Algebra 3 Quiz: Matrix Operations
12 questions · exam conditions
0:00
Matrix OperationsQuestion 1 of 12

Let A=[2−103]A = \begin{bmatrix}2&-1\\0&3\end{bmatrix} and B=[12−14]B = \begin{bmatrix}1&2\\-1&4\end{bmatrix}. Which matrix equals (A+B)(A−B)(A+B)(A-B)?

[4−106−4]\begin{bmatrix}4&-10\\6&-4\end{bmatrix}
[5−155−5]\begin{bmatrix}5&-15\\5&-5\end{bmatrix}
[6−204−6]\begin{bmatrix}6&-20\\4&-6\end{bmatrix}
[4−1064]\begin{bmatrix}4&-10\\6&4\end{bmatrix}
← Back to quizzes

Algebra 3 Quiz

Algebra 3 Quiz: Matrix Operations

Practice Matrix Operations in Algebra 3 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Matrix Operations, giving you a quick way to practice the rules, question types, and explanations that matter most for Algebra 3.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Let A=[2−103]A = \begin{bmatrix}2&-1\\0&3\end{bmatrix} and B=[12−14]B = \begin{bmatrix}1&2\\-1&4\end{bmatrix}. Which matrix equals (A+B)(A−B)(A+B)(A-B)?

  1. [4−106−4]\begin{bmatrix}4&-10\\6&-4\end{bmatrix} (correct answer)
  2. [5−155−5]\begin{bmatrix}5&-15\\5&-5\end{bmatrix}
  3. [6−204−6]\begin{bmatrix}6&-20\\4&-6\end{bmatrix}
  4. [4−1064]\begin{bmatrix}4&-10\\6&4\end{bmatrix}
Explanation: When you see an expression like (A+B)(A−B)(A+B)(A-B) with matrices, the critical thing to remember is that matrix multiplication is not commutative: expanding gives A2−AB+BA−B2A^2-AB+BA-B^2, not simply A2−B2A^2-B^2. That trap is built into the choices. Start by computing the two factors directly: A+B=[31−17],A−B=[1−31−1].A+B=\begin{bmatrix}3&1\\-1&7\end{bmatrix}, \qquad A-B=\begin{bmatrix}1&-3\\1&-1\end{bmatrix}. Then multiply: =\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.

Question 2

Let A=[1−2304−1]A = \begin{bmatrix}1&-2&3\\0&4&-1\end{bmatrix} and B=[20−1341]B = \begin{bmatrix}2&0\\-1&3\\4&1\end{bmatrix}. What is the entry (AB)2,1(AB)_{2,1}?

  1. 1616
  2. −8-8 (correct answer)
  3. 1111
  4. −3-3
Explanation: Whenever you see a matrix product like ABAB, you're working with row‑by‑column multiplication: entry (AB)i,j(AB)_{i,j} is the dot product of row ii from AA with column jj from BB. The question asks for (AB)2,1(AB)_{2,1}, so you need row 2 of AA and column 1 of BB. Row 2 of AA is [0,4,−1][0,4,-1], column 1 of BB is [2,−1,4]T[2,-1,4]^T. Multiply and sum: 0⋅2+4⋅(−1)+(−1)⋅4=0−4−4=−80\cdot 2 + 4\cdot(-1) + (-1)\cdot 4 = 0 -4 -4 = -8. That's the correct answer. Now, the other choices are the other entries of the product. Computing the full 2×22\times2 matrix gives: AB=[1⋅2+(−2)(−1)+3⋅41⋅0+(−2)3+3⋅10⋅2+4(−1)+(−1)40⋅0+4⋅3+(−1)1]=[16−3−811].AB = \begin{bmatrix} 1\cdot2+(-2)(-1)+3\cdot4 & 1\cdot0+(-2)3+3\cdot1\\ 0\cdot2+4(-1)+(-1)4 & 0\cdot0+4\cdot3+(-1)1 \end{bmatrix} = \begin{bmatrix} 16 & -3\\ -8 & 11 \end{bmatrix}. So 1616 is (AB)1,1(AB)_{1,1} — a common slip where you pick the first row instead of the second. 1111 is (AB)2,2(AB)_{2,2} — mixing up row and column indices. −3-3 is (AB)1,2(AB)_{1,2} — again, the wrong cell. Each distractor is simply a different entry of the product, not a computational error. Study tip: Before doing any arithmetic, write down which row and column you need. Circle them in the matrices. This prevents the most frequent mistake on this type of question — grabbing the wrong entry from the final product.

Question 3

For which value(s) of xx is the matrix A=[x23x]A = \begin{bmatrix}x&2\\3&x\end{bmatrix} not invertible?

  1. x=6x=\sqrt{6}
  2. x=−6x=-\sqrt{6}
  3. x=±6x=\pm\sqrt{6} (correct answer)
  4. x=±3x=\pm 3
Explanation: Whenever you see "not invertible" for a square matrix, your first thought should be determinant equal to zero. For a 2×22\times2 matrix [abcd]\begin{bmatrix}a&b\\c&d\end{bmatrix}, the determinant is ad−bcad-bc. Here, det⁡(A)=x⋅x−2⋅3=x2−6\det(A)=x\cdot x-2\cdot3=x^2-6. Setting det⁡(A)=0\det(A)=0 gives x2−6=0x^2-6=0, so x2=6x^2=6. Taking square roots of both sides gives x=±6x=\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=6x=\sqrt{6} or only x=-\sqrt{6}} treats the equation as if a square root had only one sign; but since x2=6x^2=6 has two solutions, you must include both. Choosing x=±3x=\pm3 likely comes from mistakenly writing the determinant as x2−9x^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=3x=3 gives determinant 9−6=39-6=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 2×22\times2, write the determinant, set it equal to zero, and solve the resulting equation completely. When you solve x2=positive numberx^2=\text{positive number}, never forget the ±\pm — both signs are needed unless the problem gives a domain restriction.

Question 4

Let A=[0110]A = \begin{bmatrix}0&1\\1&0\end{bmatrix}. Which form must a matrix B=[abcd]B=\begin{bmatrix}a&b\\c&d\end{bmatrix} have so that AB=BAAB=BA?

  1. [ab−ba]\begin{bmatrix}a&b\\-b&a\end{bmatrix}
  2. [abba]\begin{bmatrix}a&b\\b&a\end{bmatrix} (correct answer)
  3. [abab]\begin{bmatrix}a&b\\a&b\end{bmatrix}
  4. \begin{bmatrix}a&b\\c&d\end{bmatrix} $$ with $a=c$ and $b=d$.
Explanation: 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 AA simply swaps coordinates, so comparing ABAB and BABA will force certain entries of BB to match. . Computing directly:AB=[0110][abcd]=[cdab],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},whileBA=[abcd][0110]=[badc].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=bc=b, d=ad=a, and also a=da=d, b=cb=c—these are the same conditions. So the general commuter must have form[abba]. \begin{bmatrix}a&b\\b&a\end{bmatrix}. The candidate [ab−ba]\begin{bmatrix}a&b\\-b&a\end{bmatrix} would require c=−bc=-b and d=ad=a. Then AB=[−baab]AB=\begin{bmatrix}-b&a\\a&b\end{bmatrix} while BA=[baa−b]BA=\begin{bmatrix}b&a\\a&-b\end{bmatrix}; these match only when b=0b=0, so it is not a general solution. The repeated-row forms—both the explicit [abab]\begin{bmatrix}a&b\\a&b\end{bmatrix} and the equivalent parameter statement a=c, b=da=c,\ b=d—give AB=[abab]AB=\begin{bmatrix}a&b\\a&b\end{bmatrix} but BA=[baba]BA=\begin{bmatrix}b&a\\b&a\end{bmatrix}; equality would force a=ba=b, again only a special case. Therefore, only the matrix with matching off-diagonal entries works for all aa and bb. Study tip: if you are unsure, test a simple numerical matrix such as a=1, b=2a=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.

Question 5

Suppose AA is a 3×33\times3 matrix with det⁡(A)=5\det(A)=5. Let BB be the matrix obtained from AA by swapping two rows and then multiplying one row by 2. What is det⁡(B)\det(B)?

  1. −5-5
  2. 1010
  3. 55
  4. −10-10 (correct answer)
Explanation: Whenever you see a question about how row operations change a determinant, remember the two key rules: swapping two rows multiplies the determinant by −1-1, and multiplying one row by a constant multiplies the determinant by that constant. Here, start with det⁡(A)=5\det(A)=5. Swapping two rows gives 5⋅(−1)=−55 \cdot (-1) = -5. Then multiplying one row by 22 doubles that result, so det⁡(B)=−5⋅2=−10\det(B) = -5 \cdot 2 = -10. The choice −5-5 reflects only the row swap, forgetting the later multiplication by 22. The choice 1010 correctly multiplies by 22 but loses the sign change from the swap. The choice 55 ignores both operations entirely, as if BB had the same determinant as AA. 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 cc scales the determinant by cc.

Question 6

Let A=[10202112k]A = \begin{bmatrix}1&0&2\\0&2&1\\1&2&k\end{bmatrix}. If det⁡(A)=4\det(A)=4, what is kk?

  1. −1-1
  2. 33
  3. 55 (correct answer)
  4. −3-3
Explanation: Whenever you see a determinant equation with a variable, your job is to compute the determinant as an expression in kk, set it equal to the given value, and solve. Here, expanding along the first row is efficient because of the 00 in the middle. det⁡(A)=1⋅∣212k∣−0+2⋅∣0212∣\det(A)=1\cdot \begin{vmatrix}2&1\\2&k\end{vmatrix}-0+2\cdot \begin{vmatrix}0&2\\1&2\end{vmatrix} That becomes (2k−2)+2(0⋅2−2⋅1)=2k−2−4=2k−6.(2k-2)+2(0\cdot 2-2\cdot 1)=2k-2-4=2k-6. Set 2k−6=42k-6=4, so 2k=102k=10, and therefore k=5k=5. That is why the correct answer is 55. The other choices come from common mistakes. Choosing 33 usually means you stopped after the first 2×22\times2 determinant: 2k−2=42k-2=4 gives k=3k=3, but you forgot the −4-4 contribution from the 22 in the first row. Choosing −1-1 comes from solving 2k−6=42k-6=4 incorrectly as 2k=4−62k=4-6; you need to add 66 to both sides, not subtract it. Choosing −3-3 comes from applying the wrong sign to the first minor: using −(2k−2)+2(−2)=4-(2k-2)+2(-2)=4 gives −2k−2=4-2k-2=4, so k=−3k=-3. Remember, the (1,1)(1,1) entry has a positive sign in the cofactor expansion. Strategy: expand along a row or column with zeros, keep the alternating signs straight, and check your answer by plugging kk back into the determinant.

Question 7

Let A=[2345]A = \begin{bmatrix}2&3\\4&5\end{bmatrix}. Which matrix is A−1A^{-1}?

  1. [523221]\begin{bmatrix}\frac{5}{2}&\frac{3}{2}\\2&1\end{bmatrix}
  2. [52−322−1]\begin{bmatrix}\frac{5}{2}&-\frac{3}{2}\\2&-1\end{bmatrix}
  3. [−52−322−1]\begin{bmatrix}-\frac{5}{2}&-\frac{3}{2}\\2&-1\end{bmatrix}
  4. [−52322−1]\begin{bmatrix}-\frac{5}{2}&\frac{3}{2}\\2&-1\end{bmatrix} (correct answer)
Explanation: When you need the inverse of a 2×2 matrix, recall the formula: for A=[abcd]A = \begin{bmatrix}a&b\\c&d\end{bmatrix}, A−1=1ad−bc[d−b−ca]A^{-1} = \frac{1}{ad-bc}\begin{bmatrix}d&-b\\-c&a\end{bmatrix}. Here a=2,b=3,c=4,d=5a=2, b=3, c=4, d=5, so the determinant is 2⋅5−3⋅4=−22\cdot5-3\cdot4 = -2. Thus A−1=1−2[5−3−42]=[−52322−1]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 [523221]\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 [52−322−1]\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+2 instead of −2-2. The choice [−52−322−1]\begin{bmatrix}-\frac{5}{2}&-\frac{3}{2}\\2&-1\end{bmatrix} has the top-right negative, which arises if you forget to negate bb (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⁡1/\det. Also, swap aa and dd, negate bb and cc, then multiply. Double-check each sign—this is where most sign errors occur.

Question 8

Let AA and BB be 3×33\times3 matrices with det⁡(A)=−3\det(A)=-3 and det⁡(B)=2\det(B)=2. What is det⁡(2AB−1)\det(2AB^{-1})?

  1. 1212
  2. −48-48
  3. −12-12 (correct answer)
  4. −3-3
Explanation: 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)=cndet⁡(A)\det(cA)=c^n\det(A) for an n×nn\times n matrix, det⁡(B−1)=1/det⁡(B)\det(B^{-1})=1/\det(B), and det⁡(AB)=det⁡(A)det⁡(B)\det(AB)=\det(A)\det(B). Here AA and BB are 3×33\times3, so the scalar multiple 2A2A contributes a factor of 23=82^3=8, not just 22. Thus det⁡(2AB−1)=23det⁡(A)det⁡(B−1)=8(−3)(12)=−12.\det(2AB^{-1})=2^3\det(A)\det(B^{-1}) =8(-3)\left(\frac12\right)=-12. The choice 1212 confuses the sign: it likely drops the negative from det⁡(A)=−3\det(A)=-3. The choice −48-48 uses det⁡(B)=2\det(B)=2 directly instead of det⁡(B−1)=1/2\det(B^{-1})=1/2, effectively multiplying by 22 instead of dividing by 22. The choice −3-3 correctly handles the inverse but forgets that multiplying AA by 22 scales the determinant by 232^3, not 22. Keep the order of operations straight: take the scalar multiple to the third power, invert the determinant of BB, then multiply by det⁡(A)\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.

Question 9

Given that A−1=[1234]A^{-1}=\begin{bmatrix}1&2\\3&4\end{bmatrix} and B−1=[01−12]B^{-1}=\begin{bmatrix}0&1\\-1&2\end{bmatrix}, what is (AB)−1(AB)^{-1}?

  1. [−25411]\begin{bmatrix}-2&5\\4&11\end{bmatrix}
  2. [−25−411]\begin{bmatrix}-2&5\\-4&11\end{bmatrix}
  3. [3465]\begin{bmatrix}3&4\\6&5\end{bmatrix}
  4. [3456]\begin{bmatrix}3&4\\5&6\end{bmatrix} (correct answer)
Explanation: Whenever you see an inverse of a product, recall the defining property: (AB)−1=B−1A−1.(AB)^{-1}=B^{-1}A^{-1}. The multiplication order reverses. So compute B−1A−1=[01−12][1234]=[3456].B^{-1}A^{-1}=\begin{bmatrix}0&1\\-1&2\end{bmatrix}\begin{bmatrix}1&2\\3&4\end{bmatrix}=\begin{bmatrix}3&4\\5&6\end{bmatrix}. Thus the matrix $$ \begin{bmatrix}3&4\5&6\end{bmatrix} The choice $$ \begin{bmatrix}-2&5\\-4&11\end{bmatrix} $$ comes from computing $A^{-1}B^{-1}$ instead of $B^{-1}A^{-1}$ — a common sign that you forgot the order reversal. The choice $$ \begin{bmatrix}-2&5\\4&11\end{bmatrix} $$ is also $A^{-1}B^{-1}$ but with an additional sign error: the bottom-left entry should be $-4$, because $3(0)+4(-1)=-4$, not $4$. The choice $$ \begin{bmatrix}3&4\\6&5\end{bmatrix} $$ has the correct first row, but the bottom row entries $5$ and $6$ are swapped; this is a row/column placement error when recording $B^{-1}A^{-1}$. Strategy: immediately write $(AB)^{-1}=B^{-1}A^{-1}$ before multiplying. Then verify one entry, such as bottom-left: $-1(1)+2(3)=5$. If that entry isn't $5$, your order or arithmetic is off.

Question 10

Let AA be 3×23\times2, let BB be 2×42\times4, and let CC be 4×34\times3. Which statement is true about P=(AB)CP=(AB)C and Q=A(BC)Q=A(BC)?

  1. Both PP and QQ are 3×33\times3, but P≠QP\neq Q in general.
  2. Both PP and QQ are 3×33\times3, and P=QP=Q. (correct answer)
  3. PP is 3×43\times4, but QQ is not defined.
  4. PP is 3×43\times4, and QQ is 2×32\times3.
Explanation: 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, AA is 3×23\times2, BB is 2×42\times4, and CC is 4×34\times3. For P=(AB)CP=(AB)C, first compute ABAB: 3×23\times2 times 2×42\times4 gives a 3×43\times4 matrix. Then multiply that by CC (4×34\times3): the inner 44s match, so PP is 3×33\times3. For Q=A(BC)Q=A(BC), first BCBC is 2×42\times4 times 4×34\times3, giving a 2×32\times3 matrix. Then AA (3×23\times2) times that 2×32\times3 matrix gives a 3×33\times3 matrix. So both PP and QQ are 3×33\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=QP=Q always. The choice claiming they are 3×33\times3 but not equal in general misses this fundamental property. The choices saying PP is 3×43\times4 confuse the intermediate result ABAB with the final product, and the one saying QQ is not defined or is 2×32\times3 ignores that BCBC is 2×32\times3, which then multiplies cleanly with AA. Remember: check dimensions first, then trust associativity — it always holds.

Question 11

If 2[x3−1y]+[4−125]=[1050−1]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)(x,y)?

  1. (3, −3)(3,\,-3) (correct answer)
  2. (3, 3)(3,\,3)
  3. (−3, 3)(-3,\,3)
  4. (6, −6)(6,\,-6)
Explanation: 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[x3−1y]=[2x6−22y]2\begin{bmatrix}x&3\\-1&y\end{bmatrix}=\begin{bmatrix}2x&6\\-2&2y\end{bmatrix} . Adding the second matrix gives [2x+4502y+5]\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.

Question 12

Let A=[2513]A = \begin{bmatrix}2&5\\1&3\end{bmatrix} and C=[4723]C = \begin{bmatrix}4&7\\2&3\end{bmatrix}. If AX=CAX=C, what is the matrix XX?

  1. [260−1]\begin{bmatrix}2&6\\0&-1\end{bmatrix} (correct answer)
  2. [5−63−4]\begin{bmatrix}5&-6\\3&-4\end{bmatrix}
  3. [−2−601]\begin{bmatrix}-2&-6\\0&1\end{bmatrix}
  4. [3−5−12]\begin{bmatrix}3&-5\\-1&2\end{bmatrix}
Explanation: Whenever you see an equation like AX=CAX=C, the key is to isolate XX by multiplying both sides on the left by A−1A^{-1}: X=A−1CX = A^{-1}C. Here, AA's determinant is (2)(3)−(5)(1)=1(2)(3)-(5)(1)=1, so A−1=[3−5−12]A^{-1} = \begin{bmatrix}3&-5\\-1&2\end{bmatrix}. Then multiply: [3−5−12][4723]=[260−1]\begin{bmatrix}3&-5\\-1&2\end{bmatrix}\begin{bmatrix}4&7\\2&3\end{bmatrix} = \begin{bmatrix}2&6\\0&-1\end{bmatrix}. That matches the correct answer. The choice [5−63−4]\begin{bmatrix}5&-6\\3&-4\end{bmatrix} comes from multiplying CC by A−1A^{-1} in the opposite order (CA−1C A^{-1}), which would solve XA=CXA=C, not AX=CAX=C. The choice [−2−601]\begin{bmatrix}-2&-6\\0&1\end{bmatrix} is just the negative of the correct answer—a sign error, likely from flipping the signs in the inverse incorrectly. The choice [3−5−12]\begin{bmatrix}3&-5\\-1&2\end{bmatrix} is A−1A^{-1} itself, but you still need to apply it to CC. Always remember: matrix multiplication is not commutative, so order matters—multiply on the left when solving AX=CAX=C. Also double-check your signs when computing the inverse.