Matrix left[ {begin{array}{*{20}{c}}2&λ &{ - 4}{ - 1}&3&41&{ - 2}&{

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3 and 4 .Determinants and Matrices

easy

The matrix $\left[ {\begin{array}{*{20}{c}}2&\lambda &{ - 4}\\{ - 1}&3&4\\1&{ - 2}&{ - 3}\end{array}} \right]$is non singular, if

$\lambda \ne - 2$

$\lambda \ne 2$

$\lambda \ne 3$

$\lambda \ne - 3$

Solution

(a) The given matrix $A = \left[ {\begin{array}{*{20}{c}}2&\lambda &{ – 4}\\{ – 1}&3&4\\1&{ – 2}&{ – 3}\end{array}} \right]$ is non singular, if $|A|\, \ne 0$

$|A|\,\, = \left| {\,\begin{array}{*{20}{c}}2&\lambda &{ – 4}\\{ – 1}&3&4\\1&{ – 2}&{ – 3}\end{array}\,} \right|\,$=$\left| {\,\begin{array}{*{20}{c}}1&{\lambda + 3}&0\\{ – 1}&3&4\\1&{ – 2}&{ – 3}\end{array}\,} \right|\,$,$[{R_1} \to {R_2} + {R_1}]$

= $\left| {\,\begin{array}{*{20}{c}}1&{\lambda + 3}&0\\0&1&1\\0&{ – \lambda – 5}&{ – 3}\end{array}\,} \right|$$\left[ {\begin{array}{*{20}{c}}{{R_2} \to {R_2} + {R_3}}\\{{R_3} \to {R_3} – {R_1}}\end{array}} \right]$

= $1\,( – 3 + \lambda + 5) \ne 0$

$ \Rightarrow \lambda + 2 \ne 0$ $ \Rightarrow \lambda \,\, \ne – 2.$

Standard 12

Mathematics

Matrix $A$ satisfies $A^2 = 2A – I$ where $I$ is the identity matrix then for $n \ge 2$, $A^n$ is equal to $(n \in N)$

normal

Suppose $A=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$ is a real matrix with non-zero entries, $\alpha d-b c=0$ and $A^2=A$. Then, $a + d$ equals

hard

(KVPY-2018)

In the matrix $A=\left[\begin{array}{cccc}2 & 5 & 19 & -7 \\ 35 & -2 & \frac{5}{2} & 12 \\ \sqrt{3} & 1 & -5 & 17\end{array}\right]$, write :

$(i)$ The order of the matrix

$(ii)$ The number of elements,

$(iii)$ Write the elements $a_{13}, \,a_{21}, \,a_{33}, \,a_{24},\, a_{23}$

easy

If $A $ and $ B$ are square matrices of order $3$ such that $|A| = – 1,$ $|B| = 3,$then $|3AB|$=

easy

(IIT-1988)

Show that

$\left[ {\begin{array}{{20}{c}}
5&{ – 1} \\
6&7
\end{array}} \right]\left[ {\begin{array}{{20}{l}}
2&1 \\
3&4
\end{array}} \right]$ $ \ne \left[ {\begin{array}{{20}{l}}
2&1 \\
3&4
\end{array}} \right]\left[ {\begin{array}{{20}{l}}
5&{ – 1} \\
6&7
\end{array}} \right]$

medium

The matrix $\left[ {\begin{array}{*{20}{c}}2&\lambda &{ - 4}\\{ - 1}&3&4\\1&{ - 2}&{ - 3}\end{array}} \right]$is non singular, if

Solution

Similar Questions

Matrix $A$ satisfies $A^2 = 2A – I$ where $I$ is the identity matrix then for $n \ge 2$, $A^n$ is equal to $(n \in N)$

Suppose $A=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$ is a real matrix with non-zero entries, $\alpha d-b c=0$ and $A^2=A$. Then, $a + d$ equals

In the matrix $A=\left[\begin{array}{cccc}2 & 5 & 19 & -7 \\ 35 & -2 & \frac{5}{2} & 12 \\ \sqrt{3} & 1 & -5 & 17\end{array}\right]$, write : $(i)$ The order of the matrix $(ii)$ The number of elements, $(iii)$ Write the elements $a_{13}, \,a_{21}, \,a_{33}, \,a_{24},\, a_{23}$

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In the matrix $A=\left[\begin{array}{cccc}2 & 5 & 19 & -7 \\ 35 & -2 & \frac{5}{2} & 12 \\ \sqrt{3} & 1 & -5 & 17\end{array}\right]$, write :

$(i)$ The order of the matrix

$(ii)$ The number of elements,

$(iii)$ Write the elements $a_{13}, \,a_{21}, \,a_{33}, \,a_{24},\, a_{23}$

Show that

$\left[ {\begin{array}{{20}{c}}
5&{ – 1} \\
6&7
\end{array}} \right]\left[ {\begin{array}{{20}{l}}
2&1 \\
3&4
\end{array}} \right]$ $ \ne \left[ {\begin{array}{{20}{l}}
2&1 \\
3&4
\end{array}} \right]\left[ {\begin{array}{{20}{l}}
5&{ – 1} \\
6&7
\end{array}} \right]$