Algebra (Matrices and Determinants) Quiz - ICSE Class 12 Mathematics
Free practice quiz: Algebra (Matrices and Determinants) (ICSE Class 12 Mathematics). 62 multiple-choice questions with answers and explanations. Play online, no sign-up needed.
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If $A = \begin{pmatrix} 1 & 2 \ 3 & 4 \end{pmatrix}$ and $B = \begin{pmatrix} 2 & 0 \ 1 & 3 \end{pmatrix}$, then $A+B$ is:
- \begin{pmatrix} 3 & 2 \ 4 & 7 \end{pmatrix}
- \begin{pmatrix} 1 & 2 \ 3 & 4 \end{pmatrix}
- \begin{pmatrix} 2 & 0 \ 1 & 3 \end{pmatrix}
- \begin{pmatrix} 2 & 4 \ 3 & 1 \end{pmatrix}
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Answer: A. \begin{pmatrix} 3 & 2 \ 4 & 7 \end{pmatrix}
Matrix addition is performed by adding corresponding elements of the matrices. A+B = \begin{pmatrix} 1+2 & 2+0 \ 3+1 & 4+3 \end{pmatrix} = \begin{pmatrix} 3 & 2 \ 4 & 7 \end{pmatrix}
If $A = \begin{pmatrix} 1 & 2 \ 3 & 4 \end{pmatrix}$, then $A^T$ (Transpose of A) is:
- \begin{pmatrix} 1 & 3 \ 2 & 4 \end{pmatrix}
- \begin{pmatrix} 4 & 3 \ 2 & 1 \end{pmatrix}
- \begin{pmatrix} 1 & 2 \ 3 & 4 \end{pmatrix}
- \begin{pmatrix} 2 & 1 \ 4 & 3 \end{pmatrix}
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Answer: A. \begin{pmatrix} 1 & 3 \ 2 & 4 \end{pmatrix}
The transpose of a matrix is obtained by interchanging its rows and columns. The first row becomes the first column and the second row becomes the second column.
The system of linear equations $ax + by = c$ and $dx + ey = f$ can be represented in matrix form as:
- $\begin{pmatrix} a & b \ d & e \end{pmatrix} \begin{pmatrix} x \ y \end{pmatrix} = \begin{pmatrix} c \ f \end{pmatrix}$
- $\begin{pmatrix} a & d \ b & e \end{pmatrix} \begin{pmatrix} x \ y \end{pmatrix} = \begin{pmatrix} c \ f \end{pmatrix}$
- $\begin{pmatrix} a & c \ d & f \end{pmatrix} \begin{pmatrix} x \ y \end{pmatrix} = \begin{pmatrix} b \ e \end{pmatrix}$
- $\begin{pmatrix} a & b \ c & d \end{pmatrix} \begin{pmatrix} x \ y \end{pmatrix} = \begin{pmatrix} f \ e \end{pmatrix}$
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Answer: A. $\begin{pmatrix} a & b \ d & e \end{pmatrix} \begin{pmatrix} x \ y \end{pmatrix} = \begin{pmatrix} c \ f \end{pmatrix}$
By performing matrix multiplication on the left side, we get $ax + by = c$ and $dx + ey = f$, which matches the given system of equations.
Let $A = \begin{pmatrix} a & b \ c & d \end{pmatrix}$. Then $A$ is invertible if and only if:
- $ad - bc \neq 0$
- $ad - bc = 0$
- $a+d \neq 0$
- $a+d = 0$
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Answer: A. $ad - bc \neq 0$
A square matrix is invertible if and only if its determinant is non-zero. The determinant of a 2x2 matrix $\begin{pmatrix} a & b \ c & d \end{pmatrix}$ is $ad - bc$.
If $A = \begin{pmatrix} 1 & 2 \ 3 & 4 \end{pmatrix}$, which of the following is equal to $A^{-1}$?
- $\frac{1}{-2}\begin{pmatrix} 4 & -2 \ -3 & 1 \end{pmatrix}$
- $\frac{1}{10}\begin{pmatrix} 4 & -2 \ -3 & 1 \end{pmatrix}$
- $\frac{1}{-2}\begin{pmatrix} 1 & 2 \ 3 & 4 \end{pmatrix}$
- $\frac{1}{10}\begin{pmatrix} 1 & 2 \ 3 & 4 \end{pmatrix}$
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Answer: A. $\frac{1}{-2}\begin{pmatrix} 4 & -2 \ -3 & 1 \end{pmatrix}$
The determinant of $A$ is $(1 imes 4) - (2 imes 3) = 4 - 6 = -2$. The adjugate of $A$ is $\begin{pmatrix} 4 & -2 \ -3 & 1 \end{pmatrix}$. Therefore, $A^{-1} = \frac{1}{\det(A)} adj(A) = \frac{1}{-2}\begin{pmatrix} 4 & -2 \ -3 & 1 \end{pmatrix}$.