1. Problem: Find the unknown matrix $X$ if $$X + \begin{bmatrix}4 & 8 \\ 1 & -2 \end{bmatrix} + \begin{bmatrix}1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix}6 & 7 \\ 1 & 2 \end{bmatrix}.$$ Step 1: Combine the known matrices on the left side: $$\begin{bmatrix}4 & 8 \\ 1 & -2 \end{bmatrix} + \begin{bmatrix}1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix}4+1 & 8+0 \\ 1+0 & -2+1 \end{bmatrix} = \begin{bmatrix}5 & 8 \\ 1 & -1 \end{bmatrix}.$$ Step 2: Rewrite the equation as $$X + \begin{bmatrix}5 & 8 \\ 1 & -1 \end{bmatrix} = \begin{bmatrix}6 & 7 \\ 1 & 2 \end{bmatrix}.$$ Step 3: Solve for $X$ by subtracting the known matrix: $$X = \begin{bmatrix}6 & 7 \\ 1 & 2 \end{bmatrix} - \begin{bmatrix}5 & 8 \\ 1 & -1 \end{bmatrix} = \begin{bmatrix}6-5 & 7-8 \\ 1-1 & 2 - (-1) \end{bmatrix} = \begin{bmatrix}1 & -1 \\ 0 & 3 \end{bmatrix}.$$ Step 4: Compare with given options. None exactly match this matrix but choice (C) is close with second row (0,4) instead of (0,3). 2. Problem: If $P$ is a scalar matrix, then what is the transpose $P^T$? Step 1: A scalar matrix is a diagonal matrix with all diagonal entries equal to the same scalar. Step 2: Transpose of a diagonal matrix is the same matrix since elements off-diagonal are zero and diagonal elements stay on diagonal. Answer: $P^T = P$. So correct is (A). 3. Problem: Given $X_{p imes q} \cdot Y_{r imes s} = (XY)_{p imes s}$, find the condition. Step 1: Matrix multiplication is defined if the number of columns of first matrix equals number of rows of second matrix, so $q = r$. Answer: (A) $q = r$. 4. Problem: Which ordered pairs lie on y-axis? Pairs: I. $(0,3)$ and $(0,-3)$ II. $(3,0)$ and $(-3,0)$ III. $(0,0)$. Step 1: Points on y-axis have $x=0$. Step 2: Pairs in I and III have $x=0$. II points have $y=0$ (on x-axis). Answer: (A) I only. 5. Problem: Find x-intercept of line $3x + y =6$. Step 1: x-intercept is point where $y=0$. Step 2: Substitute $y=0$, so $3x = 6$, $$x = \frac{6}{3} = 2.$$ Answer: (C) 2. 6. Problem: Which point does NOT lie on line $y - 4x = -1$? Step 1: Test each point. For example, point (1,3): $$3 - 4(1) = 3 - 4 = -1,$$ which satisfies the equation. Point (-1,-3): $$-3 - 4(-1) = -3 + 4 = 1 \neq -1,$$ so point (-1,-3) does not lie on the line. Answer: (A) (-1,-3).