44 \end{bmatrix}$ .

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 $2 \times 2$ matrix (as $R$ is $2 \times 2$ and $RS$ must also be $2 \times 2$ 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.

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(b) Given that P=[[2,−1],[3,4]],Q=[[5,2],[7,4]]P = [[2, -1], [3, 4]], Q = [[5, 2], [7, 4]]P=[[2,−1],[3,4]],Q=[[5,2],[7,4]] and R=[[2,5],[3,8]]R = [[2, 5], [3, 8]]R=[[2,5],[3,8]] find a matrix S such that PQ - RS is a null matrix.