Matrix Operations - Algebra 3

1. Let $A=\begin{pmatrix}2&0&0\\1&3&0\\0&1&4\end{pmatrix}$ and $B=\begin{pmatrix}1&2&0\\0&1&0\\0&0&3\end{pmatrix}$. What is $\det(A^2B^{-1})$?
   Answer: $192$ The matrix $A$ is lower triangular, so $\det(A)=2\cdot3\cdot4=24$. The matrix $B$ is upper triangular, so $\det(B)=1\cdot1\cdot3=3$. Therefore $\det(A^2B^{-1})=\det(A)^2\det(B^{-1})=24^2\cdot\frac13=192$. Choice 8 is $\det(A)/\det(B)$, choice 48 comes from using $2\det(A)$, and choice 72 uses $\det(A)\det(B)$ without accounting for the inverse or the square.
2. Let $A$ and $B$ be $3\times3$ matrices with $\det(A)=-3$ and $\det(B)=2$. What is $\det(2AB^{-1})$?
   Answer: $-12$ Whenever you see a determinant problem involving products, inverses, and scalar multiples, the key is to recall how each operation affects the determinant: $\det(cA)=c^n\det(A)$ for an $n\times n$ matrix, $\det(B^{-1})=1/\det(B)$, and $\det(AB)=\det(A)\det(B)$. Here $A$ and $B$ are $3\times3$, so the scalar multiple $2A$ contributes a factor of $2^3=8$, not just $2$. Thus $$\det(2AB^{-1})=2^3\det(A)\det(B^{-1}) =8(-3)\left(\frac12\right)=-12.$$ The choice $12$ confuses the sign: it likely drops the negative from $\det(A)=-3$. The choice $-48$ uses $\det(B)=2$ directly instead of $\det(B^{-1})=1/2$, effectively multiplying by $2$ instead of dividing by $2$. The choice $-3$ correctly handles the inverse but forgets that multiplying $A$ by $2$ scales the determinant by $2^3$, not $2$. Keep the order of operations straight: take the scalar multiple to the third power, invert the determinant of $B$, then multiply by $\det(A)$. A useful habit is to write the determinant expression completely before plugging in numbers. That way you will remember both the dimension-based scalar power and the reciprocal determinant for the inverse.
3. Let $A = \begin{bmatrix}0&1\\1&0\end{bmatrix}$. Which form must a matrix $B=\begin{bmatrix}a&b\\c&d\end{bmatrix}$ have so that $AB=BA$?
   Answer: $\begin{bmatrix}a&b\\b&a\end{bmatrix}$ Whenever you see a question asking for the form of a matrix that commutes with a given matrix, the key is to compute both products and equate corresponding entries. Here $A$ simply swaps coordinates, so comparing $AB$ and $BA$ will force certain entries of $B$ to match. . Computing directly:$AB=\begin{bmatrix}0&1\\1&0\end{bmatrix}\begin{bmatrix}a&b\\c&d\end{bmatrix}=\begin{bmatrix}c&d\\a&b\end{bmatrix},$while$BA=\begin{bmatrix}a&b\\c&d\end{bmatrix}\begin{bmatrix}0&1\\1&0\end{bmatrix}=\begin{bmatrix}b&a\\d&c\end{bmatrix}.$Equating the two matrices gives $c=b$, $d=a$, and also $a=d$, $b=c$—these are the same conditions. So the general commuter must have form$\begin{bmatrix}a&b\\b&a\end{bmatrix}.$ The candidate $\begin{bmatrix}a&b\\-b&a\end{bmatrix}$ would require $c=-b$ and $d=a$. Then $AB=\begin{bmatrix}-b&a\\a&b\end{bmatrix}$ while $BA=\begin{bmatrix}b&a\\a&-b\end{bmatrix}$; these match only when $b=0$, so it is not a general solution. The repeated-row forms—both the explicit $\begin{bmatrix}a&b\\a&b\end{bmatrix}$ and the equivalent parameter statement $a=c,\ b=d$—give $AB=\begin{bmatrix}a&b\\a&b\end{bmatrix}$ but $BA=\begin{bmatrix}b&a\\b&a\end{bmatrix}$; equality would force $a=b$, again only a special case. Therefore, only the matrix with matching off-diagonal entries works for all $a$ and $b$. Study tip: if you are unsure, test a simple numerical matrix such as $a=1,\ b=2$. The only form that survives the commutation check is the one with equal off-diagonal entries—so remember that commuting with a swap matrix forces symmetry across the main diagonal.
4. Let $A = \begin{bmatrix}2&-1\\0&3\end{bmatrix}$ and $B = \begin{bmatrix}1&2\\-1&4\end{bmatrix}$. Which matrix equals $(A+B)(A-B)$?
   Answer: $\begin{bmatrix}4&-10\\6&-4\end{bmatrix}$ When you see an expression like $(A+B)(A-B)$ with matrices, the critical thing to remember is that matrix multiplication is not commutative: expanding gives $A^2-AB+BA-B^2$, not simply $A^2-B^2$. That trap is built into the choices. Start by computing the two factors directly: $A+B=\begin{bmatrix}3&1\\-1&7\end{bmatrix}, \qquad A-B=\begin{bmatrix}1&-3\\1&-1\end{bmatrix}.$ Then multiply: $(A+B)(A-B)=\begin{bmatrix}3&1\\-1&7\end{bmatrix}\begin{bmatrix}1&-3\\1&-1\end{bmatrix} =\begin{bmatrix}4&-10\\6&-4\end{bmatrix}.$ Row 1 uses $3(1)+1(1)=4$ and $3(-3)+1(-1)=-10$; row 2 uses $-1(1)+7(1)=6$ and $-1(-3)+7(-1)=3-7=-4$. The choice $\begin{bmatrix}5&-15\\5&-5\end{bmatrix}$ is exactly $A^2-B^2$, the result you get if you incorrectly assume $AB=BA$ and use the familiar binomial shortcut. The choice $\begin{bmatrix}6&-20\\4&-6\end{bmatrix}$ does not equal either product order; its second row likely comes from a computation slip in the row-column products. The choice $\begin{bmatrix}4&-10\\6&4\end{bmatrix}$ matches the correct matrix except for the bottom-right entry: it uses $3+7=4$ instead of $3-7=-4$, so it is a sign error. Study tip: with matrix expressions, compute the actual matrices first, multiply row-by-column carefully, and re-check every sign in each dot product. Don't use the binomial shortcut unless you know the matrices commute.
5. Let $A=\begin{pmatrix}x+1&3\\1&0\end{pmatrix}$, and let $I$ be the $2\times2$ identity matrix. For which value of $x$ is $A^2-2A+I$ singular?
   Answer: -3 Since $I$ commutes with $A$, $A^2-2A+I=(A-I)^2$. Also $A-I=\begin{pmatrix}x&3\\1&-1\end{pmatrix}$, so $\det(A-I)=-x-3$. Therefore $\det((A-I)^2)=(-x-3)^2=(x+3)^2$, which is $0$ only when $x=-3$. The $3$ option is the sign error $(x-3)^2$; the $-1$ option comes from computing the determinant as $-x-1$; the $0$ option comes from setting an entry of $A-I$, rather than its determinant, to $0$.
6. Let $A=\begin{pmatrix}1&2\\0&1\end{pmatrix}$. Which matrix is $(A^{25})^{-1}$?
   Answer: $\begin{pmatrix}1&-50\\0&1\end{pmatrix}$ For this shear matrix, $A^n=\begin{pmatrix}1&2n\\0&1\end{pmatrix}$, so $A^{25}=\begin{pmatrix}1&50\\0&1\end{pmatrix}$. Since its determinant is 1, its inverse is obtained by changing the sign of the shear entry: $(A^{25})^{-1}=\begin{pmatrix}1&-50\\0&1\end{pmatrix}$. The other choices confuse the exponent with the shear entry or forget the negative sign.
7. Let $A=\begin{pmatrix}1&2\\0&1\end{pmatrix}$, $B=\begin{pmatrix}1&0\\-1&1\end{pmatrix}$, and $C=\begin{pmatrix}0&1\\1&0\end{pmatrix}$. If $AXB=C$, what is $X$?
   Answer: $\begin{pmatrix}-1&1\\1&0\end{pmatrix}$ To isolate $X$, multiply on the left by $A^{-1}$ and on the right by $B^{-1}$: $X=A^{-1}CB^{-1}$. Here $A^{-1}=\begin{pmatrix}1&-2\\0&1\end{pmatrix}$ and $B^{-1}=\begin{pmatrix}1&0\\1&1\end{pmatrix}$. Thus $A^{-1}CB^{-1}=\begin{pmatrix}-1&1\\1&0\end{pmatrix}$. The other choices result from multiplying the inverses in the wrong order on the left or right.
8. Suppose $A$ is a $3\times3$ matrix with $\det(A)=5$. Let $B$ be the matrix obtained from $A$ by swapping two rows and then multiplying one row by 2. What is $\det(B)$?
   Answer: $-10$ Whenever you see a question about how row operations change a determinant, remember the two key rules: swapping two rows multiplies the determinant by $-1$, and multiplying one row by a constant multiplies the determinant by that constant. Here, start with $\det(A)=5$. Swapping two rows gives $5 \cdot (-1) = -5$. Then multiplying one row by $2$ doubles that result, so $\det(B) = -5 \cdot 2 = -10$. The choice $-5$ reflects only the row swap, forgetting the later multiplication by $2$. The choice $10$ correctly multiplies by $2$ but loses the sign change from the swap. The choice $5$ ignores both operations entirely, as if $B$ had the same determinant as $A$. None of those captures the combined effect of both row operations. A good study habit is to process row operations one at a time, updating the determinant after each step. This prevents mixing up the order and helps you catch whether the sign changed or a scaling factor was applied. Also remember: swapping rows always flips the sign, while multiplying a row by $c$ scales the determinant by $c$.
9. For a $2\times 2$ matrix $A$, $\det(A)=4$. What is $\det\left(3(2A)^{-1}\right)$?
   Answer: $\frac{9}{16}$ First, $(2A)^{-1}=\frac12 A^{-1}$, so $3(2A)^{-1}=\frac32 A^{-1}$. For a $2\times 2$ matrix, the scalar factor is squared, so $\det\left(3(2A)^{-1}\right)=\left(\frac32\right)^2\det(A^{-1})$. Since $\det(A^{-1})=\frac14$, the result is $\frac94\cdot\frac14=\frac{9}{16}$. Choice $\frac{3}{16}$ treats the outer scalar 3 linearly rather than squaring it; $\frac{3}{8}$ uses $\frac32\det(A^{-1})$ without squaring the scalar factor; $\frac{27}{8}$ comes from cubing the scalar factor and omitting the factor $\frac14$.
10. Let $A$ be $3\times2$, let $B$ be $2\times4$, and let $C$ be $4\times3$. Which statement is true about $P=(AB)C$ and $Q=A(BC)$?
    Answer: Both $P$ and $Q$ are $3\times3$, and $P=Q$. When you see a product of multiple matrices, your first instinct should be to check dimensions at each step. For matrix multiplication, the inner dimensions must match, and the result takes the outer dimensions. Here, $A$ is $3\times2$, $B$ is $2\times4$, and $C$ is $4\times3$. For $P=(AB)C$, first compute $AB$: $3\times2$ times $2\times4$ gives a $3\times4$ matrix. Then multiply that by $C$ ($4\times3$): the inner $4$s match, so $P$ is $3\times3$. For $Q=A(BC)$, first $BC$ is $2\times4$ times $4\times3$, giving a $2\times3$ matrix. Then $A$ ($3\times2$) times that $2\times3$ matrix gives a $3\times3$ matrix. So both $P$ and $Q$ are $3\times3$. But are they equal? Yes — matrix multiplication is associative, meaning grouping does not change the result as long as the product is defined. So $P=Q$ always. The choice claiming they are $3\times3$ but not equal in general misses this fundamental property. The choices saying $P$ is $3\times4$ confuse the intermediate result $AB$ with the final product, and the one saying $Q$ is not defined or is $2\times3$ ignores that $BC$ is $2\times3$, which then multiplies cleanly with $A$. Remember: check dimensions first, then trust associativity — it always holds.
11. Let $A = \begin{bmatrix}2&3\\4&5\end{bmatrix}$. Which matrix is $A^{-1}$?
    Answer: $\begin{bmatrix}-\frac{5}{2}&\frac{3}{2}\\2&-1\end{bmatrix}$ When you need the inverse of a 2×2 matrix, recall the formula: for $A = \begin{bmatrix}a&b\\c&d\end{bmatrix}$, $A^{-1} = \frac{1}{ad-bc}\begin{bmatrix}d&-b\\-c&a\end{bmatrix}$. Here $a=2, b=3, c=4, d=5$, so the determinant is $2\cdot5-3\cdot4 = -2$. Thus $A^{-1} = \frac{1}{-2}\begin{bmatrix}5&-3\\-4&2\end{bmatrix} = \begin{bmatrix}-\frac{5}{2}&\frac{3}{2}\\2&-1\end{bmatrix}$. That is the matrix with those exact entries. Now check the other options. The matrix $\begin{bmatrix}\frac{5}{2}&\frac{3}{2}\\2&1\end{bmatrix}$ has all positive entries and a 1 in the bottom right; it likely comes from forgetting to divide by the negative determinant or ignoring the sign changes. The option $\begin{bmatrix}\frac{5}{2}&-\frac{3}{2}\\2&-1\end{bmatrix}$ has the correct off-diagonal signs but a positive top-left, which happens if you accidentally used a determinant of $+2$ instead of $-2$. The choice $\begin{bmatrix}-\frac{5}{2}&-\frac{3}{2}\\2&-1\end{bmatrix}$ has the top-right negative, which arises if you forget to negate $b$ (the 3) before multiplying by the reciprocal determinant. Strategy: Always compute the determinant first. If it's negative, remember every term gets a negative sign when you multiply by $1/\det$. Also, swap $a$ and $d$, negate $b$ and $c$, then multiply. Double-check each sign—this is where most sign errors occur.
12. For which value(s) of $x$ is the matrix $A = \begin{bmatrix}x&2\\3&x\end{bmatrix}$ not invertible?
    Answer: $x=\pm\sqrt{6}$ Whenever you see "not invertible" for a square matrix, your first thought should be determinant equal to zero. For a $2\times2$ matrix $\begin{bmatrix}a&b\\c&d\end{bmatrix}$, the determinant is $ad-bc$. Here, $\det(A)=x\cdot x-2\cdot3=x^2-6$. Setting $\det(A)=0$ gives $x^2-6=0$, so $x^2=6$. Taking square roots of both sides gives $x=\pm\sqrt{6}$. Both values make the rows (or columns) proportional, so either one alone is incomplete. Each wrong answer represents a common trap. Choosing only $x=\sqrt{6}$ or only x=-\sqrt{6}} treats the equation as if a square root had only one sign; but since $x^2=6$ has two solutions, you must include both. Choosing $x=\pm3$ likely comes from mistakenly writing the determinant as $x^2-9$ or confusing the 2 and3 entries with a product of 6 and square root of 6. It is not supported by the determinant calculation; plugging in $x=3$ gives determinant $9-6=3$, not zero, so the matrix is actually invertible there. For your study strategy, remember: a matrix is singular (not invertible) exactly when its determinant is zero. For a simple $2\times2$, write the determinant, set it equal to zero, and solve the resulting equation completely. When you solve $x^2=\text{positive number}$, never forget the $\pm$ — both signs are needed unless the problem gives a domain restriction.
13. Let $A=\begin{pmatrix}1&2\\0&1\end{pmatrix}$, $B=\begin{pmatrix}2&0\\1&3\end{pmatrix}$, and $I$ be the $2\times 2$ identity matrix. If $AX+B=I$, what is $X$?
    Answer: $\begin{pmatrix}1&4\\-1&-2\end{pmatrix}$ Since $AX+B=I$, we have $AX=I-B=\begin{pmatrix}-1&0\\-1&-2\end{pmatrix}$. Thus $X=A^{-1}(I-B)$. With $A^{-1}=\begin{pmatrix}1&-2\\0&1\end{pmatrix}$, multiplying gives $X=\begin{pmatrix}1&4\\-1&-2\end{pmatrix}$. Multiplying on the wrong side gives $(I-B)A^{-1}$, using the wrong inverse gives a sign error, and computing $A^{-1}B-I$ gives the other distractor.
14. If $A^{-1}=\begin{pmatrix}1&2\\0&1\end{pmatrix}$ and $B^{-1}=\begin{pmatrix}1&0\\-1&2\end{pmatrix}$, which of the following is the first row of $(AB)^{-1}$?
    Answer: $\begin{pmatrix}1&2\end{pmatrix}$ For invertible matrices, $(AB)^{-1}=B^{-1}A^{-1}$. So compute $B^{-1}A^{-1}=\begin{pmatrix}1&0\\-1&2\end{pmatrix}\begin{pmatrix}1&2\\0&1\end{pmatrix}=\begin{pmatrix}1&2\\-1&0\end{pmatrix}$. The first row is $\begin{pmatrix}1&2\end{pmatrix}$. The row $\begin{pmatrix}-1&4\end{pmatrix}$ comes from reversing the product order, and the other distractors are single rows copied with the order or sign confused.
15. Let $A=\begin{pmatrix}2&0\\0&3\end{pmatrix}$ and $B=\begin{pmatrix}0&1\\-1&0\end{pmatrix}$. What is $\det((A+B)(A-B))$?
    Answer: $49$ Compute directly: $A+B=\begin{pmatrix}2&1\\-1&3\end{pmatrix}$ has determinant $2(3)-1(-1)=7$, and $A-B=\begin{pmatrix}2&-1\\1&3\end{pmatrix}$ has determinant $2(3)-(-1)(1)=7$. Since the determinant of a product is the product of the determinants, $\det((A+B)(A-B))=7\cdot7=49$. Choice 50 comes from incorrectly replacing the product with $A^2-B^2$; choice 36 is simply $(\det A)^2$, omitting the contribution of $B$; choice 35 comes from applying the false determinant sum/difference identities to both factors, $(\det A+\det B)(\det A-\det B)=(6+1)(6-1)$.
16. If $2\begin{bmatrix}x&3\\-1&y\end{bmatrix}+\begin{bmatrix}4&-1\\2&5\end{bmatrix}=\begin{bmatrix}10&5\\0&-1\end{bmatrix}$, what is the solution $(x,y)$?
    Answer: $(3,\,-3)$ Whenever you see a matrix equation, remember that matrix addition and scalar multiplication are done entry by entry. So your first move is to simplify the left side into a single matrix, then set matching entries equal. Here, $2\begin{bmatrix}x&3\\-1&y\end{bmatrix}=\begin{bmatrix}2x&6\\-2&2y\end{bmatrix}$. Adding the second matrix gives $\begin{bmatrix}2x+4&5\\0&2y+5\end{bmatrix}$. Now equate this to $\begin{bmatrix}10&5\\0&-1\end{bmatrix}$. From the top-left entry: $2x+4=10$, so $2x=6$ and $x=3$. From the bottom-right entry: $2y+5=-1$, so $2y=-6$ and $y=-3$. Thus the solution is $(3,-3)$. The choice $(3,3)$ gets the x-value right but makes a sign error on y: if y were 3, then $2y+5=11$, not $-1$. The choice $(-3,3)$ misses both signs; substituting x=-3 gives $2(-3)+4=-2$, not 10. The choice $(6,-6)$ comes from solving $x+4=10$ instead of $2x+4=10$ — a common mistake of forgetting to divide by the scalar 2. It also gives $2(-6)+5=-7$, which does not match $-1$. Study tip: for matrix equations, always simplify the entire left side first, then write one small equation for each corresponding entry. Check your arithmetic for both the scalar multiplication and the sign when moving constants across the equation.
17. Let $A(k)=\begin{pmatrix}1&2&k\\0&1&1\\k&0&2\end{pmatrix}$. How many positive real values of $k$ make $A(k)$ not invertible?
    Answer: $1$ Expanding along the first row, $\det(A(k))=1(2)-2(-k)+k(-k)=2+2k-k^2$. Setting this equal to zero gives $k^2-2k-2=0$, so $k=1\pm\sqrt3$. Only $1+\sqrt3$ is positive, so there is exactly one positive real value. The choice 2 would come from counting both roots without checking the positivity condition.
18. Let $A=\begin{pmatrix}2&5\\1&3\end{pmatrix}$, $B=\begin{pmatrix}1&2\\-1&0\end{pmatrix}$, and $C=\begin{pmatrix}3&7\\0&-2\end{pmatrix}$. If $AX+B=C$, what is the entry in row 1, column 2 of $X$?
    Answer: 25 First isolate the matrix product: $AX=C-B$. Since $\det(A)=1$, $A^{-1}=\begin{pmatrix}3&-5\\-1&2\end{pmatrix}$. Left-multiplying gives $X=A^{-1}(C-B)=\begin{pmatrix}1&25\\0&-9\end{pmatrix}$, so the requested entry is $25$. The $0$ option comes from computing $(C-B)A^{-1}$ instead; the $5$ option comes from sign errors in $A^{-1}$; the $31$ option comes from solving $AX=C$ and ignoring $+B$.
19. Let $A=\begin{pmatrix}x&2\\3&4\end{pmatrix}$ and $B=\begin{pmatrix}1&-1\\2&0\end{pmatrix}$. If $AB=\begin{pmatrix}6&-2\\11&-3\end{pmatrix}$, what is the entry in row 1, column 2 of $A^{-1}$?
    Answer: $-1$ From the first row of the product, $x+4=6$, so $x=2$. Then $\det(A)=2(4)-2(3)=2$. Using the inverse formula for a 2x2 matrix, $A^{-1}=\frac12\begin{pmatrix}4&-2\\-3&2\end{pmatrix}$, so row 1, column 2 is $-1$. The choice $-\frac12$ comes from using the wrong determinant; 0 and 1 come from sign or position errors in the inverse.
20. Let $A=\begin{pmatrix}1&2\\3&4\end{pmatrix}$ and $B=\begin{pmatrix}0&1\\1&0\end{pmatrix}$. What is $(A+B)(A-B)-(A^2-B^2)$?
    Answer: $\begin{pmatrix}1&3\\-3&-1\end{pmatrix}$ Expanding gives $(A+B)(A-B)=A^2-AB+BA-B^2$, so the difference is $BA-AB$ because matrix multiplication is not commutative. Here $AB=\begin{pmatrix}2&1\\4&3\end{pmatrix}$ and $BA=\begin{pmatrix}3&4\\1&2\end{pmatrix}$, so $BA-AB=\begin{pmatrix}1&3\\-3&-1\end{pmatrix}$. The zero matrix would be true only if $AB=BA$, and the negative choice is the reversed difference $AB-BA$.

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