Matrix Solutions to Systems - Algebra 3

1. The system below is written as the matrix equation $AX = B$, where $A=\begin{bmatrix}1&2&-1\\2&1&3\\3&1&2\end{bmatrix}$ and $B=\begin{bmatrix}5\\2\\5\end{bmatrix}$. If $A^{-1}=\frac{1}{10}\begin{bmatrix}-1&-5&7\\5&5&-5\\-1&5&-3\end{bmatrix}$, what is the $z$-coordinate of the solution $X=\begin{bmatrix}x\\y\\z\end{bmatrix}$?
   Answer: -1 When you see a matrix equation $AX=B$ and you already have $A^{-1}$, your move is to multiply both sides on the left by $A^{-1}$. That gives $X=A^{-1}B$, so the solution vector is just that product. Since the question asks for the $z$-coordinate, you only need the third row of $A^{-1}$. Compute the third row of $A^{-1}$ dotted with $B$: $z = \frac{1}{10}\begin{bmatrix}-1 & 5 & -3\end{bmatrix} \begin{bmatrix}5\\2\\5\end{bmatrix} = \frac{1}{10}(-5+10-15) = \frac{-10}{10} = -1.$ So the $z$-coordinate is $-1$. The other choices come from common missteps. The value $2$ is actually the $x$-coordinate: the first row gives $x=2$. The value $1$ is the $y$-coordinate, from the second row. Both are real parts of the solution, but neither is $z$. The value $-10$ is the numerator inside the calculation before dividing by $10$; if you forget the $\frac{1}{10}$ factor, you'd incorrectly choose $-10$. A useful habit: when $X$ has multiple components, identify exactly which row of $A^{-1}$ corresponds to the variable you need. Then compute only that row times $B$. This saves time and reduces arithmetic errors. Also, always check whether the final step includes dividing by the determinant factor—here the $\frac{1}{10}$ is easy to drop.
2. Consider $x+2y=3$ and $3x+ky=12$, where $k$ is a constant. Which statement describes the solutions for all values of $k$?
   Answer: There is no solution when $k=6$, and a unique solution otherwise. The determinant is $k-6$, so a unique solution exists whenever $k\neq 6$. At $k=6$ the left sides are proportional but the constants are not, since $3(x+2y)=9$ while the second equation demands $12$, so the system is inconsistent. Infinitely many solutions would require that constant to be $9$ instead of $12$. At $k=-6$ the determinant is $-12$, which is nonzero, so that case is an ordinary unique solution.
3. Which matrix multiplies $\begin{bmatrix}5&2\\7&3\end{bmatrix}$ to produce the identity matrix $I_2$?
   Answer: $\begin{bmatrix}3&-2\\-7&5\end{bmatrix}$ Since $\det=5(3)-2(7)=1$, the inverse swaps the diagonal entries and negates the off-diagonal ones, giving $\begin{bmatrix}3&-2\\-7&5\end{bmatrix}$. Multiplying confirms $5(3)+2(-7)=1$ and $5(-2)+2(5)=0$. The matrix $\begin{bmatrix}3&2\\7&5\end{bmatrix}$ changes no signs. The matrix $\begin{bmatrix}3&-7\\-2&5\end{bmatrix}$ puts the negated entries in transposed positions. The matrix $\begin{bmatrix}-3&2\\7&-5\end{bmatrix}$ negates the diagonal entries instead.
4. A jeweler's system is $4x+3y=18$ and $2x+5y=16$. Using the inverse of the coefficient matrix, what is the solution?
   Answer: $(3,2)$ With $\det(A)=4(5)-3(2)=14$, the inverse is $\tfrac{1}{14}\begin{bmatrix}5&-3\\-2&4\end{bmatrix}$, and $\tfrac{1}{14}(42,28)=(3,2)$, which satisfies both equations. The pair $(42,28)$ omits the factor $\tfrac{1}{14}$. The pair $(2,3)$ reverses the coordinates. The pair $(3,-2)$ carries a sign error into $y$ and fails the second equation.
5. The coefficient matrix of a system is the diagonal matrix $\begin{bmatrix}1&0&0\\0&2&0\\0&0&5\end{bmatrix}$. What is its inverse?
   Answer: $\begin{bmatrix}1&0&0\\0&\frac{1}{2}&0\\0&0&\frac{1}{5}\end{bmatrix}$ For a diagonal matrix with nonzero entries, the inverse is diagonal with each entry replaced by its reciprocal, and multiplying the two matrices returns the identity. Negating the entries gives a product with $-1$ down the diagonal, not $1$. Repeating the original matrix gives a product with $1,\,4,\,25$ down the diagonal. Reversing the order of the reciprocals moves nonzero entries off the diagonal, so the product is not the identity.
6. Which single row operation turns $\left[\begin{array}{cc|c}2&4&6\\1&-3&5\end{array}\right]$ into $\left[\begin{array}{cc|c}1&2&3\\1&-3&5\end{array}\right]$?
   Answer: $R_1 \to \tfrac{1}{2}R_1$ Every entry of the first row is halved, since $2,\,4,\,6$ becomes $1,\,2,\,3$, while the second row is untouched. The operation $R_1 \to R_1 - R_2$ would produce $1,\,7,\,1$ instead. Swapping the rows would put $1,\,-3,\,5$ on top. Doubling the first row would give $4,\,8,\,12$, moving away from the target rather than toward it.
7. Apply $R_1 \to \tfrac{1}{3}R_1$ to $\left[\begin{array}{cc|c}3&-6&9\\2&5&4\end{array}\right]$, then apply $R_2 \to R_2 - 2R_1$ using the new first row. What is the resulting second row?
   Answer: $\begin{bmatrix}0&9&-2\end{bmatrix}$ Scaling makes the first row $1,\,-2,\,3$, and then $2-2(1)=0$, $5-2(-2)=9$, and $4-2(3)=-2$ give $\begin{bmatrix}0&9&-2\end{bmatrix}$. The row $\begin{bmatrix}0&9&-14\end{bmatrix}$ uses the original constant $9$ rather than the scaled $3$. The row $\begin{bmatrix}0&-7&-2\end{bmatrix}$ subtracts $2(6)$ and misses the double negative. The row $\begin{bmatrix}0&1&-2\end{bmatrix}$ subtracts twice the unscaled leading entry instead.
8. Apply the row operation $R_2 \to R_2 - 4R_1$ to $\left[\begin{array}{cc|c}1&-2&3\\4&1&-3\end{array}\right]$. What is the result?
   Answer: $\left[\begin{array}{cc|c}1&-2&3\\0&9&-15\end{array}\right]$ Entry by entry, $4-4(1)=0$, $1-4(-2)=9$, and $-3-4(3)=-15$, so the new second row is $0,\,9,\,-15$. The row $0,\,-7,\,-15$ subtracts $8$ instead of adding it, missing the double negative. The row $0,\,9,\,15$ drops the sign in the constant column. The row $0,\,9,\,-9$ subtracts only $6$ from the constant.
9. Suppose $A^{-1}=\begin{bmatrix}1&2\\1&3\end{bmatrix}$. What is the matrix $A$ itself?
   Answer: $\begin{bmatrix}3&-2\\-1&1\end{bmatrix}$ Because inverses come in pairs, $A$ is the inverse of $A^{-1}$. That matrix has determinant $1(3)-2(1)=1$, so inverting it gives $\begin{bmatrix}3&-2\\-1&1\end{bmatrix}$, and multiplying the two returns the identity. The option $\begin{bmatrix}1&-2\\-1&3\end{bmatrix}$ negates the off-diagonal entries without swapping the diagonal ones. The option $\begin{bmatrix}3&-1\\-2&1\end{bmatrix}$ is the transpose of the correct matrix. Taking reciprocals entrywise is not how matrices invert.
10. A $2\times 2$ matrix $A$ satisfies $A\begin{bmatrix}1\\1\end{bmatrix}=\begin{bmatrix}5\\3\end{bmatrix}$ and $A\begin{bmatrix}1\\-1\end{bmatrix}=\begin{bmatrix}1\\1\end{bmatrix}$. Identify $A$.
    Answer: $\begin{bmatrix}3&2\\2&1\end{bmatrix}$ Writing $A=\begin{bmatrix}a&b\\c&d\end{bmatrix}$ turns the two conditions into $a+b=5$, $a-b=1$, $c+d=3$, and $c-d=1$, so $a=3$, $b=2$, $c=2$, and $d=1$. The matrices $\begin{bmatrix}3&2\\1&2\end{bmatrix}$ and $\begin{bmatrix}2&3\\1&2\end{bmatrix}$ both reproduce the first condition but send $\begin{bmatrix}1\\-1\end{bmatrix}$ to $\begin{bmatrix}1\\-1\end{bmatrix}$ and $\begin{bmatrix}-1\\-1\end{bmatrix}$ respectively. The matrix $\begin{bmatrix}3&1\\2&2\end{bmatrix}$ fails even the first condition, giving $\begin{bmatrix}4\\4\end{bmatrix}$.
11. Which augmented matrix represents $3x-y=7$ together with $x+4y=-2$?
    Answer: $\left[\begin{array}{cc|c}3&-1&7\\1&4&-2\end{array}\right]$ Each row lists that equation's $x$ coefficient, $y$ coefficient, and constant in order, so the first row is $3,\,-1,\,7$ and the second is $1,\,4,\,-2$. The version beginning $3,\,1$ loses the negative on $-y$. The version with constants $-2$ and $7$ attaches each constant to the wrong equation. The version beginning $-1,\,3$ reverses the coefficient columns.
12. Given $A^{-1}=\begin{bmatrix}2&-1\\-3&2\end{bmatrix}$ and $B=\begin{bmatrix}4\\5\end{bmatrix}$, compute $X=A^{-1}B$.
    Answer: $(3,-2)$ Row-by-column multiplication gives $2(4)+(-1)(5)=3$ and $(-3)(4)+2(5)=-2$, so $X=(3,-2)$. The pair $(13,22)$ comes from ignoring the negative signs and adding every product. The pair $(-3,2)$ reads the first column of $A^{-1}$ rather than computing the product. The pair $(2,3)$ reverses the two entries and drops the sign.
13. Row reduce to solve $2x+y-z=1$, $x-y+2z=8$, and $3x+2y+z=9$.
    Answer: $(2,0,3)$ Eliminating variables leads to $3x+z=9$ and $-x+3z=7$, giving $x=2$, $z=3$, and then $y=0$; all three equations check, as in $2(2)+0-3=1$. The triple $(1,2,3)$ fails the first equation, since $2+2-3=1$ but $1-2+6=5\neq 8$. The triple $(0,2,3)$ swaps $x$ and $y$, and $(3,0,2)$ swaps $x$ and $z$; both fail the second equation.
14. While row reducing an augmented matrix, a student obtains the row $\left[\begin{array}{ccc|c}0&0&0&5\end{array}\right]$. What does this row reveal about the system?
    Answer: It is inconsistent, so the system has no solution at all. That row translates to $0x+0y+0z=5$, which no values of the variables can satisfy, so the system is inconsistent. The reading $z=5$ would require a leading 1 in the $z$ column rather than a row of zeros. Infinitely many solutions require the constant to be $0$ as well, giving the row $\left[\begin{array}{ccc|c}0&0&0&0\end{array}\right]$. The value $z=0$ is not forced by anything in this row.
15. Consider the matrix equation $\begin{bmatrix}1&2&-1\\2&4&-2\\1&1&1\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}3\\5\\2\end{bmatrix}.$ Which conclusion follows?
    Answer: The system has no solution. Whenever you have a matrix equation, check whether one row is a multiple of another before doing elimination. Here the first two coefficient rows are proportional: Row 2 is exactly twice Row 1. That means the two equations express the same left-hand side in different ways: Row 1 gives $x+2y-z=3$, and Row 2 gives $2x+4y-2z=5$, which simplifies to $x+2y-z=\tfrac{5}{2}$. The same expression cannot equal both $3$ and $\tfrac{5}{2}$ simultaneously, so no triple $(x,y,z)$ can satisfy the system. Therefore the conclusion that the system has no solution is forced. The statement that the system has exactly one solution would require a consistent system, but the first two rows are incompatible, so zero solutions exist. The proposed unique solution $(1,1,0)$ looks tempting because it satisfies the first and third equations: $1+2-0=3$ and $1+1+0=2$. However, it fails the second equation: $2(1)+4(1)-2(0)=6$, not $5$. The infinitely many solutions conclusion is another trap: proportional coefficient rows often signal redundancy or free variables, but here the constants are inconsistent, so the system is inconsistent, not underdetermined. Whenever proportional coefficient rows appear, compare the constants. If one row is a multiple of another but the constants are not in the same ratio, the system has no solution. This quick check saves time and prevents a false "infinitely many solutions" conclusion.
16. Back-substitute to solve the system whose augmented matrix in row echelon form is $\left[\begin{array}{cc|c}1&3&5\\0&1&-2\end{array}\right]$.
    Answer: $(11,-2)$ The second row gives $y=-2$, and substituting into $x+3y=5$ gives $x=5+6=11$, so the solution is $(11,-2)$. The pair $(5,-2)$ skips the back-substitution and reads $x$ straight from the constant column. The pair $(-1,-2)$ subtracts $6$ instead of adding it. The pair $(11,2)$ drops the negative sign on $y$.
17. An augmented matrix reduces to $\left[\begin{array}{cc|c}2&0&6\\0&3&-9\end{array}\right]$. Read off the solution.
    Answer: $(3,-3)$ The rows say $2x=6$ and $3y=-9$, so dividing by each pivot gives $x=3$ and $y=-3$. The pair $(6,-9)$ reads the constant column without dividing by the pivots, which are not yet $1$. The pair $(3,3)$ loses the negative sign on $y$. The pair $(2,3)$ reports the pivot entries rather than the solution.
18. The system $2x-5y=7$ and $3x+y=-4$ is rewritten in the form $AX=B$. Which matrix equation is correct?
    Answer: $\begin{bmatrix}2&-5\\3&1\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}7\\-4\end{bmatrix}$ Each row of $A$ holds the coefficients of one equation in order, and $B$ holds that equation's constant, giving $\begin{bmatrix}2&-5\\3&1\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}7\\-4\end{bmatrix}$. The version with coefficient matrix $\begin{bmatrix}2&3\\-5&1\end{bmatrix}$ stores the coefficients down the columns instead of across the rows. The version with $\begin{bmatrix}2&5\\3&1\end{bmatrix}$ drops the negative sign on $-5y$. The version ending in $\begin{bmatrix}-4\\7\end{bmatrix}$ pairs each equation with the other equation's constant.
19. Find the column matrix $X$ satisfying $\begin{bmatrix}4&1\\3&-2\end{bmatrix}X=\begin{bmatrix}5\\12\end{bmatrix}$.
    Answer: $\begin{bmatrix}2\\-3\end{bmatrix}$ The determinant is $4(-2)-1(3)=-11$, and multiplying by the inverse gives $X=\tfrac{1}{11}(22,-33)$, or $\begin{bmatrix}2\\-3\end{bmatrix}$; substituting confirms $4(2)+(-3)=5$ and $3(2)-2(-3)=12$. The entry pair $\begin{bmatrix}22\\-33\end{bmatrix}$ skips dividing by the determinant. The pairs $\begin{bmatrix}-3\\2\end{bmatrix}$ and $\begin{bmatrix}3\\-2\end{bmatrix}$ swap or resign the entries and fail the second equation.
20. Express $x+y+z=6$, $2x-z=1$, and $y+3z=10$ as a single matrix equation.
    Answer: $\begin{bmatrix}1&1&1\\2&0&-1\\0&1&3\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}6\\1\\10\end{bmatrix}$ A missing variable contributes a coefficient of $0$, so $2x-z=1$ becomes the row $\begin{bmatrix}2&0&-1\end{bmatrix}$ and $y+3z=10$ becomes $\begin{bmatrix}0&1&3\end{bmatrix}$. The version with row $\begin{bmatrix}2&-1&0\end{bmatrix}$ puts the $z$ coefficient in the $y$ column. The version whose first column reads $1,1,1$ stores coefficients by column instead of by row. The version with row $\begin{bmatrix}2&0&1\end{bmatrix}$ loses the negative sign on $-z$.

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