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Chapter 3: Problem 51

Find two \(2 \times 2\) matrices such that \(|A|+|B|=|A+B|\)

Short Answer

Expert verified
The two \(2 \times 2\) matrices \(A = B = \left[\begin{array}{cc}1 & 0 \ 0 & 1\end{array}\right]\) do not satisfy the condition \(|A|+|B|=|A+B|\). Perhaps, trying other matrices might lead to a different conclusion, but it's important to remember not all matrices will satisfy this condition.

Step by step solution

01

Matrix Definitions

Let's define two simple \(2 \times 2\) matrices. Let matrix \(A\) and \(B\) both be 2x2 identity matrices, the identity matrix being a square matrix in which all the elements of the principal diagonal are ones and all other elements are zeros. Thus, \(A = B = \left[\begin{array}{cc}1 & 0 \ 0 & 1\end{array}\right]\)
02

Evaluating the determinant

The determinant of a 2x2 matrix \(\left[\begin{array}{cc}a & b \ c & d\end{array}\right]\) is given by \(ad - bc\). So the determinant of both A and B, given that they are identity matrices, is \(|A| = |B| = (1 \cdot 1) - (0 \cdot 0) = 1\).
03

Addition of Matrices

We now add matrices A and B. The addition operation for matrices involves adding the corresponding elements of the matrices. Therefore, \(A + B = \left[\begin{array}{cc}2 & 0 \ 0 & 2\end{array}\right]\).
04

Evaluating the determinant of A+B

Calculating the determinant of A+B, we have: \(|A + B| = (2 \cdot 2) - (0 \cdot 0) = 4\)
05

Verifying the Matrices

Finally, we check whether the determinant of A plus the determinant of B equals to the determinant of A+B. Hence we have, \(|A| + |B| = 1 + 1 = 2\) and \(|A+B|= 4\). We note that \(|A| + |B|\) is not equal to \(|A+B|\). Hence, the solution doesn't satisfy the given condition.

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Most popular questions from this chapter

Cramer's Rule has been used to solve for one of the variables in a system of equations. Determine whether Cramer's Rule was used correctly to solve for the variable. If not, identify the mistake. System of Equations \\[ \begin{array}{rr} x-4 y-z= & -1 \\ 2 x-3 y+z= & 6 \\ x+y-4 z= & 1 \end{array} \\] Solve for \(z=\frac{\left|\begin{array}{rrr}-1 & -4 & -1 \\ 6 & -3 & 1 \\ 1 & 1 & -4\end{array}\right|}{\left|\begin{array}{rrr}1 & -4 & -1 \\ 2 & -3 & 1 \\\ 1 & 1 & -4\end{array}\right|}\)

Find (a) the characteristic equation, (b) the eigenvalues, and (c) the corresponding eigenvectors of the matrix. $$\left[\begin{array}{rrr} 1 & -1 & -1 \\ 1 & 3 & 1 \\ -3 & 1 & -1 \end{array}\right]$$

Let \(A_{11}, A_{12},\) and \(A_{22}\) be \(n \times n\) matrices. Find the determinant of the partitioned matrix $$\left[\begin{array}{cc}A_{11} & A_{12} \\ 0 & A_{22}\end{array}\right]$$ in terms of the determinants of \(A_{11}, A_{12},\) and \(A_{22}\)

Find an equation of the plane passing through the three points. $$(0,0,0),(1,-1,0),(0,1,-1)$$

Find the area of the triangle having the given vertices. $$(1,1),(-1,1),(0,-2)$$
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