Let $a\in\mathbb{R}$ and $A$ be a matrix of order $3\times3$ such that $\det(A)=-4$ and $$A+I=\begin{pmatrix}1 & a & 1\\2 & 1 & 0\\a & 1 & 2\end{pmatrix},$$ where $I$ is the $3\times3$ identity. If $$\det\bigl((a+1)\operatorname{adj}((a-1)A)\bigr)=2^m3^n,\quad m,n\in\{0{,}1{,}2,\dots,20\},$$ then $m+n$ is equal to:

Step 1: Calculate the determinant of (A+I)

First, we need to find the value of $a$. We are given the matrix $A+I$. We can calculate its determinant using the cofactor expansion method. Expanding along the second row is convenient due to the zero entry.

Step 2: Solve for 'a'

We calculate the determinant of $A+I$ as $3a$. We know that $\det(A+I) = \det(A) + \text{sum of principal minors} + \det(I)$ is not generally true. Instead, we use the property that if $A+I = B$, then $\det(B) = 3a$. We are given $\det(A) = -4$. However, we need to find $A$ first to relate $\det(A+I)$ to $\det(A)$. This is a common mistake. Let's re-evaluate. The given matrix is $A+I$. We need to find $A$ from $A+I$. Then we can calculate $\det(A)$ from the expression for $A$. But we are already given $\det(A)=-4$. This means we need to find $a$ such that $\det(A)=-4$. Let's find $A$ first.

Step 3: Determine matrix A and its determinant

Since $A+I = \begin{pmatrix}1 & a & 1\\2 & 1 & 0\\a & 1 & 2\end{pmatrix}$, we can find $A$ by subtracting the identity matrix $I = \begin{pmatrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{pmatrix}$ from $A+I$. This gives us the matrix $A$. Now we can calculate $\det(A)$.

Step 4: Calculate $\det(A)$ and solve for 'a'

We calculate the determinant of $A$ by expanding along the first row. We get $\det(A) = -2a+2$. We are given that $\det(A) = -4$. Equating these two expressions, we can solve for $a$.

Step 5: Find the value of 'a'

Solving the equation $-2a+2 = -4$ for $a$, we find that $a=3$. Now we have the value of $a$, which is crucial for the next part of the problem.

Step 6: Evaluate the determinant expression

We use the property $\det(kA) = k^n \det(A)$ for an $n \times n$ matrix $A$. Here, $k = (a+1)$ and the matrix is $\operatorname{adj}((a-1)A)$. Since $A$ is a $3 \times 3$ matrix, $\operatorname{adj}((a-1)A)$ is also a $3 \times 3$ matrix, so $n=3$.

Step 7: Apply adjoint determinant property

We use the property $\det(\operatorname{adj}(B)) = (\det(B))^{n-1}$ for an $n \times n$ matrix $B$. Here, $B = (a-1)A$ and $n=3$. So, $\det(\operatorname{adj}((a-1)A)) = (\det((a-1)A))^2$.

Step 8: Simplify further using determinant properties

Again, we use the property $\det(kA) = k^n \det(A)$. Here, $k = (a-1)$ and the matrix is $A$. Since $A$ is a $3 \times 3$ matrix, $n=3$. So, $\det((a-1)A) = (a-1)^3 \det(A)$. Substituting this into the expression, we get the simplified form.

Step 9: Substitute values and calculate the final result

Now we substitute the value $a=3$ and $\det(A)=-4$ into the simplified expression. We then simplify the powers of 2 to get the final result in the form $2^m 3^n$. Since there is no factor of 3, $n=0$.

Step 10: Determine m and n, and calculate m+n

From the calculation, we have $2^{16}$. Comparing this with $2^m 3^n$, we find $m=16$ and $n=0$. Therefore, $m+n = 16+0 = 16$.

More from this chapter

• Evaluate the determinant 1 & 0 & 4 3 & 5 & -1 0 & 1 & 2 (i) with the help of second row. (ii) with the help of third column.
• If A = [[1, 2], [3, 4]] find det(A) and adj(A).
• The value of 1 & 2 & 3 4 & 5 & 6 7 & 8 & 9 is:
• Using Cramer's rule, show that the following system of linear equations is consistent and hence solve it: x + y + z = 1, 2x + 3y + 2z = 2, x + y + 2z = 4.
• Let A be a 3 × 3 matrix such that |adj (adj(adj A))| = 81. If S = n Z: |adj(adj A)|^(((n-1)²)/(2)) = |A|³n² - 5n - 4, then Σ_n S | A^(n² + n) | is equal to:
• Find the equation of the line joining A(1,3) and B(0,0) using determinants and find k if D(k, 0) is a point such that area of triangle ABD is 3 sq units.
• If the system of linear equations 3x + y + β z = 3, 2x + α y - z = -3, x + 2y + z = 4 has infinitely many solutions, then the value of 22β - 9α is:
• Let A = 2 & 2 + p & 2 + p + q 4 & 6 + 2p & 8 + 3p + 2q 6 & 12 + 3p & 20 + 6p + 3q. If det(adj(adj(3A))) = 2^m · 3^n, m, n N, then m + n is equal to:

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