Q.If $A = \begin{bmatrix}1 & 5\\ 7 & 12\end{bmatrix}$ and $B = \begin{bmatrix}9 & 1\\ 7 & 8\end{bmatrix}$, find a matrix $C$ such that $3A + 5B + 2C$ is a null matrix.

The key here is not to multiply $A$ five times directly — that would be messy and error-prone. Instead, we look for a pattern in powers of $A$. Many matrices in such problems satisfy a simple recurrence, often because their characteristic polynomial is of low degree. The trick is to compute $A^2$ and $A^3$ first, and see if a cycle emerges.

1. Compute $A^2$. Multiply $A$ by itself:

Row by row:

• First row: $(1)(1)+(-1)(2)+(1)(1) = 1-2+1=0$; $(1)(-1)+(-1)(-1)+(1)(0) = -1+1+0=0$; $(1)(1)+(-1)(0)+(1)(0) = 1+0+0=1$.
• Second row: $(2)(1)+(-1)(2)+(0)(1) = 2-2+0=0$; $(2)(-1)+(-1)(-1)+(0)(0) = -2+1+0=-1$; $(2)(1)+(-1)(0)+(0)(0) = 2+0+0=2$.
• Third row: $(1)(1)+(0)(2)+(0)(1) = 1$; $(1)(-1)+(0)(-1)+(0)(0) = -1$; $(1)(1)+(0)(0)+(0)(0) = 1$.

So

2. Compute $A^3$. Multiply $A^2$ by $A$:

• First row: $(0)(1)+(0)(2)+(1)(1)=1$; $(0)(-1)+(0)(-1)+(1)(0)=0$; $(0)(1)+(0)(0)+(1)(0)=0$.
• Second row: $(0)(1)+(-1)(2)+(2)(1) = 0-2+2=0$; $(0)(-1)+(-1)(-1)+(2)(0) = 0+1+0=1$; $(0)(1)+(-1)(0)+(2)(0)=0$.
• Third row: $(1)(1)+(-1)(2)+(1)(1) = 1-2+1=0$; $(1)(-1)+(-1)(-1)+(1)(0) = -1+1+0=0$; $(1)(1)+(-1)(0)+(1)(0)=1$.

Hence

We are given

and told that

We need $m+n$.

Concept and intuition

The adjugate (or classical adjoint) of a matrix is the transpose of its cofactor matrix. That is, the $(i,j)$ entry of $\operatorname{adj}(A)$ is the cofactor $C_{ji}$ (note the swapped indices). So if we compute the cofactors of $A$ directly, we can match them to the given entries and solve for $m$ and $n$.

A cofactor $C_{ij} = (-1)^{i+j} M_{ij}$, where $M_{ij}$ is the determinant of the submatrix obtained by deleting row $i$ and column $j$.

Step-by-step solution

1. Find the entry in position (1,1) of $\operatorname{adj}(A)$ This is $C_{11}$ (cofactor of $a_{11}$). Delete row 1, column 1:

Since $(-1)^{1+1}=1$, $C_{11}=5$. This matches the given (1,1) entry 5 — good check.

2. Find the entry in position (1,2) of $\operatorname{adj}(A)$ This is $C_{21}$ (cofactor of $a_{21}$). Delete row 2, column 1:

$(-1)^{2+1} = -1$, so $C_{21} = (-1)(-4) = 4$.

The given (1,2) entry is $m$. Hence $m = 4$. …

The problem gives a $3 \times 3$ matrix and tells you the cofactors of two specific entries. A cofactor is not just the minor — it includes a sign based on the position. So the first step is to recall exactly what a cofactor means.

For an entry $a_{ij}$ in a matrix, its cofactor $C_{ij}$ is $(-1)^{i+j}$ times the determinant of the submatrix obtained by deleting the $i$-th row and $j$-th column. That sign pattern is crucial: positions alternate signs like a chessboard, starting with $+$ in the top-left.

Here the matrix is:

We are told:

• The cofactor of $-6$ (which is $a_{23}$) is $22$.
• The cofactor of $-7$ (which is $a_{31}$) is $27$.

Let’s work through each.

1. Cofactor of $-6$ (entry $a_{23}$) Position: row 2, column 3. So $i=2$, $j=3$, sign factor $(-1)^{2+3} = (-1)^5 = -1$. The minor is the determinant of the matrix left after deleting row 2 and column 3:

Therefore the cofactor is:

And we know $C_{23} = 22$, so:

2. Cofactor of $-7$ (entry $a_{31}$) Position: row 3, column 1. So $i=3$, $j=1$, sign factor $(-1)^{3+1} = (-1)^4 = +1$. Delete row 3 and column 1: $\text{Minor} = \det\begin{bmatrix} x & 3 \\ -5 & -6 \end{bmatrix} = (x)(-6) - (3)(-5) = -6x + 15$ …

Setup. The coefficient vector $(2,5,3)$ does not match any single equation (equation 2 is $5x+2y+3z$), so we solve the system outright.

Step 1 - Eliminate $x$ using (3): $x=-3-2y+3z$.

Into (1): $2(-3-2y+3z)-3y+5z=12\Rightarrow -7y+11z=18.$

Into (2): $5(-3-2y+3z)+2y+3z=11\Rightarrow -8y+18z=26\Rightarrow -4y+9z=13.$

Step 2 - Solve the reduced pair.

$7y-11z=-18,\qquad 4y-9z=-13.$ …

Concept and Intuition

A system of linear equations represents a set of conditions that multiple variables must satisfy simultaneously. For a system of three linear equations with three variables ($x, y, z$), a unique solution $(\alpha, \beta, \gamma)$ means there is exactly one point in 3D space where all three planes (represented by the equations) intersect.

The most common and often simplest method to solve such a system is elimination. The core idea is to systematically remove one variable at a time from pairs of equations until you are left with a simpler system (e.g., two equations with two variables, then one equation with one variable). Once one variable's value is found, you substitute it back into the simpler equations to find the others.

In this problem, we are specifically asked for the value of $\beta$, which is the $y$-coordinate of the solution. We will use elimination to isolate $y$.

Step-by-step Solution

1. Write down the given system of equations:

   The given equations are:

   (1) $5x - 7y + 3z = 0$

   (2) $7x + 10y - 8z = 3$

   (3) $2x + 3y - 4z + 4 = 0 \implies 2x + 3y - 4z = -4$

2. Eliminate $z$ from equations (1) and (3):

   To eliminate $z$, we need the coefficients of $z$ in both equations to be additive inverses. The coefficients are $3$ and $-4$. The least common multiple of $3$ and $4$ is $12$.

   Multiply equation (1) by $4$:

   $4 \times (5x - 7y + 3z) = 4 \times 0$

   $20x - 28y + 12z = 0$ (Equation 4)

   Multiply equation (3) by $3$:

   $3 \times (2x + 3y - 4z) = 3 \times (-4)$

   $6x + 9y - 12z = -12$ (Equation 5)

   Now, add Equation 4 and Equation 5:

   $(20x - 28y + 12z) + (6x + 9y - 12z) = 0 + (-12)$

   $26x - 19y = -12$ (Equation 6)

   This is our first equation with only $x$ and $y$.

3. Eliminate $z$ from equations (1) and (2):

   The coefficients of $z$ in equations (1) and (2) are $3$ and $-8$. The least common multiple of $3$ and $8$ is $24$.

   Multiply equation (1) by $8$:

   $8 \times (5x - 7y + 3z) = 8 \times 0$

   $40x - 56y + 24z = 0$ (Equation 7)

   Multiply equation (2) by $3$:

   $3 \times (7x + 10y - 8z) = 3 \times 3$

   $21x + 30y - 24z = 9$ (Equation 8)

   Now, add Equation 7 and Equation 8:

   $(40x - 56y + 24z) + (21x + 30y - 24z) = 0 + 9$

   $61x - 26y = 9$ (Equation 9)

   This is our second equation with only $x$ and $y$. …

The key insight is that the relation $A^2 + I = 2A$ is not just an algebraic curiosity — it tells us that $A$ satisfies a quadratic polynomial. For matrices, such a polynomial lets us reduce any higher power to a linear combination of $A$ and $I$ (since the degree is 2). This is like having a recurrence: every time we see $A^2$, we can replace it with $2A - I$. Then we can compute $A^3$, $A^4$, … up to $A^9$ by repeated substitution, or better yet, find a pattern.

Let’s work it out step by step.

1. Rewrite the given condition

   $A^2 + I = 2A \implies A^2 = 2A - I$.

   This is our reduction rule: any $A^2$ can be replaced by $2A - I$.

2. Compute $A^3$

   $A^3 = A \cdot A^2 = A(2A - I) = 2A^2 - A$.

   Now replace $A^2$ again: $2(2A - I) - A = 4A - 2I - A = 3A - 2I$.

3. Compute $A^4$

   $A^4 = A \cdot A^3 = A(3A - 2I) = 3A^2 - 2A$.

   Substitute $A^2$: $3(2A - I) - 2A = 6A - 3I - 2A = 4A - 3I$.

4. Spot the pattern

   From the results:

   $A^1 = 1A + 0I$

   $A^2 = 2A - 1I$

   $A^3 = 3A - 2I$

   $A^4 = 4A - 3I$

   It looks like $A^n = nA - (n-1)I$ for $n \ge 1$. Let’s test for $n=5$ to be sure:

   $A^5 = A \cdot A^4 = A(4A - 3I) = 4A^2 - 3A = 4(2A - I) - 3A = 8A - 4I - 3A = 5A - 4I$. Pattern holds.

5. Prove the pattern (optional but satisfying)

   Assume $A^k = kA - (k-1)I$. Then

   $A^{k+1} = A \cdot A^k = A(kA - (k-1)I) = kA^2 - (k-1)A$

   $= k(2A - I) - (k-1)A = 2kA - kI - (k-1)A = (k+1)A - kI$. …

Concept and Intuition

The problem asks for the trace of matrix $A$, which is the sum of its diagonal elements: $\alpha + \beta + \gamma$. We are given the matrix $A$ with these unknown diagonal elements and the product $AA^T$. The key idea is to use the properties of matrix multiplication, specifically how the elements of $AA^T$ are formed, to find $\alpha, \beta,$ and $\gamma$.

When we multiply a matrix $A$ by its transpose $A^T$, each element $(AA^T)_{ij}$ is the dot product of the $i$-th row of $A$ and the $j$-th row of $A$.

• For the diagonal elements, $(AA^T)_{ii}$ is the dot product of the $i$-th row of $A$ with itself. This means $(AA^T)_{ii}$ is the sum of the squares of the elements in the $i$-th row of $A$. This property is very useful because it directly gives us equations involving $\alpha^2, \beta^2,$ and $\gamma^2$.
• For the off-diagonal elements, $(AA^T)_{ij}$ (where $i \neq j$) is the dot product of the $i$-th row of $A$ and the $j$-th row of $A$. These elements will help us determine the signs of $\alpha$ and $\beta$ once their squares are known.

Let's apply this understanding to find the unknown values and then the trace.

Step-by-Step Solution

1. Identify the matrices and the goal:

   We are given:

   $A = \begin{bmatrix} \alpha & 3 & 1 \\ 2 & \beta & 6 \\ -2 & -1 & \gamma \end{bmatrix}$

   $AA^T = \begin{bmatrix} 35 & 28 & -13 \\ 28 & 56 & -8 \\ -13 & -8 & 5 \end{bmatrix}$

   Our goal is to find the Trace of $A$, which is $\alpha + \beta + \gamma$.

2. Calculate the diagonal elements of $AA^T$ using the rows of $A$:

   The element $(AA^T)_{11}$ is the dot product of the first row of $A$ with itself:

   $(AA^T)_{11} = (\alpha)( \alpha) + (3)(3) + (1)(1) = \alpha^2 + 9 + 1 = \alpha^2 + 10$.

   From the given $AA^T$ matrix, $(AA^T)_{11} = 35$.

   So, $\alpha^2 + 10 = 35 \implies \alpha^2 = 25$.

   The element $(AA^T)_{22}$ is the dot product of the second row of $A$ with itself:

   $(AA^T)_{22} = (2)(2) + (\beta)(\beta) + (6)(6) = 4 + \beta^2 + 36 = \beta^2 + 40$.

   From the given $AA^T$ matrix, $(AA^T)_{22} = 56$.

   So, $\beta^2 + 40 = 56 \implies \beta^2 = 16$.

   The element $(AA^T)_{33}$ is the dot product of the third row of $A$ with itself:

   $(AA^T)_{33} = (-2)(-2) + (-1)(-1) + (\gamma)(\gamma) = 4 + 1 + \gamma^2 = \gamma^2 + 5$.

   From the given $AA^T$ matrix, $(AA^T)_{33} = 5$.

   So, $\gamma^2 + 5 = 5 \implies \gamma^2 = 0$.

3. Determine the possible values for $\alpha, \beta, \gamma$:

   From the equations above:

   $\alpha^2 = 25 \implies \alpha = \pm 5$.

   $\beta^2 = 16 \implies \beta = \pm 4$.

   $\gamma^2 = 0 \implies \gamma = 0$.

   We have two possibilities for $\alpha$ and two for $\beta$. We need to use an off-diagonal element of $AA^T$ to determine the correct signs.

4. Use an off-diagonal element of $AA^T$ to find the specific values of $\alpha$ and $\beta$: …

The key insight is that $A$ is not diagonalizable in the usual sense, but it has a very special structure: it is a shear matrix (a Jordan block with eigenvalue 1). For such matrices, powers follow a neat pattern that lets us sum the series without multiplying twelve matrices.

Let’s see why. A shear matrix like $A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$ represents a transformation that leaves the x-axis fixed but shifts points parallel to it. When you apply it repeatedly, the off-diagonal entry grows linearly with the power.

1. Find a pattern for $A^n$. Compute the first few powers manually:

The pattern is clear: $A^n = \begin{bmatrix} 1 & n \\ 0 & 1 \end{bmatrix}$.

You can prove this by induction: if $A^k = \begin{bmatrix} 1 & k \\ 0 & 1 \end{bmatrix}$, then $A^{k+1} = A^k \cdot A = \begin{bmatrix} 1 & k \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & k+1 \\ 0 & 1 \end{bmatrix}$.

2. Write $S$ as a sum of these powers.

3. Compute each sum.

   • The sum of twelve 1’s is $12$.
   • The sum of the first 12 natural numbers is $\frac{12 \times 13}{2} = 78$.
   • The sum of twelve 0’s is $0$.

   So: …

Write $A$ as a rotation matrix: