If mathrm{A}=left[begin{array}{rrr}-1 & 2 & 3 5 & 7 & 9 -2 & 1 & 1end{array}

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3 and 4 .Determinants and Matrices

easy

If $\mathrm{A}=\left[\begin{array}{rrr}-1 & 2 & 3 \\ 5 & 7 & 9 \\ -2 & 1 & 1\end{array}\right]$ and $\mathrm{B}=\left[\begin{array}{rrr}-4 & 1 & -5 \\ 1 & 2 & 0 \\ 1 & 3 & 1\end{array}\right],$ then verify that $(A-B)^{\prime}=A^{\prime}-B^{\prime}$

Option A

Option B

Option C

Option D

Solution

$A-B=\left[\begin{array}{ccc}-1 & 2 & 3 \\ 5 & 7 & 9 \\ -2 & 1 & 1\end{array}\right]-\left[\begin{array}{ccc}-4 & 1 & -5 \\ 1 & 2 & 0 \\ 1 & 3 & 1\end{array}\right]=\left[\begin{array}{ccc}3 & 1 & 8 \\ 4 & 5 & 9 \\ -3 & -2 & 0\end{array}\right]$

$\therefore(A-B)^{\prime}=\left[\begin{array}{ccc}3 & 4 & -3 \\ 1 & 5 & -2 \\ 8 & 9 & 0\end{array}\right]$

$A^{\prime}-B^{\prime}=\left[\begin{array}{ccc}-1 & 5 & -2 \\ 2 & 7 & 1 \\ 3 & 9 & 1\end{array}\right]-\left[\begin{array}{ccc}-4 & 1 & 1 \\ 1 & 2 & 3 \\ -5 & 0 & 1\end{array}\right]=\left[\begin{array}{ccc}3 & 4 & -3 \\ 1 & 5 & -2 \\ 8 & 9 & 0\end{array}\right]$

Hence, we have verified that $(A-B)^{\prime}=A^{\prime}-B^{\prime}$

Standard 12

Mathematics

If $ A$ is a square matrix, then which of the following matrices is not symmetric

easy

For any square matrix $A$, $A{A^T}$ is a

normal

The trace of a square matrix is defined to be the sum of its diagonal entries. If $A$ is a $2 \times 2$ matrix such that the trace of $A$ is $3$ and the trace of $A^3$ is -$18$ , then the value of the determinant of $A$ is. . . . .

normal

(IIT-2020)

If $A=\left[\begin{array}{lll}3 & \sqrt{3} & 2 \\ 4 & 2 & 0\end{array}\right]$ and $B=\left[\begin{array}{rrr}2 & -1 & 2 \\ 1 & 2 & 4\end{array}\right],$ verify that $(k B)^{\prime}=k B^{\prime},$ where $k$ is any constant.

medium

If $\mathrm{A}^{\prime}=\left[\begin{array}{cc}3 & 4 \\ -1 & 2 \\ 0 & 1\end{array}\right]$ and $\mathrm{B}=\left[\begin{array}{ccc}-1 & 2 & 1 \\ 1 & 2 & 3\end{array}\right],$ then verify that $(\mathrm{A}+\mathrm{B})^{\prime}=\mathrm{A}^{\prime}+\mathrm{B}^{\prime}$