1. **Stating the problem:** Determine which statements about matrices are always true and for which values of $a$ and $b$ the given system has more than one solution. 2. **Matrix statements analysis:** - Statement: "If $A \neq 0$ is a square matrix, then $A$ is invertible." This is **false** because a nonzero matrix can be singular. - Statement: "If $A$ and $B$ are invertible, then $A+B$ is invertible." This is **false**; sum of invertible matrices need not be invertible. - Statement: "$A + A^T$ is skew-symmetric." This is **false**; $A + A^T$ is symmetric. - Statement: "If $A$ and $B$ are invertible, then $(A^{-1}B)^T$ is invertible." This is **true** because transpose and inverse preserve invertibility. - Statement: "If the $(3,1)$-entry of $A$ is 5, then the $(1,3)$-entry of $A^T$ is $-5$." This is **false**; transpose swaps entries without changing sign. 3. **System of equations:** Given: $$\begin{cases} -x + 3y + 2z = -8 \\ x + 0y + z = 2 \\ 2x + 2y + az = b \end{cases}$$ 4. **Row operations and matrix form:** $$\left[\begin{array}{ccc|c} -1 & 3 & 2 & -8 \\ 1 & 0 & 1 & 2 \\ 2 & 2 & a & b \end{array}\right]$$ After operations: $$\left[\begin{array}{ccc|c} 1 & -3 & -2 & 8 \\ 0 & 3 & 3 & -6 \\ 0 & 8 & a+4 & b+16 \end{array}\right]$$ 5. **Reduced form:** Divide $R_2$ by 3: $$\left[\begin{array}{ccc|c} 1 & -3 & -2 & 8 \\ 0 & 1 & 1 & -2 \\ 0 & 8 & a+4 & b+16 \end{array}\right]$$ 6. **Eliminate $y$ from $R_3$:** $$R_3 \to R_3 - 8R_2$$ $$\left[\begin{array}{ccc|c} 1 & -3 & -2 & 8 \\ 0 & 1 & 1 & -2 \\ 0 & 0 & a+4 - 8 & b+16 - 8(-2) \end{array}\right] = \left[\begin{array}{ccc|c} 1 & -3 & -2 & 8 \\ 0 & 1 & 1 & -2 \\ 0 & 0 & a - 4 & b + 32 \end{array}\right]$$ 7. **Condition for infinite solutions:** The system has more than one solution if the last equation is dependent or zero row: $$a - 4 = 0 \implies a = 4$$ and $$b + 32 = 0 \implies b = -32$$ 8. **Check options:** - (A) $a=4$, $b \neq 0$ (false, must have $b=-32$) - (B) $a \neq -4$, $b \neq 0$ (false) - (C) $a = -4$, $b \neq 0$ (false) - (D) $a=4$, $b=0$ (false, must be $b=-32$) - (E) $a \neq 4$, $b \neq 0$ (false) **Correct answer for infinite solutions:** $a=4$ and $b=-32$ (not listed exactly). **Summary:** - Always true matrix statement: "If $A$ and $B$ invertible, then $(A^{-1}B)^T$ invertible." - System has more than one solution only if $a=4$ and $b=-32$.