Class xii matrices, determinants worksheet 2 | PDF - Free Printable

Let’s solve each problem one by one, carefully and step by step.

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Problem 7:

We are given a 3×3 matrix M:

M =
[ -1 ? ? ]
[ 2 ? ? ]
[ 0 5 ? ]

And we’re told:

- M · [1; 1; 0] = [-1; 1; 5]
- M · [1; 0; -1] = [0; 1; 12]

We need to find the sum of the diagonal entries of M — that is, if M = [a_ij], then sum = a₁₁ + a₂₂ + a₃₃.

We already know from the matrix shown:
a₁₁ = -1 (top-left)
a₂₁ = 2
a₃₁ = 0
a₃₂ = 5

Let’s denote the unknowns:

Let M =
[ -1 x y ]
[ 2 p q ]
[ 0 5 r ]

So diagonal entries: -1, p, r → sum = -1 + p + r

Now use the first equation:

M · [1; 1; 0] =
Row1: (-1)(1) + x(1) + y(0) = -1 + x
Row2: (2)(1) + p(1) + q(0) = 2 + p
Row3: (0)(1) + 5(1) + r(0) = 5

This equals [-1; 1; 5]

So:

-1 + x = -1 → x = 0
2 + p = 1 → p = -1
5 = 5 ✓ (checks out)

Now second equation:

M · [1; 0; -1] =
Row1: (-1)(1) + x(0) + y(-1) = -1 - y
Row2: (2)(1) + p(0) + q(-1) = 2 - q
Row3: (0)(1) + 5(0) + r(-1) = -r

This equals [0; 1; 12]

So:

-1 - y = 0 → y = -1
2 - q = 1 → q = 1
-r = 12 → r = -12

Now we have all diagonal entries:

a₁₁ = -1
a₂₂ = p = -1
a₃₃ = r = -12

Sum = -1 + (-1) + (-12) = -14

Wait — but looking at options:

(a) 1
(b) 3
(c) 6
(d) 9

None is -14. Did I make a mistake?

Wait — let me double-check.

From first multiplication:

M * [1,1,0]^T = [-1,1,5]^T

Row1: (-1)*1 + x*1 + y*0 = -1 + x = -1 → x=0 ✔️
Row2: 2*1 + p*1 + q*0 = 2 + p = 1 → p = -1 ✔️
Row3: 0*1 + 5*1 + r*0 = 5 ✔️

Second multiplication:

M * [1,0,-1]^T = [0,1,12]^T

Row1: (-1)*1 + x*0 + y*(-1) = -1 - y = 0 → y = -1 ✔️
Row2: 2*1 + p*0 + q*(-1) = 2 - q = 1 → q = 1 ✔️
Row3: 0*1 + 5*0 + r*(-1) = -r = 12 → r = -12 ✔️

Diagonal: -1 (from given), p=-1, r=-12 → sum = -14

But -14 not in options. Hmm.

Wait — maybe the matrix was written with some entries missing? Let me re-read the problem.

It says: “Let M be a 3×3 matrix satisfying” and shows:

M [ -1 ? ? ]
[ 2 ? ? ]
[ 0 5 ? ]

Actually, looking again — perhaps the first column is given as [-1, 2, 0]^T? That matches what I used.

But answer isn’t among options. Maybe I misread the target vectors?

First product: M * [1;1;0] = [-1;1;5] — yes.

Second: M * [1;0;-1] = [0;1;12] — yes.

Wait — perhaps the question is asking for something else? “Sum of the diagonal entries” — trace.

Maybe there's a typo in my calculation? Or in the problem?

Alternatively — perhaps the matrix is transposed? No.

Wait — let me compute the full matrix now:

M =
[ -1 0 -1 ]
[ 2 -1 1 ]
[ 0 5 -12 ]

Check first product:

[ -1 0 -1 ] [1] = -1*1 + 0*1 + (-1)*0 = -1
[ 2 -1 1 ] [1] = 2*1 + (-1)*1 + 1*0 = 1
[ 0 5 -12 ] [0] = 0*1 + 5*1 + (-12)*0 = 5 → correct.

Second product:

[ -1 0 -1 ] [1] = -1*1 + 0*0 + (-1)*(-1) = -1 + 1 = 0
[ 2 -1 1 ] [0] = 2*1 + (-1)*0 + 1*(-1) = 2 -1 = 1
[ 0 5 -12 ] [-1] = 0*1 + 5*0 + (-12)*(-1) = 12 → correct.

So matrix is correct. Diagonal: -1, -1, -12 → sum = -14.

But options don't include -14. This suggests either:

- There’s a typo in the problem or options.
- Or I misread the matrix structure.

Wait — look back at original image description (even though I shouldn’t describe it, but for solving):

The user wrote:

“M [ -1 ? ? ]
[ 2 ? ? ]
[ 0 5 ? ]”

But perhaps the first row is [-1, ?, ?], second [?, ?, ?], third [?, ?, ?]? No, it says “M [ -1 ... ]” etc., implying first column is -1,2,0.

Another possibility: maybe the vector multiplications are on the left? But no, standard is M times column vector.

Perhaps the question meant “sum of absolute values”? |-1| + |-1| + |-12| = 14 — still not in options.

Or maybe I miscalculated r?

Third row of second product: M_row3 • [1,0,-1] = 0*1 + 5*0 + r*(-1) = -r = 12 → r = -12. Correct.

Unless... wait! The problem says: “Then the sum of the diagonal entries of M is”

Options: (a)1 (b)3 (c)6 (d)9

All positive small numbers. My answer is -14. Something’s wrong.

Wait — let me check if the matrix is written differently.

In the text: “M [ -1 ? ? ]” — could this mean the first ROW is [-1, ?, ?]? That would change everything!

Ah! That might be it. In many texts, when they write:

M = [ a b c ]
[ d e f ]
[ g h i ]

They list rows.

But here it says: “M [ -1 ? ? ]” — likely meaning first row is [-1, ?, ?]

Similarly, next line “[ 2 ? ? ]” — second row starts with 2? But that doesn’t match typical notation.

Wait — look at the way it's written in the user input:

“M [ -1 ? ? ]
[ 2 ? ? ]
[ 0 5 ? ]”

This is ambiguous. But in standard math notation, when you write a matrix like that, each bracketed line is a row.

So probably:

Row 1: [-1, ?, ?]
Row 2: [2, ?, ?]
Row 3: [0, 5, ?]

That makes more sense with the options being small positives.

Let me reinterpret with that.

Assume:

M =
[ -1 x y ]
[ 2 p q ]
[ 0 5 r ]

Same as before? Wait, same as what I did earlier. Because I assumed first column was [-1,2,0], but if it's rows, then first row is [-1,x,y], second [2,p,q], third [0,5,r] — which is exactly what I used!

Because in my setup, I had:

Row1: -1, x, y
Row2: 2, p, q
Row3: 0, 5, r

Yes — so my interpretation was correct.

But then why is sum -14 not in options?

Unless... the problem has a different set of equations.

Let me read again:

“M [1;1;0] = [-1;1;5]” — that’s column vector.

“M [1;0;-1] = [0;1;12]”

With M as above, we got consistent results.

Perhaps the question is not asking for trace, but something else? “Sum of the diagonal entries” — definitely trace.

Another idea: maybe the matrix is symmetric? Not stated.

Or perhaps I need to find only the known diagonals? But no, we found all.

Wait — let's calculate the actual diagonal from our matrix:

M =
[ -1 0 -1 ]
[ 2 -1 1 ]
[ 0 5 -12 ]

Diagonal: position (1,1)= -1, (2,2)= -1, (3,3)= -12 → sum -14.

But since -14 is not an option, and the closest thing is perhaps I made a sign error.

In the second product, third component: M_row3 • [1,0,-1] = 0*1 + 5*0 + r*(-1) = -r

Set equal to 12 → -r = 12 → r = -12. Correct.

Unless the vector is [1;0;1]? But it says [1;0;-1].

Perhaps the result is [0;1;-12]? But it says [0;1;12].

I think there might be a typo in the problem or options. But since this is a multiple-choice and -14 isn't there, let me see if any other interpretation works.

Alternative approach: suppose we don't assume the first column, but rather the first row is given as [-1, a, b], etc.

But that's what I did.

Perhaps "diagonal entries" means something else? No.

Another thought: maybe the matrix is 3x3, but the given parts are not the first column, but specific positions.

Look at the way it's written: “M [ -1 ? ? ]” — this might mean the element at (1,1) is -1, and others unknown.

Then “[ 2 ? ? ]” — this might mean (2,1) = 2? Or (2,2)? Ambiguous.

In many worksheets, when they write:

M = [ -1 _ _ ]
[ 2 _ _ ]
[ 0 5 _ ]

It usually means:

Row 1: col1=-1, col2=?, col3=?
Row 2: col1=2, col2=?, col3=?
Row 3: col1=0, col2=5, col3=?

Which is what I used.

Perhaps the '2' is at (2,2)? Let's try that.

Suppose the given entries are:

M_11 = -1
M_22 = 2 (instead of M_21=2)
M_31 = 0
M_32 = 5

That could be possible if the formatting is off.

Let me try that interpretation.

Assume:

M =
[ -1 x y ]
[ a 2 b ] // here M_22=2, not M_21=2
[ 0 5 c ]

Now, diagonal entries: -1, 2, c → sum = 1 + c

Now use the equations.

First: M * [1;1;0] = [-1;1;5]

Compute:

Row1: (-1)(1) + x(1) + y(0) = -1 + x = -1 → x=0

Row2: a(1) + 2(1) + b(0) = a + 2 = 1 → a = -1

Row3: 0(1) + 5(1) + c(0) = 5 = 5 ✓

Second: M * [1;0;-1] = [0;1;12]

Row1: (-1)(1) + x(0) + y(-1) = -1 - y = 0 → y = -1

Row2: a(1) + 2(0) + b(-1) = a - b = 1

But a = -1, so -1 - b = 1 → b = -2

Row3: 0(1) + 5(0) + c(-1) = -c = 12 → c = -12

Now diagonal: M_11=-1, M_22=2, M_33=c=-12 → sum = -1 + 2 + (-12) = -11

Still not in options.

If M_22 is not 2, but something else.

Perhaps the '2' is M_12? Unlikely.

Another idea: perhaps the matrix is:

From the text: "M [ -1 ? ? ]" might mean the first row is [-1, ?, ?], and "[ 2 ? ? ]" means the second row is [?, 2, ?], i.e., M_22=2.

Let's try that.

Assume:

M =
[ -1 x y ]
[ a 2 b ]
[ 0 5 c ]

Same as above, got sum -11.

Not helping.

Perhaps the '0' and '5' are in row 3, columns 2 and 3? Let's try.

Suppose:

M =
[ -1 x y ]
[ a b c ]
[ d 0 5 ] // but the text says "[ 0 5 ? ]", which would be row 3: 0,5,?

I think I need to accept that with the standard interpretation, sum is -14, but since it's not an option, and this is a worksheet, perhaps there's a mistake in the problem, or in my reading.

Let me look at the options again: 1,3,6,9 — all positive, small.

Perhaps the question is to find the sum of the squares or something, but no.

Another thought: maybe "diagonal entries" means the main diagonal, but in the context, perhaps they want the sum after finding, but I have it.

Perhaps I miscalculated the second product's third component.

M * [1;0;-1] , third row: if M_row3 = [0,5,r], then dot with [1,0,-1] = 0*1 + 5*0 + r*(-1) = -r

Set to 12, so r = -12.

But if the result is [0;1;-12], then -r = -12, r=12, then diagonal sum = -1 + (-1) + 12 = 10, not in options.

If r=12, sum=10.

Still not.

Perhaps the first product's third component is not 5, but something else.

I recall that in some problems, they give two conditions to solve for unknowns, and here we have three unknowns in diagonal, but we can find them.

Perhaps the matrix is such that the diagonal sum is to be found without finding all, but we need to.

Let's denote the diagonal as d1, d2, d3.

From the matrix, d1 = -1 (given).

From first equation, when we multiply by [1,1,0], the second component is M21*1 + M22*1 + M23*0 = M21 + M22 = 1

But if M21 is given as 2, then 2 + M22 = 1, so M22 = -1.

Similarly, from second equation, third component: M31*1 + M32*0 + M33*(-1) = M31 - M33 = 12

If M31 = 0, then - M33 = 12, M33 = -12.

Same as before.

Unless M31 is not 0. In the matrix, it's written as "[ 0 5 ? ]" for the third row, so M31=0, M32=5.

I think there might be a typo in the problem or options. But since this is a test, and -14 is not there, perhaps the intended answer is based on different given.

Let me try to force the sum to be one of the options.

Suppose sum is 9, then -1 + p + r = 9, so p + r = 10.

From first equation, 2 + p = 1, so p = -1, then r = 11.

From second equation, -r = 12, so r = -12, contradiction.

If sum is 6, p + r = 7, p= -1, r=8, but -r=12 implies r= -12, no.

Perhaps the second product's third component is -12, not 12.

If M * [1;0;-1] = [0;1;-12], then -r = -12, r=12.

Then diagonal: -1, -1, 12 = 10, not in options.

If r=12, and p= -1, sum=10.

Still not.

Another idea: perhaps the '5' is M33, not M32.

Let's try that.

Assume the matrix is:

Row 1: [-1, x, y]
Row 2: [2, p, q]
Row 3: [0, s, 5] // so M33=5

Then diagonal: -1, p, 5 → sum = 4 + p

From first product: M*[1;1;0] = [-1;1;5]

Row1: -1 + x = -1 → x=0

Row2: 2 + p = 1 → p= -1

Row3: 0 + s*1 + 5*0 = s = 5 → s=5

From second product: M*[1;0;-1] = [0;1;12]

Row1: -1 + 0 + y*(-1) = -1 - y = 0 → y= -1

Row2: 2 + 0 + q*(-1) = 2 - q = 1 → q=1

Row3: 0 + s*0 + 5*(-1) = -5 = 12? Contradiction, -5 ≠ 12.

So not.

If M33=5, then in second product, third component is -5, but should be 12, impossible.

So that doesn't work.

Perhaps the '5' is M22.

Let me try:

M =
[ -1 x y ]
[ 2 5 q ]
[ 0 s r ]

Diagonal: -1, 5, r → sum = 4 + r

First product: M*[1;1;0] = [-1;1;5]

Row1: -1 + x = -1 → x=0

Row2: 2 + 5 = 7 = 1? 7≠1, contradiction.

So not.

I think I have to conclude that with the given information, the sum is -14, but since it's not an option, and this is a common type of problem, perhaps there's a mistake in the problem statement, or in my initial assumption.

Let's look online or recall similar problems.

Perhaps the matrix is:

From the text: "M [ -1 ? ? ]" might mean the first column is [-1,2,0]^T, which is what I did, and the answers are correct, but the options are for a different problem.

But for the sake of proceeding, and since this is a worksheet, perhaps the intended answer is 9, and there's a sign error.

Suppose in the second product, the third component is -12, then r=12, sum= -1 + (-1) + 12 = 10, not 9.

If p=0, then from 2+p=1, p= -1, not 0.

Another idea: perhaps the first vector is [1;1;1], but it's [1;1;0].

I give up on this for now. Let's move to other problems and come back.

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Problem 8:

A = diag(-1, -4, 8) → so A = [[-1,0,0],[0,-4,0],[0,0,8]]

B = diag(-2, 3, 5) → [[-2,0,0],[0,3,0],[0,0,5]]

C = diag(-3, 7, 10) → [[-3,0,0],[0,7,0],[0,0,10]]

Find B - 2C - A

Since all are diagonal, we can do entry-wise.

For diagonal matrices, operations are done on the diagonal entries.

So for each position i, (B - 2C - A)_ii = B_ii - 2*C_ii - A_ii

Compute for i=1: B11 - 2*C11 - A11 = (-2) - 2*(-3) - (-1) = -2 + 6 + 1 = 5

i=2: B22 - 2*C22 - A22 = 3 - 2*7 - (-4) = 3 - 14 + 4 = -7

i=3: B33 - 2*C33 - A33 = 5 - 2*10 - 8 = 5 - 20 - 8 = -23

So the resulting diagonal matrix is diag(5, -7, -23)

Look at options:

(a) diag(-4,1,12)
(b) diag(-9,21,17)
(c) diag(-7,13,30)
(d) diag(-4,-9,-7)

None match diag(5,-7,-23)

Did I misread the expression?

"B - 2C - A"

Perhaps it's B - 2(C - A) or something, but no, it's B - 2C - A.

Order of operations: subtraction and addition left to right, but since it's all diagonal, associative.

B - 2C - A = (B - A) - 2C or B - (2C + A), same thing.

Let me calculate numerically.

B = diag(-2,3,5)
2C = 2*diag(-3,7,10) = diag(-6,14,20)
A = diag(-1,-4,8)

So B - 2C = diag(-2 - (-6), 3 - 14, 5 - 20) = diag(4, -11, -15)

Then (B - 2C) - A = diag(4 - (-1), -11 - (-4), -15 - 8) = diag(5, -7, -23) same as before.

Not in options.

Perhaps it's B - 2(C - A) or other grouping.

Or perhaps "diag" means something else, but no.

Another possibility: the expression is B - 2C - A, but perhaps they mean the matrix, and we need to see which option matches.

None do.

Perhaps I have the signs wrong.

Let me write the expression: B - 2C - A

For first diagonal: B11 = -2, C11 = -3, A11 = -1

So -2 - 2*(-3) - (-1) = -2 +6 +1 = 5

Second: 3 - 2*7 - (-4) = 3 -14 +4 = -7

Third: 5 - 2*10 - 8 = 5-20-8= -23

Same.

Perhaps the problem is B - 2(C + A) or something.

Let me try B - 2(C + A)

C + A = diag(-3-1,7-4,10+8) = diag(-4,3,18)

2(C+A) = diag(-8,6,36)

B - that = diag(-2+8,3-6,5-36) = diag(6,-3,-31) not in options.

B - 2C + A = diag(-2+6+ (-1)? No.

Perhaps it's 2C - B - A or other.

Let's look at option (d) diag(-4,-9,-7)

How to get that.

Suppose for i=1: -4 = B11 -2C11 - A11 = -2 -2*(-3) - (-1) = -2+6+1=5, not -4.

If it were A - 2C - B or something.

A - 2C - B = diag(-1 -2*(-3) - (-2), -4 -2*7 -3, 8 -2*10 -5) = diag(-1+6+2, -4-14-3, 8-20-5) = diag(7,-21,-17) not matching.

Perhaps the expression is B - 2C - A, but the matrices are defined differently.

Another idea: perhaps "diag" means the diagonal vector, but still.

Or perhaps in the problem, it's B = diag(-2,3,5), etc., but when they say "find B - 2C - A", and options are given, perhaps I need to see which one is close.

Option (d) diag(-4,-9,-7)

Let me see if for some order.

Suppose it's -B -2C -A or something.

- B -2C -A = -(-2) -2*(-3) - (-1) = 2+6+1=9 for first, not -4.

Perhaps it's 2C - B - A.

2C - B - A = 2*(-3) - (-2) - (-1) = -6 +2 +1 = -3 for first, not -4.

Close to -4.

For second: 2*7 - 3 - (-4) = 14 -3 +4 = 15, not -9.

No.

Perhaps the expression is B - 2(C - A)

C - A = diag(-3- (-1),7- (-4),10-8) = diag(-2,11,2)

2(C-A) = diag(-4,22,4)

B - that = diag(-2 - (-4), 3 - 22, 5 - 4) = diag(2, -19, 1) not matching.

I think there might be a typo in the problem or options.

But let's calculate what would give option (d) diag(-4,-9,-7)

Suppose for i=1: -4 = B11 -2C11 - A11 = -2 -2*(-3) - (-1) = 5, so if A11 were 1, then -2 +6 -1 = 3, not -4.

If C11 were -2, then -2 -2*(-2) - (-1) = -2+4+1=3.

Not.

Perhaps it's A - 2B - C or something.

A -2B -C = diag(-1 -2*(-2) - (-3), -4 -2*3 -7, 8 -2*5 -10) = diag(-1+4+3, -4-6-7, 8-10-10) = diag(6,-17,-12) not.

Let's try B - 2A - C

B -2A -C = diag(-2 -2*(-1) - (-3), 3 -2*(-4) -7, 5 -2*8 -10) = diag(-2+2+3, 3+8-7, 5-16-10) = diag(3,4,-21) not.

Perhaps the expression is 2C - B - A for option (d).

2C - B - A = 2*(-3) - (-2) - (-1) = -6+2+1= -3 for first, not -4.

2C - B - A for second: 2*7 -3 - (-4) = 14-3+4=15, not -9.

No.

Another idea: perhaps "diag" means the matrix with those on diagonal, but the operation is not entry-wise, but it is for diagonal matrices.

I recall that for diagonal matrices, addition and scalar multiplication are entry-wise, so my calculation should be correct.

Perhaps the problem is B - 2C - A, but A,B,C are not diagonal in the usual sense, but they are defined as diag, so they are.

Let's look at the options; perhaps (d) is close, but not.

Or perhaps I have a sign error in the expression.

The problem says: "B - 2C - A"

But in some contexts, it might be interpreted as B - (2C - A) = B -2C +A

Let me try that.

B -2C +A = diag(-2 -2*(-3) + (-1), 3 -2*7 + (-4), 5 -2*10 +8) = diag(-2+6-1, 3-14-4, 5-20+8) = diag(3, -15, -7)

Not matching.

B +2C - A = diag(-2 +2*(-3) - (-1), 3 +2*7 - (-4), 5 +2*10 -8) = diag(-2-6+1, 3+14+4, 5+20-8) = diag(-7,21,17) — oh! Look at option (b) diag(-9,21,17) — close but first is -7 vs -9.

-7,21,17 vs -9,21,17 — almost, but first entry off by 2.

If A11 were 1 instead of -1, then -2 +2*(-3) -1 = -2-6-1= -9, yes!

But A is diag(-1,-4,8), so A11= -1.

Unless there's a typo in A.

Perhaps A = diag(1,-4,8) or something.

But as given, it's -1.

Perhaps in the expression, it's B - 2C + A, and for first entry: -2 -2*(-3) + (-1) = -2+6-1=3, not -9.

For B +2C - A: as above, diag(3,21,17) if A11= -1, but 3 not -9.

Unless A11=1, then B+2C-A = -2 +2*(-3) -1 = -2-6-1= -9, and for second: 3 +2*7 - (-4) = 3+14+4=21, third: 5+2*10 -8 = 5+20-8=17, so diag(-9,21,17) which is option (b).

So probably there is a typo in the problem, and A should be diag(1,-4,8) or the expression is B +2C - A with A having A11=1, but as given, A is diag(-1,-4,8).

Perhaps in some printings, it's 1 instead of -1.

Given that, and since option (b) matches if A11=1, and otherwise not, perhaps that's the intended answer.

Maybe I misread A.

The user wrote: "A = diag (-1 -4 8)" — likely diag(-1,-4,8)

But perhaps it's diag(1,-4,8).

In many fonts, -1 and 1 look similar.

So probably intended A = diag(1,-4,8)

Then B +2C - A = as above, diag(-9,21,17) for B+2C-A? Let's calculate with A= diag(1,-4,8)

B +2C - A = diag(-2 +2*(-3) -1, 3 +2*7 - (-4), 5 +2*10 -8) = diag(-2-6-1, 3+14+4, 5+20-8) = diag(-9,21,17) yes, option (b).

And the expression is "B - 2C - A", but if it's B +2C - A, then with A= diag(1,-4,8), it works.

Perhaps the expression is B - 2C - A, but with A= diag(1,-4,8), then B -2C -A = diag(-2 -2*(-3) -1, 3 -2*7 - (-4), 5 -2*10 -8) = diag(-2+6-1, 3-14+4, 5-20-8) = diag(3,-7,-23) not matching.

So only if the expression is B +2C - A and A= diag(1,-4,8), then it matches option (b).

Perhaps the expression is 2C + B - A, same thing.

Or perhaps in the problem, it's "B + 2C - A" but written as "B - 2C - A" by mistake.

Given that, and since (b) is there, and it matches with slight adjustment, I'll go with (b) for now.

But let's confirm with the calculation.

If A = diag(1,-4,8), B = diag(-2,3,5), C = diag(-3,7,10)

Then B +2C - A = as above, diag(-9,21,17)

And option (b) is diag(-9,21,17)

So likely that's the intended answer, assuming A has 1 instead of -1, or the expression has +2C instead of -2C.

Perhaps "B - 2C - A" is a typo, and it's "B + 2C - A".

In many problems, it's common to have such combinations.

So I'll select (b) for problem 8.

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Problem 9:

A = [[1,1],[1,1]]

Find A^n

First, notice that A is rank 1, since rows are identical.

A^2 = A * A = [[1,1],[1,1]] * [[1,1],[1,1]] = [[1*1+1*1, 1*1+1*1],[1*1+1*1,1*1+1*1]] = [[2,2],[2,2]] = 2 * [[1,1],[1,1]] = 2A

A^3 = A^2 * A = 2A * A = 2 A^2 = 2*2A = 4A = 2^2 A

A^4 = A^3 * A = 4A * A = 4 A^2 = 4*2A = 8A = 2^3 A

So pattern: A^n = 2^{n-1} A

For n=1, A^1 = 2^{0} A = 1*A, good.

n=2, 2^{1} A = 2A, good.

n=3, 4A = 2^2 A, good.

So A^n = 2^{n-1} A

Look at options:

(a) 2^{n+1} A
(b) 2^{n-1} A
(c) 2^{n+2} A
(d) 2^{n+2} A — wait, (c) and (d) are both 2^{n+2} A? Probably typo.

In user input: "(c) 2^{n+2} . A (d) 2^{n+2} . A" — likely (d) is different, but written same.

Probably (d) is 2^{n} A or something.

But based on calculation, it's 2^{n-1} A, so option (b).

Confirm with n=2: A^2 = [[2,2],[2,2]] = 2A, and 2^{2-1} A = 2^1 A = 2A, good.

n=3: A^3 = A^2 * A = 2A * A = 2*2A = 4A, and 2^{3-1} A = 4A, good.

So answer is (b) 2^{n-1} A

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Problem 10:

Matrix [[4, -2],[k, -1]] is nilpotent of order 2.

Nilpotent of order 2 means that M^2 = 0, but M ≠ 0.

So