1. Problem 18: Find matrix $X$ such that $$X + \begin{bmatrix}4 & 8 \\ 1 & -2\end{bmatrix} \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix} = \begin{bmatrix}6 & 7 \\ 1 & 2\end{bmatrix}$$ Since multiplication with the identity matrix does not change the other matrix, $$\begin{bmatrix}4 & 8 \\ 1 & -2\end{bmatrix} \times \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix} = \begin{bmatrix}4 & 8 \\ 1 & -2\end{bmatrix}$$ So, $$X + \begin{bmatrix}4 & 8 \\ 1 & -2\end{bmatrix} = \begin{bmatrix}6 & 7 \\ 1 & 2\end{bmatrix}$$ Subtracting, $$X = \begin{bmatrix}6 & 7 \\ 1 & 2\end{bmatrix} - \begin{bmatrix}4 & 8 \\ 1 & -2\end{bmatrix} = \begin{bmatrix}6-4 & 7-8 \\ 1-1 & 2-(-2)\end{bmatrix} = \begin{bmatrix}2 & -1 \\ 0 & 4\end{bmatrix}$$ Answer: Option C 2. Problem 19: If $P$ is a scalar matrix, then all elements outside the main diagonal are zero and all diagonal elements are equal. Transpose $P^T$ swaps rows and columns. For scalar matrices, transpose does not change matrix, so $$P^T = P$$ Answer: A 3. Problem 20: If $X_{p \times q} \cdot Y_{r \times s} = (X.Y)_{p \times s}$, then matrix multiplication is defined iff the number of columns of $X$ equals the number of rows of $Y$. Therefore, $$q = r$$ Answer: A 4. Problem 21: Points lie on y-axis if their $x$ coordinate is zero. I. (0, 3) and (0, -3) lie on y-axis II. (3, 0) and (-3, 0) lie on x-axis, not y-axis III. (0, 0) lies on y-axis Correct options: I and III Answer: Not given directly but closest is A (I only) or D (II and III), but since II points have nonzero x, choice is A 5. Problem 22: Equation: $$3x + y = 6$$ To find x-intercept, set $y=0$: $$3x + 0 = 6 \implies x = 2$$ Answer: C 6. Problem 23: Check which points satisfy equation $$y - 4x = -1$$ Check each: - $(-1, -3)$: $-3 - 4(-1) = -3 + 4 = 1 \neq -1$ - $(0, -1)$: $-1 - 4(0) = -1 = -1$ ✅ - $(3, 11)$: $11 - 4(3) = 11 - 12 = -1$ ✅ - $(1, 3)$: $3 - 4(1) = 3 -4 = -1$ ✅ Point that does NOT lie on line is $(-1, -3)$ Answer: A