Find a matrix X such that 2A + B + X = 0 , where \(A=\begin{bmatrix} -1 & 2 \\\ 3 & 4 \end{bmatrix} \ \text{and} \;\rm B =\ \begin{bmatrix} 3 & -2 \\\ 1 & 5 \end{bmatrix} \ ?\)
Finding Matrix X: Step-by-Step Solution
The problem asks us to find a matrix X based on the given equation \(2A + B + X = 0\), where matrices \(A=\begin{bmatrix} -1 & 2 \\ 3 & 4 \end{bmatrix}\) and \(B =\begin{bmatrix} 3 & -2 \\ 1 & 5 \end{bmatrix}\) are provided.
Rearranging the Matrix Equation
To find matrix X, we first rearrange the given equation: \(2A + B + X = 0\) Subtracting \(2A\) and \(B\) from both sides, we get: \(X = - (2A + B)\)
Calculating 2A
First, we need to perform scalar multiplication on matrix A. We multiply each element of matrix A by 2: $$ 2A = 2 \times \begin{bmatrix} -1 & 2 \\ 3 & 4 \end{bmatrix} $$ $$ 2A = \begin{bmatrix} 2 \times (-1) & 2 \times 2 \\ 2 \times 3 & 2 \times 4 \end{bmatrix} $$ $$ 2A = \begin{bmatrix} -2 & 4 \\ 6 & 8 \end{bmatrix} $$
Calculating 2A + B
Next, we add the resulting matrix \(2A\) to matrix B. Matrix addition involves adding corresponding elements: $$ 2A + B = \begin{bmatrix} -2 & 4 \\ 6 & 8 \end{bmatrix} + \begin{bmatrix} 3 & -2 \\ 1 & 5 \end{bmatrix} $$ $$ 2A + B = \begin{bmatrix} (-2 + 3) & (4 + (-2)) \\ (6 + 1) & (8 + 5) \end{bmatrix} $$ $$ 2A + B = \begin{bmatrix} 1 & 2 \\ 7 & 13 \end{bmatrix} $$
Calculating Matrix X
Finally, we find matrix X by taking the negative of the result from the previous step \((2A + B)\). This means multiplying each element of \((2A + B)\) by -1: $$ X = - (2A + B) $$ $$ X = - \begin{bmatrix} 1 & 2 \\ 7 & 13 \end{bmatrix} $$ $$ X = \begin{bmatrix} -1 \times 1 & -1 \times 2 \\ -1 \times 7 & -1 \times 13 \end{bmatrix} $$ $$ X = \begin{bmatrix} -1 & -2 \\ -7 & -13 \end{bmatrix} $$
Final Result Verification
The calculated matrix X is \(\begin{bmatrix} -1 & -2 \\ -7 & -13 \end{bmatrix}\). This matches one of the provided options.
Important Questions from Operations on Matrices
The adjoint of matrix \(\left[ {\begin{array}{*{20}{c}} a&b\\ c&d \end{array}} \right]\)is
If \({\rm{E}}\left( {\rm{\theta }} \right) = \left[ {\begin{array}{*{20}{c}} {\cos {\rm{\theta }}}&{\sin {\rm{\theta }}}\\ { - \sin {\rm{\theta \;}}}&{\cos {\rm{\theta }}} \end{array}} \right]\) then E(α) E(β) is equal to
Consider the following in respect of matrices A, B and C of same order:
1) (A + B + C)' = A' + B’ + C’
2) (AB)’ = A’B’
3) (ABC)’ = C’B’A’
Where A’ is the transpose of the matrix A.
Which of the above are correct?If \({\rm{A}} = \left[ {\begin{array}{*{20}{c}} 1&0&{ - 2}\\ 2&{ - 3}&4 \end{array}} \right]\) , then the matrix X for which 2X + 3A = 0 holds true is
The solution of the matrix equation \(\left[ {\begin{array}{*{20}{c}} 2&{ - 1}&3\\ 1&1&1\\ 1&{ - 1}&1 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} x\\ y\\ z \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 9\\ 6\\ 2 \end{array}} \right]\) is:
